diff options
Diffstat (limited to '1019')
145 files changed, 2689 insertions, 0 deletions
diff --git a/1019/CH2/EX2.1/Example_2_1.sce b/1019/CH2/EX2.1/Example_2_1.sce new file mode 100644 index 000000000..49aea3495 --- /dev/null +++ b/1019/CH2/EX2.1/Example_2_1.sce @@ -0,0 +1,15 @@ +//Example 2.1
+clear;
+clc;
+
+//Given
+p1=10;//initial pressure in atm
+p2=510;//final pressure in atm
+R=0.082;// gas constant in L atm K^-1 mol^-1
+T=300;// temperature in K
+
+// To determine enthalpy change delH
+delH=(0.039-((2*1.34)/(R*T)))*(p2-p1);//using the given equation for joule-thomson coefficient in dm^3 atm mol^-1
+delH1=delH*(1.01325*100);// enthalpy change in joule mol^-1
+mprintf('change in enthalpy = %f Joule mol^-1',delH1);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.10/Example_2_10.sce b/1019/CH2/EX2.10/Example_2_10.sce new file mode 100644 index 000000000..6d1a69337 --- /dev/null +++ b/1019/CH2/EX2.10/Example_2_10.sce @@ -0,0 +1,24 @@ +//Example 2.10
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+t1=298;// temperature in K
+p1=30000;// initial pressure in N m^-2
+p2=10000;// final pressure in N m^-2
+Cv=20.8;// heat capacity of CO at constant volume in J K^-1 mol^-1
+W=0.1;// weight of CO taken in kg
+
+// To determine t2,w,delH and delE
+Cp=Cv+R;//heat capacity at constant pressure
+n=W/0.028;//moles of CO
+t2=t1*((Cv+((p2/p1)*R))/Cp);//final temperature in K
+delE=n*Cv*(t2-t1);//change in internal energy in J
+w=delE;//work done in J
+delH=n*Cp*(t2-t1);//enthalpy change in J
+mprintf('Final Temperature = %f K',t2);
+mprintf('\n delE = %f J',delE);
+mprintf('\n delH = %f J',delH);
+mprintf('\n w = %f J',w);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.11/Example_2_11.sce b/1019/CH2/EX2.11/Example_2_11.sce new file mode 100644 index 000000000..9fcf6131f --- /dev/null +++ b/1019/CH2/EX2.11/Example_2_11.sce @@ -0,0 +1,34 @@ +//Example 2.11
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+t1=300;// temperature in K
+p1=1000;// initial pressure in Mpa
+p2=100;// final pressure in Mpa
+Cv=1.5*R;// heat capacity at constant volume in J K^-1 mol^-1
+W=0.1;// weight of CO taken in kg
+n=1;//moles of the gas
+
+// To determine q,w,delH and delE
+Cp=Cv+R;//heat capacity at constant pressure
+//(a) isothermal reversible
+w=(-1)*R*t1*log(p1/p2);//work done in J
+q=(-1)*w;//heat in J
+//(b) isothermal irreversible
+w1=(-1)*R*t1*(1-(p2/p1));//work done in J
+q1=(-1)*w1;//heat in J
+//(c) adiabatic reversible
+t2=t1*((p2/p1)^(R/Cp));
+delE=n*Cv*(t2-t1);//change in internal energy in J
+delH=n*Cp*(t2-t1);//change in enthalpy in J
+//(d) adiabatic irreversible
+T2=t1*((Cv+((p2/p1)*R))/Cp);//final temperature in K
+delE1=n*Cv*(T2-t1);//change in internal energy in J
+delH1=n*Cp*(T2-t1);//change in enthalpy in J
+mprintf('(a) w = %f J mol^-1,delH = 0 J mol^-1,q= %f J mol^-1 and delE = 0 J mol^-1',w,q);
+mprintf('\n (b) w = %f J mol^-1,delH = 0 J mol^-1,q= %f J mol^-1 and delE = 0 J mol^-1',w1,q1);
+mprintf('\n (c) T2 = %f K,w = %f J mol^-1,delH = %f J mol^-1,q= 0 J mol^-1 and delE = %f J mol^-1',t2,delE,delH,delE);
+mprintf('\n (d) T2 = %f K,w = %f J mol^-1,delH = %f J mol^-1,q= 0 J mol^-1 and delE = %f J mol^-1',T2,delE1,delH1,delE1);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.12/Example_2_12.sce b/1019/CH2/EX2.12/Example_2_12.sce new file mode 100644 index 000000000..001d3070f --- /dev/null +++ b/1019/CH2/EX2.12/Example_2_12.sce @@ -0,0 +1,21 @@ +//Example 2.12
+clear;
+clc;
+
+//Given
+R=0.082;// gas constant in atm dm^3 K^-1 mol^-1
+v1=1;// initial volume in dm^3 mol^-1
+v2=50;// fina volume in dm^3 mol^-1
+T=273;// temperature in K
+a=6.5;// van der waals constant inatm dm^6 mol^-2
+b=0.056;//atm dm^3 K^-1 mol^-1
+n=1; //moles of given gas
+
+// To determine t2,w,delH and delE
+w=(-1)*101.325*R*T*log(v2/v1);//work done in J mol^-1
+W=101.325*(-1)*((R*T*log((v2-(n*b))/(v1-(n*b))))+(a*n*n*((1/v2)-(1/v1))));//work done in J in terms of van der waals gas
+delE=(-1)*101.325*a*((1/v2)-(1/v1));//change in internal energy in J mol^-1
+q=delE-W;
+mprintf('(i) For ideal gas, W = %f J mol^-1, delE = 0,q = 0',w);
+mprintf('\n (ii) For Van der Waals gas, W = %f J mol^-1,delE = %f J mol^-1,q = %f J mol^-1',W,delE,q);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.13/Example_2_13.sce b/1019/CH2/EX2.13/Example_2_13.sce new file mode 100644 index 000000000..58deac87b --- /dev/null +++ b/1019/CH2/EX2.13/Example_2_13.sce @@ -0,0 +1,16 @@ +//Example 2.13
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+p1=10;// initial pressure in atm
+p2=1;// final pressure in atm
+T=300;// temperature in K
+
+// To determine Wexp and Wcomp
+Wexp=R*T*(1-(p2/p1));//work done in J mol^-1 during expansion
+Wcomp=(-1)*R*T*(1-(p1/p2));//work done in J mol^-1 during compression
+mprintf('(i) Work done by the system in expansion process = %f J mol^-1',Wexp);
+mprintf('\n (ii) Work done on the system during compression process = %f J mol^-1',Wcomp);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.14/Example_2_14.sce b/1019/CH2/EX2.14/Example_2_14.sce new file mode 100644 index 000000000..b60fff1c7 --- /dev/null +++ b/1019/CH2/EX2.14/Example_2_14.sce @@ -0,0 +1,18 @@ +//Example 2.14
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+p1=2;// initial pressure in atm
+v1=0.5;// initial volume in dm^3
+v2=2;// final volume in dm^3
+V=1.4;// coefficient of adiabatic expansion (gamma)
+
+// To determine Work done
+p2=p1*((v1/v2)^V);//final pressure in atm
+Wad=(-1)*(((p1*101*v1)-(p2*v2*101))/(V-1));//work done in adiabatic process in J
+Wiso=p1*v1*101*log(v2/v1);////work done in isothermal process in J
+mprintf('(i) Work done in adiabatic process = %f J',Wad);
+mprintf('\n (ii) Work done during isothermal process = %f J',Wiso);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.15/Example_2_15.sce b/1019/CH2/EX2.15/Example_2_15.sce new file mode 100644 index 000000000..22479de3d --- /dev/null +++ b/1019/CH2/EX2.15/Example_2_15.sce @@ -0,0 +1,15 @@ +//Example 2.15
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+d1=0.25; //initial gas density in kg mol m^-3
+d2=0.083; //final gas density in kg mol m^-3
+a=138; //van der waals constant in N m^4 mol^-2
+Cv=20.8; //specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+
+// To determine the change in temperature
+delT=(a/Cv)*(d2-d1);//temperature change in K
+mprintf('Change in temperature = %f K',delT);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.16/Example_2_16.sce b/1019/CH2/EX2.16/Example_2_16.sce new file mode 100644 index 000000000..fdd4ead92 --- /dev/null +++ b/1019/CH2/EX2.16/Example_2_16.sce @@ -0,0 +1,23 @@ +//Example 2.16
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+p1=5; //initial gas pressure in atm
+p2=2; //final gas pressure in atm
+Pext=1; //external pressure in atm
+Cv=2.5*R; //specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+n=1.6;//moles of the gas
+T1=300; //initial temperature in K
+
+// To determine q,w,delE and delH
+T2=(60+750)/3;//final temperature in K
+Cp=Cv+R;//specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+delE=n*Cv*(T2-T1);//internal energy change in J
+delH=n*Cp*(T2-T1);//enthalpy change in J
+mprintf('q = 0 J');
+mprintf('\n delE = %f J',delE);
+mprintf('\n delH = %f J',delH);
+mprintf('\n w = %f J',delE);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.17/Example_2_17.sce b/1019/CH2/EX2.17/Example_2_17.sce new file mode 100644 index 000000000..5a18e776e --- /dev/null +++ b/1019/CH2/EX2.17/Example_2_17.sce @@ -0,0 +1,28 @@ +//Example 2.17
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+v1=11.2; //initial gas volume in dm^3
+Cv=2.5*R; //specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+n=2;//moles of the gas
+T1=273; //initial temperature in K
+T2=373; //final temperature in K
+
+// To determine q,w,delE and delH
+Cp=Cv+R;//specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+delE=n*Cv*(T2-T1);//internal energy change in J
+delH=delE+(n*R*(T2-T1));//enthalpy change in J
+q=delE;//heat absorbed in J
+w=n*R*(T2-T1);//work done in J
+mprintf('(a) For isochoric process,');
+mprintf('\n delE = %f J',delE);
+mprintf('\n delH = %f J',delH);
+mprintf('\n Heat absorbed,q = %f J',delE);
+mprintf('\n (b) For isobaric process,');
+mprintf('\n delE = %f J',delE);
+mprintf('\n delH = %f J',delH);
+mprintf('\n Work done by the system,w = %f J',w);
+mprintf('\n Heat absorbed,q = %f J',delE);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.18/Example_2_18.sce b/1019/CH2/EX2.18/Example_2_18.sce new file mode 100644 index 000000000..22553e015 --- /dev/null +++ b/1019/CH2/EX2.18/Example_2_18.sce @@ -0,0 +1,24 @@ +//Example 2.18
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+q=1675;//heat absorbed in J mol^-1
+Cp=2.5*R; //specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+n=1;//moles of the gas
+T1=273; //initial temperature in K
+delH=2078.5;//enthalpy change in J mol^-1
+P1=1;//initial pressure in atm
+
+// To determine P2,w,delE and T2
+Cv=Cp-R;//specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+T2=T1+(delH/Cp);//final temperature in K
+delE=n*Cv*(T2-T1);//internal energy change in J mol^-1
+w=q-delE;//work done in J mol^-1
+P2=(P1*T2)/(T1*2);//final pressure in atm
+mprintf('delE = %f J mol^-1',delE);
+mprintf('\n work done = %f J mol^-1',w);
+mprintf('\n Final temperature = %f K',T2);
+mprintf('\n Final pressure = %f atm',P2);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.19/Example_2_19.sce b/1019/CH2/EX2.19/Example_2_19.sce new file mode 100644 index 000000000..33ef78828 --- /dev/null +++ b/1019/CH2/EX2.19/Example_2_19.sce @@ -0,0 +1,18 @@ +//Example 2.19
+clear;
+clc;
+
+//Given
+T1=273; //initial temperature in K
+T2=1073; //final temperature in K
+w=1;//weight of aluminium taken in kg
+mp=931;//melting point of aluminium in K
+delHm=362.3;//enthalpy change during melting process in kJ kg^-1
+
+// To determine delH
+delH1=(0.9121*(mp-T1))+(2.0083*0.00005*((mp+T1)*(mp-T1)))-(2.0083*0.0001*273*(mp-T1));//enthalpy change in 1st step in kJ
+delHf=delHm*w;//enthalpy change during melting in kJ
+delH3=1.0836*(T2-mp);//enthalpy change in last step in kJ
+delH=delH1+delHf+delH3;
+mprintf('delH = %f kJ ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.2/Example_2_2.sce b/1019/CH2/EX2.2/Example_2_2.sce new file mode 100644 index 000000000..845f54630 --- /dev/null +++ b/1019/CH2/EX2.2/Example_2_2.sce @@ -0,0 +1,21 @@ +//Example 2.2
+clear;
+clc;
+
+//Given
+v1=2.28;//initial volume in m^3
+v2=4.56;//final volume in m^3
+R=8.314;// gas constant in J K^-1 mol^-1
+T=300.15;// temperature in K
+n=1;//moles
+
+// To determine delH,delE,q,w
+w=(-1)*R*T* log(v2/v1);// w in joule mol^-1
+delE=0;//for reversible process
+q=delE-w;//by using 1st law
+delH=0;//as del(PV)=0
+mprintf('w = %f Joule mol^-1',w);
+mprintf('\n q = %f Joule mol^-1',q);
+mprintf('\n delH = %f Joule mol^-1',delH);
+mprintf('\n delE = %f Joule mol^-1 since it is a reversible process',delE);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.20/Example_2_20.sce b/1019/CH2/EX2.20/Example_2_20.sce new file mode 100644 index 000000000..1bc3f6b9a --- /dev/null +++ b/1019/CH2/EX2.20/Example_2_20.sce @@ -0,0 +1,14 @@ +//Example 2.20
+clear;
+clc;
+
+//Given
+T1=300; //initial temperature in K
+T2=400; //final temperature in K
+W=1000; //work obtained in J
+
+// To determine heat withdrawn from the reservoir
+n=1-(T1/T2);//efficiency of the engine
+q=W/n;//heat absorbed in J
+mprintf('heat withdrawn from the reservoir = %f',q);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.21/Example_2_21.sce b/1019/CH2/EX2.21/Example_2_21.sce new file mode 100644 index 000000000..3c0bc980f --- /dev/null +++ b/1019/CH2/EX2.21/Example_2_21.sce @@ -0,0 +1,25 @@ +//Example 2.21
+clear;
+clc;
+
+//Given
+T1=300; //initial temperature in K
+T2=600; //final temperature in K
+R=8.314;// gas constant in J K^-1 mol^-1
+Cv=25; //specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+n=1;//moles of the gas
+P1=100; //initial gas pressure in kN m^-2
+P2=1000; //final gas pressure in kN m^-2
+
+// To determine net work done and efficiency
+w1=R*T1*log(P2/P1);//work done in 1st step in J
+w2=Cv*(T2-T1);//work done in 2nd step in J
+w3=R*T2*log(P1/P2);//work done in 3rd step in J
+q2=(-1)*w3;//heat taken up in J
+w4=Cv*(T1-T2);//work done in final step in J
+W=-(w1+w2+w3+w4);//total work done step in J
+N=W/q2;//efficiency
+mprintf('Net work done = %f',W);
+mprintf('\n Heat absorbed = %f',q2);
+mprintf('\n efficiency = %f',N);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.22/Example_2_22.sce b/1019/CH2/EX2.22/Example_2_22.sce new file mode 100644 index 000000000..145225218 --- /dev/null +++ b/1019/CH2/EX2.22/Example_2_22.sce @@ -0,0 +1,24 @@ +//Example 2.22
+clear;
+clc;
+
+//Given
+T1=373; //initial temperature in K
+R=8.314;// gas constant in J K^-1 mol^-1
+Cv=2.5*R; //specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+n=1;//moles of the gas
+V=1.4;// coefficient of adiabatic expansion (gamma)
+
+// To determine net work done and efficiency
+w1=(-1)*R*T1*log(2);//work done in 1st step in J
+q=-w1;//heat absorbed in J
+T2=T1*((2/3)^(V-1));//final temperature in K
+w2=Cv*(T2-T1);//work done in 2nd step in J
+w3=(-1)*R*T2*log(1/2);//work done in 3rd step in J
+w4=Cv*(T1-T2);//work done in final step in J
+W=w1+w2+w3+w4;//total work done in J
+N=-100*W/q;//efficiency in percent
+mprintf('Net work done = %f',W);
+mprintf('\n delE = 0 since it is a cyclic process');
+mprintf('\n efficiency = %f percent',N);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.3/Example_2_3.sce b/1019/CH2/EX2.3/Example_2_3.sce new file mode 100644 index 000000000..a079f91d5 --- /dev/null +++ b/1019/CH2/EX2.3/Example_2_3.sce @@ -0,0 +1,15 @@ +//Example 2.3
+clear;
+clc;
+
+//Given
+p1=1013.25;//initial pressure in N m^-2
+p2=101325;//final pressure in N m^-2
+R=8.314;// gas constant in J K^-1 mol^-1
+T=300;// temperature in K
+n=0.5;//moles of oxygen present
+
+// To determine minimum work required
+w=n*R*T* log(p2/p1);// w in joule
+mprintf('minimum work required (w) = %f Joule',w);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.4/Example_2_4.sce b/1019/CH2/EX2.4/Example_2_4.sce new file mode 100644 index 000000000..129a9631c --- /dev/null +++ b/1019/CH2/EX2.4/Example_2_4.sce @@ -0,0 +1,19 @@ +//Example 2.4
+clear;
+clc;
+
+//Given
+pH=7.4;//pH of the body fluid
+v=3;//volume of gastric juice produced per day in dm^3
+R=8.314;// gas constant in J K^-1 mol^-1
+T=310;// temperature in K
+ph=1;// pH of the gastric juice produced
+
+// To determine minimum work required
+c1=10^((-1)*pH);// hydrogen ion concentration of body fluid in mol dm^-3
+c2=10^((-1)*ph);//hydrogen ion concentration of the gastric juice in mol dm^-3
+n=c2*v;//moles of hydrogen ions in given volume of gastric juice
+w=n*R*T* log(c2/c1);// w in joule
+W=w*0.001;//w in kJ
+mprintf('minimum work required (w) = %f kJ',W);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.5/Example_2_5.sce b/1019/CH2/EX2.5/Example_2_5.sce new file mode 100644 index 000000000..9a36b5b97 --- /dev/null +++ b/1019/CH2/EX2.5/Example_2_5.sce @@ -0,0 +1,16 @@ +//Example 2.5
+clear;
+clc;
+
+//Given
+n=2;//moles of glucose dissolved
+v=1;//volume of glucose solution in dm^3
+R=8.314;// gas constant in J K^-1 mol^-1
+T=298;// temperature in K
+c2=0.2;//concentration to which the solution was diluted in mol dm^-3
+
+// To determine work done
+c1=n/v;;// initial concentration of glucose solution in mol dm^-3
+w=n*R*T*log(c1/c2);// w in joule
+mprintf('Work done (w) = %f J',w);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.6/Example_2_6.sce b/1019/CH2/EX2.6/Example_2_6.sce new file mode 100644 index 000000000..8f8cc9319 --- /dev/null +++ b/1019/CH2/EX2.6/Example_2_6.sce @@ -0,0 +1,21 @@ +//Example 2.6
+clear;
+clc;
+
+//Given
+n=1;//moles of benzene
+Lv=395;//heat of vapourization of benzene in J g^-1
+R=8.314;// gas constant in J K^-1 mol^-1
+T=353.2;// temperature in K
+m=78;//molecular weight of benzene in g mol^-1
+
+// To determine q,w,delH and DelE
+q=Lv*m;//heat supplied in J mol^-1
+w=(-1)*R*T;// w in joule
+delE=q+w;
+delH=delE-w;
+mprintf('w = %f J mol^-1',w);
+mprintf('\n q = %f J mol^-1',q);
+mprintf('\n delE = %f J mol^-1',delE);
+mprintf('\n delH = %f J mol^-1',delH);
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.7/Example_2_7.sce b/1019/CH2/EX2.7/Example_2_7.sce new file mode 100644 index 000000000..bc8f98c60 --- /dev/null +++ b/1019/CH2/EX2.7/Example_2_7.sce @@ -0,0 +1,19 @@ +//Example 2.7
+clear;
+clc;
+
+//Given
+n=2;//moles of ideal gas
+R=8.314;// gas constant in J K^-1 mol^-1
+T=273;// temperature in K
+p1=10;// initial pressure in atm
+p2=0.4;// final pressure in atm
+
+// To determine q,w,delH and DelE
+w=(-1)*n*R*T*(1-(p2/p1));// w in joule
+q=(-1)*w;
+mprintf('q = %f J',q);
+mprintf('\n w = %f J',w);
+mprintf('\n delE = 0 J,since it is an isothermal process');
+mprintf('\n delH = 0 J,since it is an isothermal process');
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.8/Example_2_8.sce b/1019/CH2/EX2.8/Example_2_8.sce new file mode 100644 index 000000000..bd1e91f52 --- /dev/null +++ b/1019/CH2/EX2.8/Example_2_8.sce @@ -0,0 +1,20 @@ +//Example 2.8
+clear;
+clc;
+
+//Given
+n=5;//moles of ideal gas
+R=8.314;// gas constant in J K^-1 mol^-1
+T=300;// temperature in K
+p1=10;// initial pressure in atm
+p2=4;// final pressure in atm
+P=1;// external pressure in atm
+
+// To determine q,w,delH and DelE
+w=(-1)*n*R*T*((P/p2)-(P/p1));// w in joule
+q=(-1)*w;//q in J
+mprintf('q = %f J',q);
+mprintf('\n w = %f J',w);
+mprintf('\n delE = 0 J,since it is an isothermal process');
+mprintf('\n delH = 0 J,since it is an isothermal process');
+//end
\ No newline at end of file diff --git a/1019/CH2/EX2.9/Example_2_9.sce b/1019/CH2/EX2.9/Example_2_9.sce new file mode 100644 index 000000000..9bbf918bc --- /dev/null +++ b/1019/CH2/EX2.9/Example_2_9.sce @@ -0,0 +1,26 @@ +//Example 2.9
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+t1=298;// temperature in K
+p1=303975;// initial pressure in Pa
+v1=1.43;//initial volume in dm^3
+v2=2.86;// final volume in dm^3
+Cp=29.1;// heat capacity at constant pressure in J mol^-1 K^-1
+
+// To determine t2,p2,w,delH and DelE
+Cv=Cp-R;//heat capacity at constant volume
+t2=t1*((v1/v2)^(R/Cv));//final temperature in K
+p2=p1*((v1/v2)^(Cp/Cv));//final pressure in Pa
+n=(p1*v1*0.001)/(R*t1);//number of moles
+delE=n*Cv*(t2-t1);//delE in J
+w=delE;//work done in J
+delH=n*Cp*(t2-t1);//enthalpy change in J
+mprintf('Final Temperature = %f K',t2);
+mprintf('\n Final Pressure = %f Pa',p2);
+mprintf('\n delE = %f J',delE);
+mprintf('\n delH = %f J',delH);
+mprintf('\n w = %f J',w);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.1/Example_3_1.sce b/1019/CH3/EX3.1/Example_3_1.sce new file mode 100644 index 000000000..f34af9008 --- /dev/null +++ b/1019/CH3/EX3.1/Example_3_1.sce @@ -0,0 +1,14 @@ +//Example 3.1
+clear;
+clc;
+
+//Given
+delE = -97030;//change in internal energy in Joule
+R = 8.314;//R is gas constant in J K^-1 mol^-1
+T = 298;//T is temperature in K
+
+//To determine the heat of reaction
+delv= 1-(1+(1/2));//change in moles
+delH = delE + (delv*R*T);//H is the heat of the reaction in Joule (1st law of thermodynamics)
+mprintf('Heat of the reaction= %f J',delH);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.10/Example_3_10.sce b/1019/CH3/EX3.10/Example_3_10.sce new file mode 100644 index 000000000..df0cf5b63 --- /dev/null +++ b/1019/CH3/EX3.10/Example_3_10.sce @@ -0,0 +1,19 @@ +//Example 3.10
+clear;
+clc;
+
+//Given
+delHfHCl = -92.3;//heat of formation in kJ of HCl
+delHfH2O = -285.8;//heat of formation in kJ of water
+delHfNaCl = -411.0;//heat of formation in kJ of NaCl
+delHfNaOH = -426.7;//heat of formation in kJ of NaOH
+delH1 = -75.7;//heat of reaction(i)
+delH2 = -40.9;//heat of reaction (ii)
+delH3 = 4.26;//heat of reaction (iii)
+//To determine delH
+delHf1=delH1+delHfHCl;//delH for (i) in kJ
+delHf2=delH2+delHfNaOH;//delH for (ii) in kJ
+delHf3=delH3+delHfNaCl;//delH for (iii) in kJ
+delH=delHf3+delHfH2O-(delHf1+delHf2);//delH in kJ
+mprintf('delH = %f kJ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.11/Example_3_11.sce b/1019/CH3/EX3.11/Example_3_11.sce new file mode 100644 index 000000000..4a1303972 --- /dev/null +++ b/1019/CH3/EX3.11/Example_3_11.sce @@ -0,0 +1,12 @@ +//Example 3.11
+clear;
+clc;
+
+//Given
+delHfs = -57.36;//heat of neutralization in kJ of strong acid and strong base
+delHfw = -42.02;//heat of neutralization in kJ of weak acid and weak base
+
+//To determine the heat of ionization of weak acid and weak base
+delHi = delHfw-delHfs;//heat of ionization in kJ
+mprintf('heat of ionization of weak acid and weak base = %f kJ',delHi);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.12/Example_3_12.sce b/1019/CH3/EX3.12/Example_3_12.sce new file mode 100644 index 000000000..a4a19fb40 --- /dev/null +++ b/1019/CH3/EX3.12/Example_3_12.sce @@ -0,0 +1,12 @@ +//Example 3.12
+clear;
+clc;
+
+//Given
+delHfHCl = -168.0;//heat of formation in kJ of HCl
+delHf = 0;//heat of formation of H+ ions in kJ
+
+//To determine the heat of formation of chloride ions
+delHi = delHfHCl-delHf;//heat of formation of chloride ions in kJ
+mprintf('heat of formation of chloride ions = %f kJ',delHi);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.13/Example_3_13.sce b/1019/CH3/EX3.13/Example_3_13.sce new file mode 100644 index 000000000..7bb3f27c2 --- /dev/null +++ b/1019/CH3/EX3.13/Example_3_13.sce @@ -0,0 +1,12 @@ +//Example 3.13
+clear;
+clc;
+
+//Given
+delHfNaOH = -470.7;//heat of formation in kJ of NaOH
+delHfOH = -228.8;//heat of formation of OH- ions in kJ
+
+//To determine the heat of formation of sodium ions
+delHNa = delHfNaOH-delHfOH;//heat of formation of sodium ions in kJ
+mprintf('heat of formation of sodium ions = %f kJ',delHNa);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.14/Example_3_14.sce b/1019/CH3/EX3.14/Example_3_14.sce new file mode 100644 index 000000000..964e117d5 --- /dev/null +++ b/1019/CH3/EX3.14/Example_3_14.sce @@ -0,0 +1,26 @@ +//Example 3.14
+clear;
+clc;
+
+//Given
+delHsubLi = 161;//heat of sublimation of Li in kJ
+delHsubNa = 109;//heat of sublimation of Na in kJ
+delHd = 122;//heat of dissociation of chlorine in kJ
+delHaff = -350;//electron affinity of Cl in kJ
+delHipLi = 520;// ionization potential of Li in kJ
+delHipNa = 496;//ioization potential of Na in kJ
+delHfLiCl = -410;//heat of formation of LiCl in kJ
+delHfNaCl = -411;//heat of formation of NaCl in kJ
+delHLisol= -35.1;//heat of solution of LiCl in kJ
+delHNasol= 4.3;//heat of solution of NaCl in kJ
+
+//To determine the (a) enthalpy change (b) heat of hydration
+delHcLiCl = delHsubLi+delHd+delHipLi+delHaff-delHfLiCl;//born haber cycle
+mprintf('(a) heat of formation of LiCl crystal = %f kJ',delHcLiCl);
+delHcNaCl = delHsubNa+delHd+delHipNa+delHaff-delHfNaCl;//born haber cycle
+mprintf('\n heat of formation of NaCl crystal = %f kJ',delHcNaCl);
+delHLiCl = delHLisol-delHcLiCl;//heat of hydration of LiCl in kJ
+delHNaCl = delHNasol-delHcNaCl;//heat of hydration of NaCl in kJ
+mprintf('\n \n (b) heat of hydration of LiCl = %f kJ',delHLiCl);
+mprintf('\n heat of hydration of NaCl = %f kJ',delHNaCl);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.15/Example_3_15.sce b/1019/CH3/EX3.15/Example_3_15.sce new file mode 100644 index 000000000..0c6afd041 --- /dev/null +++ b/1019/CH3/EX3.15/Example_3_15.sce @@ -0,0 +1,40 @@ +//Example 3.15
+clear;
+clc;
+
+//Given
+delHfC2H6=-84.5;//formation enthalpy of ethane in kJ
+delHfC2H4=52.6;//formation enthalpy of ethene in kJ
+delHfC2H2=226.9;//formation enthalpy of acetylene in kJ
+delHfCH3CHO=-166.3;//formation enthalpy of ethanal in kJ
+delHfH2Og=-241.8;//formation enthalpy of water(gas) in kJ
+delHfCH3OHg=-201.3;//formation enthalpy of methanol(gas) in kJ
+DHO2=-249.17;//half of the bond energy in oxygen molecule in kJ
+DHH2=217.97;//half of the bond energy in hydrogen molecule in kJ
+DHC=716.68;//energy required to obtain free carbon atom from graphite in kJ
+DHCH=413;//bond energy of carbon hydrogen bond
+
+//(i) To determine bond energy of carbon single bond
+delHCC=(2*DHC)-delHfC2H6+(6*DHH2)-(6*DHCH);//bond energy of carbon single bond in kJ
+mprintf('(i) bond energy of carbon single bond = %f kJ',delHCC);
+
+//(ii) To determine bond energy of carbon double bond
+delHC2C=(2*DHC)-delHfC2H4+(4*DHH2)-(4*DHCH);//bond energy of carbon double bond in kJ
+mprintf('\n (ii) bond energy of carbon double bond = %f kJ',delHC2C);
+
+//(iii) To determine bond energy of carbon triple bond
+delHC3C=(2*DHC)-delHfC2H2+(2*DHH2)-(2*DHCH);//bond energy of carbon triple bond in kJ
+mprintf('\n (iii) bond energy of carbon triple bond = %f kJ',delHC3C);
+
+//(iv) To determine the bond energy of O-H bond
+delHOH=((2*DHH2)-delHfH2Og-DHO2)/2;//bond energy of O-H bond in kJ
+mprintf('\n (iv) bond energy of O-H bond = %f kJ',delHOH);
+
+//(v) To determine the bond energy of C=O bond
+delHC2O=(2*DHC)-(4*DHCH)-delHfCH3CHO+(delHCC*2);//bond energy of C=O bond in kJ
+mprintf('\n (v) bond energy of C=O bond = %f kJ',delHC2O);
+
+//(vi) To determine the bond energy of C-O bond
+delHCO=DHC-(3*DHCH)-delHfCH3OHg+delHOH-DHO2;//bond energy of C-O bond in kJ
+mprintf('\n (vi) bond energy of O-H bond = %f kJ',delHCO);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.16/Example_3_16.sce b/1019/CH3/EX3.16/Example_3_16.sce new file mode 100644 index 000000000..0b7ab234a --- /dev/null +++ b/1019/CH3/EX3.16/Example_3_16.sce @@ -0,0 +1,19 @@ +//Example 3.16
+clear;
+clc;
+
+//Given
+DHOH=463;//bond energy of O-H bond in kJ
+DHCO=351;//bond energy of C-O bond in kJ
+DHCC=348;//bond energy of C-C bond in kJ
+DHCH=413;//bond energy of C-H bond in kJ
+DHOO=249.2;//half of the bond energy in oxygen molecule in kJ
+DHHH=217.97;//half of the bond energy in hydrogen molecule in kJ
+DHC=716.68;//energy required to obtain free carbon atom from graphite in kJ
+
+//To determine enthalpy of formation of ethyl alcohol
+BE=(5*DHCH)+DHCC+DHCO+DHOH;//total bond enthalpy in kJ
+EA=(6*DHHH)+(2*DHC)+DHOO;//enthalpy of atomization in kJ
+delHf=EA-BE;//enthalpy of formation in kJ
+mprintf('Enthalpy change = %f kJ',delHf);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.17/Example_3_17.sce b/1019/CH3/EX3.17/Example_3_17.sce new file mode 100644 index 000000000..41de7fdcb --- /dev/null +++ b/1019/CH3/EX3.17/Example_3_17.sce @@ -0,0 +1,14 @@ +//Example 3.17
+clear;
+clc;
+
+//Given
+DHCC=348;//bond energy of C-C bond in kJ
+DHCH=413;//bond energy of C-H bond in kJ
+DHHH=436;//half of the bond energy in hydrogen molecule in kJ
+DHC2C=610;//bond energy of C=C bond in kJ
+
+//To determine enthalpy change
+delHf=DHC2C+DHHH-(2*DHCH)-DHCC;//enthalpy change in kJ
+mprintf('Enthalpy change = %f kJ',delHf);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.18/Example_3_18.sce b/1019/CH3/EX3.18/Example_3_18.sce new file mode 100644 index 000000000..617bb0481 --- /dev/null +++ b/1019/CH3/EX3.18/Example_3_18.sce @@ -0,0 +1,19 @@ +//Example 3.18
+clear;
+clc;
+
+//Given
+DHOH=463;//bond energy of O-H bond in kJ
+DHCO=350;//bond energy of C-O bond in kJ
+DHCC=348;//bond energy of C-C bond in kJ
+DHCH=413;//bond energy of C-H bond in kJ
+DHOO=249.17;//half of the bond energy in oxygen molecule in kJ
+DHHH=217.94;//half of the bond energy in hydrogen molecule in kJ
+DHC=716.7;//energy required to obtain free carbon atom from graphite in kJ
+
+//To determine enthalpy of formation of dimethyl ether
+BE=(6*DHCH)+(2*DHCO);//total bond enthalpy in kJ
+EA=(6*DHHH)+(2*DHC)+DHOO;//enthalpy of atomization in kJ
+delHf=EA-BE;//enthalpy of formation in kJ
+mprintf('Enthalpy of formation of dimethyl ether = %f kJ',delHf);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.19/Example_3_19.sce b/1019/CH3/EX3.19/Example_3_19.sce new file mode 100644 index 000000000..09c456cc5 --- /dev/null +++ b/1019/CH3/EX3.19/Example_3_19.sce @@ -0,0 +1,16 @@ +//Example 3.19
+clear;
+clc;
+
+//Given
+CpH2=28.83;//specific heat at constant pressure of hydrogen in J K^-1 mol^-1
+CpO2=29.12;//specific heat at constant pressure of oyygen in J K^-1 mol^-1
+CpH2O=33.56;//specific heat at constant pressure of water in J K^-1 mol^-1
+delHdH2O=241750;//heat of dissociation of water at 291 K of water in J
+delT=341-291;//change in temperature (K)
+
+//To determine heat of dissociation of water at 341 K
+Cp=CpH2+(CpO2/2)-CpH2O;//specific heat at constant pressure in J K^-1 mol^-1
+delHd=delHdH2O+(Cp*delT);//heat of dissociation of water at 341 K in J mol^-1
+mprintf('heat of dissociation of water at 341 K = %f J mol^-1',delHd);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.2/Example_3_2.sce b/1019/CH3/EX3.2/Example_3_2.sce new file mode 100644 index 000000000..5040abc6b --- /dev/null +++ b/1019/CH3/EX3.2/Example_3_2.sce @@ -0,0 +1,26 @@ +//Example 3.2
+clear;
+clc;
+
+//Given
+delH1 = -393.5//H1 is the heat of reaction in the formation of CARBON DIOXIDE in kJ (i)
+delH2 = -110.5//H2 is the heat of reaction in the formation of CARBON MONOXIDE in kJ (ii)
+delH3 = -890.35//H3 is the heat of reaction in the combustion of METHANE in kJ (iii)
+delH4 = -85.3//H4 is the heat of reaction in the formation of SILVER CHLORIDE in kJ (iv)
+R = 0.008314;//R is gas constant in kJ K^-1 mol^-1
+T = 298;//T is temperature in K
+
+//To determine the heat of formation
+delv1= 1-(1);//change in moles in reaction (i)
+delE1 = delH1 - (delv1*R*T);//E1 is the internal energy (i) in kJ (1st law of thermodynamics)
+mprintf('(i) change in internal energy = %f kJ',delE1);
+delv2=1-(0.5);//change in moles in reaction (ii)
+delE2 = delH2 - (delv2*R*T);//E2 is the internal energy (ii) in kJ (1st law of thermodynamics)
+mprintf('\n (ii) change in internal energy = %f kJ',delE2);
+delv3=1-(1+2);//change in moles in reaction (iii)
+delE3 = delH3 - (delv3*R*T);//E3 is the internal energy (iii) in kJ (1st law of thermodynamics)
+mprintf('\n (iii) change in internal energy = %f kJ',delE3);
+delv4=0-(1);//change in moles in reaction (iv)
+delE4 = delH4 - (delv4*R*T);//E4 is the internal energy (iv) in kJ (1st law of thermodynamics)
+mprintf('\n (iv) change in internal energy = %f kJ',delE4);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.21/Example_3_21.sce b/1019/CH3/EX3.21/Example_3_21.sce new file mode 100644 index 000000000..eb12f0451 --- /dev/null +++ b/1019/CH3/EX3.21/Example_3_21.sce @@ -0,0 +1,13 @@ +//Example 3.21
+clear;
+clc;
+
+//Given
+delE=-240500;//energy liberated in J mol^-1
+T1=300;//initial temperature in K
+Cv=24.15;//specific heat of water at constant volume in J K^-1 mol^-1
+
+//To determine maximum temperature
+T2=(-delE/Cv)+300;//maximum temperature in K
+mprintf('The maximum temperature = %i K',T2);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.23/Example_3_23.sce b/1019/CH3/EX3.23/Example_3_23.sce new file mode 100644 index 000000000..1b8b9e4a5 --- /dev/null +++ b/1019/CH3/EX3.23/Example_3_23.sce @@ -0,0 +1,15 @@ +//Example 3.23
+clear;
+clc;
+
+//Given
+delHfH2O=-286;//enthalpy of formation of liquid water in kJ mol^-1
+delHfCO2=-394;//enthalpy of formation of carbon dioxide in kJ mol^-1
+delHfC6H12O6=-1260;//enthalpy of formation of glucose in water in kJ mol^-1
+
+//To determine power output of the brain
+delH=(6*delHfH2O)+(6*delHfCO2)-(delHfC6H12O6);//heat of combustion in kJ
+q=delH/18;//heat supplied by 10 g glucose in kJ
+p=-(1000*q)/3600;//power in Watt
+mprintf('Power output of the brain = %f W',p);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.25/Example_3_25.sce b/1019/CH3/EX3.25/Example_3_25.sce new file mode 100644 index 000000000..ecdae6900 --- /dev/null +++ b/1019/CH3/EX3.25/Example_3_25.sce @@ -0,0 +1,24 @@ +//Example 3.25
+clear;
+clc;
+
+//Given
+delH1=-242;//heat of reaction (i) in kJ mol^-1
+delH2=-640;//heat of reaction (ii) in kJ mol^-1
+delH3=-540;//heat of reaction (iii) in kJ mol^-1
+
+//To determine enthalpy changes per kg
+nH2=200;//number of moles of hydrogen in 1 kg gas (mol)
+nO2=31.25;//number of moles of oxygen in 1 kg gas (mol)
+nCH3OH=31.25;////number of moles of methanol in 1 kg (mol)
+nF2=26.3;//number of moles of flourine in 1 kg gas (mol)
+
+delH11=delH1*(2*nO2);//enthalpy of reaction (i)
+mprintf('enthalpy of reaction (i) = %i kJ',delH11);
+
+delH22=delH2*(20.8);//enthalpy of reaction (ii)
+mprintf('\n enthalpy of reaction (ii) = %i kJ',delH22);
+
+delH33=delH3*(nF2);//enthalpy of reaction (iii)
+mprintf('\n enthalpy of reaction (iii) = %i kJ',delH33);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.26/Example_3_26.sce b/1019/CH3/EX3.26/Example_3_26.sce new file mode 100644 index 000000000..14692fb39 --- /dev/null +++ b/1019/CH3/EX3.26/Example_3_26.sce @@ -0,0 +1,20 @@ +//Example 3.26
+clear;
+clc;
+
+//Given
+Cp=4.2;//specific heat at constant pressure in J/K g
+delH=10^7;//heat in J
+T1=310;//initial temperature (K)
+m=70000;//mass of the man in g
+delHvap=2405;//latent heat of vapourization of water in J/g
+//(a) To determine temterature after 24 hours
+T2=(delH+(m*Cp*T1))/(m*Cp);//temperature after 24 hours in Kelvin
+t2=T2-273;//temperature in degree Celsius
+mprintf('(a) Temperature after 24 hours = %i oC',t2);
+
+//(b) To determine the amount of water evaporated to maintain temperature
+m=(delH*1)/delHvap;//mass of water in g
+M=m/1000;//mass of water in kg
+mprintf('\n (b) mass of water evaporated to maintain temperature = %f kg',M);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.27/Example_3_27.sce b/1019/CH3/EX3.27/Example_3_27.sce new file mode 100644 index 000000000..c8149b731 --- /dev/null +++ b/1019/CH3/EX3.27/Example_3_27.sce @@ -0,0 +1,16 @@ +//Example 3.27
+clear;
+clc;
+
+//Given
+E=-21;
+c1=3;
+delE = E * 10^6;//evolved energy in J
+c = c1 * 10^8;//speed of light in m/s^2
+
+//To determine change in mass
+delm = delE/(c^2);//change in mass in kg by Einstein equation
+m=E/(c1^2);
+t=6-(2*8)
+mprintf('change in mass = %f*10^%i kg',m,t);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.3/Example_3_3.sce b/1019/CH3/EX3.3/Example_3_3.sce new file mode 100644 index 000000000..53a61b9c0 --- /dev/null +++ b/1019/CH3/EX3.3/Example_3_3.sce @@ -0,0 +1,12 @@ +//Example 3.3
+clear;
+clc;
+
+//Given
+delH1 = -78.705;//H1 is the heat of reaction (i) in kJ
+delH2 = -3.420;//H2 is the heat of reaction (ii) in kJ
+
+//To determine the heat change
+delH=delH1-delH2;//change in heat in kJ
+mprintf('change in heat = %f kJ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.4/Example_3_4.sce b/1019/CH3/EX3.4/Example_3_4.sce new file mode 100644 index 000000000..32c8af36b --- /dev/null +++ b/1019/CH3/EX3.4/Example_3_4.sce @@ -0,0 +1,13 @@ +//Example 3.4
+clear;
+clc;
+
+//Given
+delH1 = 10.72;//delH1 is the heat of reaction in kJ (i)
+delH2 = 4.68;//delH2 is the heat of reaction in kJ (ii)
+delH3 = -1.16;//delH3 is the heat of reaction in kJ (iii)
+
+//To determine the heat of required reaction
+delH = delH1-delH2+delH3;//heat of reaction in kJ
+mprintf('heat of required reaction = %f kJ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.5/Example_3_5.sce b/1019/CH3/EX3.5/Example_3_5.sce new file mode 100644 index 000000000..7fe28b261 --- /dev/null +++ b/1019/CH3/EX3.5/Example_3_5.sce @@ -0,0 +1,13 @@ +//Example 3.5
+clear;
+clc;
+
+//Given
+delH1 = -890.35;//delH1 is the heat of reaction in kJ (i)
+delH2 = -285.84;//delH2 is the heat of reaction in kJ (ii)
+delH3 = -393.51;//delH3 is the heat of reaction in kJ (iii)
+
+//To determine the heat of formation of methane
+delH = delH3+(2*delH2)-delH1;//heat of formation of methane in kJ
+mprintf('heat of formation of methane = %f kJ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.6/Example_3_6.sce b/1019/CH3/EX3.6/Example_3_6.sce new file mode 100644 index 000000000..39734026d --- /dev/null +++ b/1019/CH3/EX3.6/Example_3_6.sce @@ -0,0 +1,14 @@ +//Example 3.6
+clear;
+clc;
+
+//Given
+delHfCO2 = -393.5;//heat of formation in kJ of carbondioxide
+delHfH2O = -285.8;//heat of formation in kJ of water
+delHfC2H6 = -84.5;//heat of formation in kJ ethane
+delHfO2 = 0;//heat of formation of oxygen
+
+//To determine the enthalpy change for given reaction
+delH = (2*delHfCO2)+(3*delHfH2O)-(delHfC2H6+(3.5*delHfO2));//enthalpy change in kJ
+mprintf('enthalpy change for given reaction = %f kJ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.7/Example_3_7.sce b/1019/CH3/EX3.7/Example_3_7.sce new file mode 100644 index 000000000..75f69d72c --- /dev/null +++ b/1019/CH3/EX3.7/Example_3_7.sce @@ -0,0 +1,13 @@ +//Example 3.7
+clear;
+clc;
+
+//Given
+delHfCO2 = -393.5;//heat of formation in kJ of carbondioxide
+delHfH2O = -285.8;//heat of formation in kJ of water
+delH = -3303;//heat of reaction in kJ
+
+//To determine the heat of formation of benzene
+delHfC6H6 = (3*delHfH2O)+(6*delHfCO2)-(delH);//heat of formation of benzene in kJ
+mprintf('heat of formation of benzene = %f kJ',delHfC6H6);
+//end
\ No newline at end of file diff --git a/1019/CH3/EX3.9/Example_3_9.sce b/1019/CH3/EX3.9/Example_3_9.sce new file mode 100644 index 000000000..5124ef82f --- /dev/null +++ b/1019/CH3/EX3.9/Example_3_9.sce @@ -0,0 +1,16 @@ +//An Introduction to Chemical Thermodynamics
+//Chapter 3
+//Thermochemistry
+
+//Example 3.9
+clear;
+clc;
+
+//Given
+delHf1 = -92.3;//heat of formation in kJ of HCl
+delHf2 = -168.0;//heat of formation in kJ of HCl.100 H2O
+
+//To determine the heat of solution of HCl in 100 H20
+delH = delHf2-delHf1;//heat of solution of HCl in 100 H20 in kJ
+mprintf('heat of solution of l in 100 H2O = %f kJ',delH);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.10/Example_4_10.sce b/1019/CH4/EX4.10/Example_4_10.sce new file mode 100644 index 000000000..978b688a1 --- /dev/null +++ b/1019/CH4/EX4.10/Example_4_10.sce @@ -0,0 +1,17 @@ +//Example 4.10
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+Cp=2.5*R; //specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+V1=10;//volume of gas in m^3
+T1=300; //initial temperature in K
+T2=400;//final temperature in K
+P=101000;//pressure in N m^-2
+
+//to calculate the entropy change
+n=(P*V1*0.001)/(R*T);//moles
+delS=n*(Cp*log(T2/T1));//entropy change in J K^-1
+mprintf('Change in entropy = %f J K^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.11/Example_4_11.sce b/1019/CH4/EX4.11/Example_4_11.sce new file mode 100644 index 000000000..7253d5925 --- /dev/null +++ b/1019/CH4/EX4.11/Example_4_11.sce @@ -0,0 +1,19 @@ +//Example 4.11
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+T=300; //initial temperature in K
+r=3;//ratio of final volume to initial volume
+
+//To calculate the entropy
+q=R*T*log(r);//r=V2/V1
+n=(P*V1*0.001)/(R*T);//moles
+delSsys=q/T;//entropy of system in J K^-1 mol^-1
+delSuniv=R*log(r);//entropy of universe in J K^-1
+mprintf('(a) Entropy of the system = %f J K^-1 mol^-1',delSsys);
+mprintf('\n Entropy of the surrounding = 0 J K^-1 mol^-1');
+mprintf('\n \n (b) Entropy of the system = 0 J K^-1 mol^-1');
+mprintf('\n Entropy of the surrounding = %f J K^-1 mol^-1',delSuniv);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.13/Example_4_13.sce b/1019/CH4/EX4.13/Example_4_13.sce new file mode 100644 index 000000000..4396813f0 --- /dev/null +++ b/1019/CH4/EX4.13/Example_4_13.sce @@ -0,0 +1,19 @@ +//Example 4.13
+clear;
+clc;
+
+//Given
+R=0.082;// gas constant in dm^3 atm K^-1 mol^-1
+R1=8.314;// gas constant in J K^-1 mol^-1
+T1=298; //initial temperature in K
+T2=373; //final temperature in K
+V1=0.5;//initial volume in dm^3
+V2=1;//final volume in dm^3
+P1=1;//initial pressure in atm
+Cv=12.6;//specific heat of the gas at constant volume in J K^-1 mol^-1
+
+//To calculate the entropy change
+n=P1*V1/(R*T1);//moles
+delS=(n*Cv*log(T2/T1))+(n*R1*log(V2/V1));//entropy change in J K^-1
+mprintf('Entropy change = %f J K^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.14/Example_4_14.sce b/1019/CH4/EX4.14/Example_4_14.sce new file mode 100644 index 000000000..9dd4ccec5 --- /dev/null +++ b/1019/CH4/EX4.14/Example_4_14.sce @@ -0,0 +1,24 @@ +//Example 4.14
+clear;
+clc;
+
+//Given
+Cpw=75.42;//heat capacity of water in J K^-1 mol^-1
+T=363; //temperature of water in K
+P=1;//pressure in atm
+Cpi=37.20;//heat capacity of ice in J K^-1 mol^-1
+delHf=5980;// latent heat of fusion in J mol^-1
+mp=273;//melting point of water in K
+n=10/18;//moles of ice taken
+
+//To calculate the entropy change
+T2=306;//since q=0
+q1=n*delHf;//heat in J
+q2=n*Cpw*(T2-mp);//heat in J
+q3=n*2*Cpi*(T2-T);//heat in J
+delS1=q1/mp;//entropy change during 1st step in J K^-1
+delS2=n*Cpw*log(T2/mp);//entropy change during 2nd step in J K^-1
+delS3=2*n*Cpw*log(T2/T);//entropy change during final step in J K^-1
+delS=delS1+delS2+delS3;//total entropy change in J K^-1
+mprintf('Entropy change = %f J K^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.15/Example_4_15.sce b/1019/CH4/EX4.15/Example_4_15.sce new file mode 100644 index 000000000..f779b7e24 --- /dev/null +++ b/1019/CH4/EX4.15/Example_4_15.sce @@ -0,0 +1,30 @@ +//Example 4.15
+clear;
+clc;
+
+//Given
+Cpw=75.42;//heat capacity of water in J K^-1 mol^-1
+T=263; //temperature in K
+P=1;//pressure in atm
+Cpi=37.20;//heat capacity of ice in J K^-1 mol^-1
+delHf=6008;// latent heat of fusion in J mol^-1
+mp=273;//melting point of water in K
+n=1;//moles of ice taken
+
+//To calculate the entropy change
+delS1=Cpw*log(mp/T);//entropy change during 1st step in J K^-1
+delS2=-delHf/mp;//entropy change during 2nd step in J K^-1
+delS3=Cpi*log(T/mp);//entropy change during final step in J K^-1
+delS=delS1+delS2+delS3;//total entropy change in J K^-1
+mprintf('Entropy change = %f J K^-1 mol^-1',delS);
+delH1=Cpw*(mp-T);//enthalpy change during 1st step in J
+delH2=-delHf;//enthalpy change during 2nd step in J
+delH3=Cpi*(T-mp);//entropy change during final step in J
+delHsys=delH1+delH2+delH3;//total enthalpy change in J
+delSsurr=-delHsys/T;//entropy change of surrounding in J K^-1
+delSuni=delS+delSsurr;//entropy of universe in J K^-1
+mprintf('\n Enthalpy change of the system = %f J mol^-1',delHsys);
+mprintf('\n Enthalpy change of the surrounding = %f J',-delHsys);
+mprintf('\n Entropy change of the surrounding = %f J K^-1',delSsurr);
+mprintf('\n Entropy change of the universe = %f J K^-1',delSuni);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.17/Example_4_17.sce b/1019/CH4/EX4.17/Example_4_17.sce new file mode 100644 index 000000000..5ec59146c --- /dev/null +++ b/1019/CH4/EX4.17/Example_4_17.sce @@ -0,0 +1,17 @@ +//Example 4.14
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+T=273;//temperature in K
+V1=1;//initial volume in dm^3
+V2=50;//final volume in dm^3
+a=6.5;// Van der Waals constant in atm dm^6 mol^-2
+b=0.056;//Van der Waals constant in dm^3 mol^-1
+
+//to determine entropy change
+qrev=R*T*log((V2-b)/(V1-b));
+delS=qrev/T;//entropy change in J K^-1 mol^-1
+mprintf('Change in entropy = %f J K^-1 mol^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.18/Example_4_18.sce b/1019/CH4/EX4.18/Example_4_18.sce new file mode 100644 index 000000000..ef2c0434b --- /dev/null +++ b/1019/CH4/EX4.18/Example_4_18.sce @@ -0,0 +1,23 @@ +//Example 4.14
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+T=298;//temperature in K
+r=2.5;//rato of final volume to initial volume
+n=2;//moles of gas
+
+//To determine entropy change
+delSgas=R*n*log(r);
+qrev=n*R*T*log(r);
+delSsurr=-qrev/T;//entropy of the surrounding
+delSt=delSgas+delSsurr;//total entropy change in J K^-1
+mprintf('(a) delSgas = %f J K^-1 \n delSsurr = %f J K^-1 \n delStotal = %f J K^-1',delSgas,delSsurr,delSt);
+delSgas=R*n*log(r);
+qirr=qrev-800;
+delSsur=-qirr/T;//entropy of the surrounding
+delSto=delSgas+delSsur;//total entropy change in J K^-1
+mprintf('\n (b) delSgas = %f J K^-1 \n delSsurr = %f J K^-1 \n delStotal = %f J K^-1',delSgas,delSsur,delSto);
+mprintf('\n (c) delSgas = %f J K^-1 \n delSsurr = 0 J K^-1 \n delStotal = %f J K^-1',delSgas,delSgas);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.19/Example_4_19.sce b/1019/CH4/EX4.19/Example_4_19.sce new file mode 100644 index 000000000..fe5013fbf --- /dev/null +++ b/1019/CH4/EX4.19/Example_4_19.sce @@ -0,0 +1,20 @@ +//Example 4.19
+clear;
+clc;
+
+//Given
+delHfus=15648;//Latent heat of fusion in J mol^-1
+Tm=386.6;//melting point in K
+delHvap=25522;//latent heat of vapourization in J mol^-1
+Cp=81.588;//heat capacity in J K^-1 mol^-1
+Tb=457;//boiling point in K
+T=298;
+
+//To determine the entropy change
+delS1=(54.684*log(Tm/T))+(13.431*0.0001*(Tm-T))-(298*13.431*0.0001*log(Tm/T));//entropy change in 1st step in J K^-1
+delS2=delHfus/Tm;//entropy change in 2nd step in J K^-1
+delS3=Cp*log(Tb/Tm);//entropy change in 3rd step in J K^-1
+delS4=delHvap/Tb;//entropy change in final step in J K^-1
+delS=delS1+delS2+delS3+delS4;//total entropy change in J K^-1
+mprintf('Change in entropy = %f J K^-1 mol^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.20/Example_4_20.sce b/1019/CH4/EX4.20/Example_4_20.sce new file mode 100644 index 000000000..6fef0bf53 --- /dev/null +++ b/1019/CH4/EX4.20/Example_4_20.sce @@ -0,0 +1,14 @@ +//Example 4.20
+clear;
+clc;
+
+//Given
+SoC2H5OH=160.7;//So for ethanol
+SoC=5.7;//So for graphite
+SoH2=130.6;//So for hydrogen
+SoO2=205.1;//So for oxygen
+
+//To determine the standard entropy of formation of ethanol
+delSo=SoC2H5OH-((2*SoC)+(3*SoH2)+(0.5*SoO2));//standard entropy of formation of ethanol in J K^-1 mol^-1
+mprintf('Standard entropy of formation of ethanol = %f J K^-1 mol^-1',delSo);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.3/Example_4_3.sce b/1019/CH4/EX4.3/Example_4_3.sce new file mode 100644 index 000000000..ef65c4050 --- /dev/null +++ b/1019/CH4/EX4.3/Example_4_3.sce @@ -0,0 +1,14 @@ +//Example 4.3 (b)
+clear;
+clc;
+
+//Given
+a=1.24;//alpha at 290K and 1 atm in 10^-3 K^-1
+b=9.3;//beta at 290K and 1 atm in 10^-5 atm^-1
+T=290;//temperature in K
+delS=2.1;//entropy change in J K^-1 mol^-1
+
+//to calculate the change in molar volume
+delV=(delS*b)/(a*100*101.325);//change in molar volume in dm^3 mol^-1
+mprintf('change in molar volume = %f dm^3 mol^-1',delV);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.4/Example_4_4.sce b/1019/CH4/EX4.4/Example_4_4.sce new file mode 100644 index 000000000..65deeeacb --- /dev/null +++ b/1019/CH4/EX4.4/Example_4_4.sce @@ -0,0 +1,18 @@ +//Example 4.4
+clear;
+clc;
+
+//Given
+n=1;//moles of ice
+Tm=273;//melting temperature in K
+P=1;//pressure in atm;
+delHf=6008;// enthalpy of fusion in J mol^-1
+Tb=373;//boiling temperature in K
+delHv=40850;// enthalpy of fusion in J mol^-1
+
+//to calculate the change in entropy
+delSm=delHf/Tm;//entropy change during melting in J mol^-1 K^-1
+mprintf('(a) change in entropy during melting = %f J K^-1 mol^-1',delSm);
+delSv=delHv/Tb;//entropy change during boiling in J mol^-1 K^-1
+mprintf('\n (b) change in entropy during boiling = %f J K^-1 mol^-1',delSv);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.5/Example_4_5.sce b/1019/CH4/EX4.5/Example_4_5.sce new file mode 100644 index 000000000..3cc247a08 --- /dev/null +++ b/1019/CH4/EX4.5/Example_4_5.sce @@ -0,0 +1,18 @@ +//Example 4.5
+clear;
+clc;
+
+//Given
+n=1;//moles of ice
+Ttrans=286;//melting temperature in K
+P=1;//pressure in atm;
+delHtrans=2090;// enthalpy of transformation in J mol^-1
+Tb=373;//boiling temperature in K
+delHv=40850;// enthalpy of fusion in J mol^-1
+
+//to calculate the change in entropy
+delSv=delHv/Tb;//entropy change during boiling in J mol^-1 K^-1
+mprintf('(a) change in entropy during boiling of water = %f J K^-1 mol^-1',delSv);
+delStrans=delHtrans/Ttrans;//entropy change during transition in J mol^-1 K^-1
+mprintf('\n (b) change in entropy during phase transformation of Sn = %f J K^-1 mol^-1',delStrans);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.6/Example_4_6.sce b/1019/CH4/EX4.6/Example_4_6.sce new file mode 100644 index 000000000..ce9345d63 --- /dev/null +++ b/1019/CH4/EX4.6/Example_4_6.sce @@ -0,0 +1,21 @@ +//Example 4.6
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+Cv=20.8; //specific heat capacity at constant volume of the gas in J K^-1 mol^-1
+w=0.1;//weight of the gas in kg
+T1=298; //initial temperature in K
+P1=30;//initial pressure in atm
+P2=10;//final pressure in atm
+
+//to calculate the entropy
+Cp=Cv+R;//specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+n=(w*1000)/28;//moles
+T2=T1*((Cv+(R*(P2/P1)))/Cp);//final temperature in K
+r=(P1*T2)/(T1*P2);//r=V2/V1
+delS=(n*Cv*log(T2/T1))+(n*R*log(r));//entropy change in J K^-1
+mprintf('Entropy of the system = %f J K^-1',delS);
+mprintf('\n final temperature = %f K',T2);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.7/Example_4_7.sce b/1019/CH4/EX4.7/Example_4_7.sce new file mode 100644 index 000000000..28f422771 --- /dev/null +++ b/1019/CH4/EX4.7/Example_4_7.sce @@ -0,0 +1,15 @@ +//Example 4.7
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+Cp=23.7; //specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+n=3;//moles of gas
+T1=300; //initial temperature in K
+T2=1000;//final temperature in K
+
+//to calculate the entropy
+delS=n*Cp*log(T2/T1);//entropy in J K^-1
+mprintf('Change in Entropy of the system = %f J K^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH4/EX4.9/Example_4_9.sce b/1019/CH4/EX4.9/Example_4_9.sce new file mode 100644 index 000000000..cedce2d3f --- /dev/null +++ b/1019/CH4/EX4.9/Example_4_9.sce @@ -0,0 +1,16 @@ +//Example 4.9
+clear;
+clc;
+
+//Given
+R=8.314;// gas constant in J K^-1 mol^-1
+Cp=20.9; //specific heat capacity at constant pressure of the gas in J K^-1 mol^-1
+n=1;//moles of gas
+delSm=146;//molar entropy of the gas at 298 K in J K^-1 mol^-1
+T1=298; //initial temperature in K
+T2=500;//final temperature in K
+
+//to calculate the molar entropy at 500 K
+delS=delSm+(Cp*log(T2/T1));//molar entropy in J K^-1 mol^-1
+mprintf('Molar Entropy at 500K = %f J K^-1 mol^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.1/Example_5_1.sce b/1019/CH5/EX5.1/Example_5_1.sce new file mode 100644 index 000000000..1730cf510 --- /dev/null +++ b/1019/CH5/EX5.1/Example_5_1.sce @@ -0,0 +1,15 @@ +//Example 5.1
+clear;
+clc;
+
+//Given
+n=4;//moles of gas
+P1=2.02;//initial pressure in 10^5 N m^-2
+P2=4.04;//final pressure in 10^5 N m^-2
+R=8.314;//gas constant in J K^-1 mol^-1
+T=300//temperature in K
+
+//To determine the free energy change delG
+delG=n*R*T*log(P2/P1);//the free energy change in J
+mprintf('Free energy change = %f J',delG);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.10/Example_5_10.sce b/1019/CH5/EX5.10/Example_5_10.sce new file mode 100644 index 000000000..81e44af29 --- /dev/null +++ b/1019/CH5/EX5.10/Example_5_10.sce @@ -0,0 +1,16 @@ +//Example 5.10
+clear;
+clc;
+
+//Given
+delG=18660-(14.4*T*log10(T))-(6.07*T)+(8.24*(10^(-3))*(T^2));//delGo in J mol^-1 in terms of temperature T in K
+T1=298//temperature in K
+
+//To determine delGo delSo and delHo
+delGo=18660-(14.4*T1*log10(T1))-(6.07*T1)+(8.24*(10^(-3))*(T^2));
+delHo=18660+(6.25*T1)-(8.24*10^(-3)*(T1^2));
+delSo=(delHo-delGo)/T1;
+mprintf('(i) delGo = %f J mol^-1',delGo);
+mprintf('\n (ii) delSo = %f J K^-1 mol^-1',delSo);
+mprintf('\n (iii) delHo = %f J mol^-1',delHo);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.11/Example_5_11.sce b/1019/CH5/EX5.11/Example_5_11.sce new file mode 100644 index 000000000..190934397 --- /dev/null +++ b/1019/CH5/EX5.11/Example_5_11.sce @@ -0,0 +1,19 @@ +//Example 5.11
+clear;
+clc;
+
+//Given
+DHoHH=435;//Bond dissociation energy of H-H bond in kJ mol^-1
+DHoClCl=240;//Bond dissociation energy of Cl-Cl bond in kJ mol^-1
+DHoHCl=430;//Bond dissociation energy of H-Cl bond in kJ mol^-1
+SoH2=130.59;//Standard entropy of hydrogen molecule in J K^- mol^-1
+SoCl2=222.95;//Standard entropy of chlorine molecule J K^- mol^-1
+SoHCl=186.68;//Standard entropy of HCl molecule J K^- mol^-1
+T=298;//Temperature in K
+
+//To determine the free energy change
+delHo=DHoHH+DHoClCl-(2*DHoHCl);
+delSo=(SoHCl*2)-SoCl2-SoH2;
+delGo=delHo-(T*delSo*10^(-3));
+mprintf('Free energy change = %f kJ',delGo);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.12/Example_5_12.sce b/1019/CH5/EX5.12/Example_5_12.sce new file mode 100644 index 000000000..056a5a6f0 --- /dev/null +++ b/1019/CH5/EX5.12/Example_5_12.sce @@ -0,0 +1,21 @@ +//Example 5.12
+clear;
+clc;
+
+//Given
+HfNH4NO3=-365;//enthalpy of formation of NH4OH in kJ mol^-1
+HfH2=0;//enthalpy of formation of H2 in kJ mol^-1
+HfH2O=-242;//enthalpy of formation of H2O in kJ mol^-1
+HfN2H4=50;//enthalpy of formation of N2H4 in kJ mol^-1
+SoNH4NO3=150;//Standard entropy of NH4NO3 molecule in J K^- mol^-1
+SoH2=130;//Standard entropy of Hydrogen molecule J K^- mol^-1
+SoH2O=189;//Standard entropy of H2O molecule J K^- mol^-1
+SoN2H4=120;//Standard entropy of N2H4 molecule J K^- mol^-1
+T=298;//Temperature in K
+
+//To determine the free energy change
+delHo=(3*HfH2O)+HfN2H4-HfNH4NO3-(3*HfH2);
+delSo=(SoH2O*3)+SoN2H4-SoNH4NO3-(3*SoH2);
+delGo=delHo-(T*delSo*10^(-3));
+mprintf('Free energy change = %f kJ',delGo);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.13/Example_5_13.sce b/1019/CH5/EX5.13/Example_5_13.sce new file mode 100644 index 000000000..fdc678ded --- /dev/null +++ b/1019/CH5/EX5.13/Example_5_13.sce @@ -0,0 +1,14 @@ +//Example 5.13
+clear;
+clc;
+
+//Given
+T=300;//temperature in K
+delVg=-1.5;//moles of gaseous product-moles of gaseous reactant
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine the difference between delG and delA
+a=delG-delA;//assume
+a=delVg*R*T;//difference between delG and delA in J
+mprintf('delG - delA = %f J mol^-1',a);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.14/Example_5_14.sce b/1019/CH5/EX5.14/Example_5_14.sce new file mode 100644 index 000000000..1d1b03ce6 --- /dev/null +++ b/1019/CH5/EX5.14/Example_5_14.sce @@ -0,0 +1,12 @@ +//Example 5.14
+clear;
+clc;
+
+//Given
+delGo1=-29.2;//delGo value for hydrolysis of creatine phosphate in kJ
+delGo2=-12.4;//delGo value for hydrolysis of glucose phosphate in kJ
+
+//To determine delGo for given reaction
+delGo3=delGo2-delGo1;//gibbs free energy in kJ
+mprintf('delGo for the given reaction = %f kJ',delGo3);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.16/Example_5_16.sce b/1019/CH5/EX5.16/Example_5_16.sce new file mode 100644 index 000000000..7f59ba985 --- /dev/null +++ b/1019/CH5/EX5.16/Example_5_16.sce @@ -0,0 +1,13 @@ +//Example 5.16
+clear;
+clc;
+
+//Given
+delE=-2880;//internal energy in kJ mol^-1
+delS=182.4;//Entropy in J K^-1 mol^-1
+T=298;//Temperature in K
+
+//To determine delA
+delA=delE-(T*delS*0.001);//helmoltz free energy in kJ mol^-1
+mprintf('The amount of energy that can be extracted as heat = %f kJ mol^-1',delA);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.17/Example_5_17.sce b/1019/CH5/EX5.17/Example_5_17.sce new file mode 100644 index 000000000..f8ddc53d2 --- /dev/null +++ b/1019/CH5/EX5.17/Example_5_17.sce @@ -0,0 +1,12 @@ +//Example 5.17
+clear;
+clc;
+
+//Given
+delGo1=3.0;//delGo value for conversion of malate to fumarate in kJ
+delGo2=-15.5;//delGo value for conversion of fumarate to asparate in kJ
+
+//To determine delGo for given reaction
+delGo3=delGo2+delGo1;//net free energy change for the required reaction in kJ
+mprintf('delGo for the conversion of malate to asparate = %f kJ',delGo3);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.18/Example_5_18.sce b/1019/CH5/EX5.18/Example_5_18.sce new file mode 100644 index 000000000..5e0b4581c --- /dev/null +++ b/1019/CH5/EX5.18/Example_5_18.sce @@ -0,0 +1,16 @@ +//Example 5.18
+clear;
+clc;
+
+//Given
+delHv=40820;//latent heat of vapourization of water in J mol^-1
+Vv=30.199;//volume of vapour in dm^3 mol^-1
+Vl=0.019;//volume of liquid in dm^3 mol^-1
+T=373;//temperature in K
+
+//To determine the change in boiling point with change in 1 mm pressure
+delVm=Vv-Vl;
+a=(delHv*760)/(T*delVm*0.001*101325);//a=(dP/dT)
+b=a^(-1);//b=(dT/dP)
+mprintf('change in boiling point of water per mm change in pressure=%f K mm^-1',b);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.19/Example_5_19.sce b/1019/CH5/EX5.19/Example_5_19.sce new file mode 100644 index 000000000..e6d9dafee --- /dev/null +++ b/1019/CH5/EX5.19/Example_5_19.sce @@ -0,0 +1,15 @@ +//Example 5.19
+clear;
+clc;
+
+//Given
+delHf=128.6;//latent heat of fusion of benzene in J g^-1
+Vs=1.06;//volume of solid in cm^3 g^-1
+Vl=1.119;//volume of liquid in cm^3 g^-1
+T=278;//temperature in K
+
+//To determine the pressure to bring about a change in melting point by 1 K
+delVm=Vl-Vs;//change in volume in cm^3 g^-1
+a=(delHf*10)/(T*delVm*1.01325);//a=(dP/dT)
+mprintf('To cause an increase of 1K in melting point of benzene,atmospheric pressure change required=%f atm K^-1',a);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.2/Example_5_2.sce b/1019/CH5/EX5.2/Example_5_2.sce new file mode 100644 index 000000000..f13075a95 --- /dev/null +++ b/1019/CH5/EX5.2/Example_5_2.sce @@ -0,0 +1,13 @@ +//Example 5.2
+clear;
+clc;
+
+//Given
+n=2;//number of electrons transferred
+E=1.1;//cell potential in volt
+F=96500;//Farady charge in C
+
+//To determine the free energy change delG
+delG=-n*F*E;//the free energy change in J
+mprintf('Free energy change for Daniell Cell = %f J',delG);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.20/Example_5_20.sce b/1019/CH5/EX5.20/Example_5_20.sce new file mode 100644 index 000000000..5a49afdc4 --- /dev/null +++ b/1019/CH5/EX5.20/Example_5_20.sce @@ -0,0 +1,16 @@ +//Example 5.20
+clear;
+clc;
+
+//Given
+delHf=335;//latent heat of fusion in J g^-1
+Vs=1.0908;//volume of solid in cm^3 g^-1
+Vl=1.0002;//volume of liquid in cm^3 g^-1
+T=273;//temperature in K
+
+//To determine the decrease in melting point with increase in pressure
+delVm=Vl-Vs;//volume change in cm^3 g^-1
+a=(delHf*10)/(T*delVm*1.01325);//a=(delP/delT)
+b=a^(-1);//b=(delT/delP)
+mprintf('An increase in pressure of 1 atm lowers the freezing point by %f K atm^-1',b);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.21/Example_5_21.sce b/1019/CH5/EX5.21/Example_5_21.sce new file mode 100644 index 000000000..80ce60377 --- /dev/null +++ b/1019/CH5/EX5.21/Example_5_21.sce @@ -0,0 +1,14 @@ +//Example 5.21
+clear;
+clc;
+
+//Given
+delHtrans=13.4;//latent heat of fusion in J g^-1
+delVm=0.0126;//change in volume due to transition in cm^3 g^-1
+T=368.5;//temperature in K
+
+//To determine the increase in the transition point between 2 forms of sulphur for increase in atmospheric pressure
+a=(delHtrans*10)/(T*delVm*1.01325);//a=(delP/delT)
+b=a^(-1);//b=(delT/delP)
+mprintf('The transition point between 2 forms of sulphur should be increased by %f K atm^-1',b);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.22/Example_5_22.sce b/1019/CH5/EX5.22/Example_5_22.sce new file mode 100644 index 000000000..226a97a82 --- /dev/null +++ b/1019/CH5/EX5.22/Example_5_22.sce @@ -0,0 +1,14 @@ +//Example 5.22
+clear;
+clc;
+
+//Given
+//log10(P)=(-834.13/T)+(1.75*log(T))-(8.375*0.001*T)+5.3234 pressure in mm of Hg as a function og temperature in K
+Tb=169.25;//boiling point of ethylene in K
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine latent heat of vapourization delHv
+//delHv=R*(T^2)*(dlogP/dT) by clausius clapeyron equation
+delHv=((-R*834.13*2.303)+(1.75*R*Tb*2.303)-(8.375*2.303*0.001*R*(Tb^2)))*0.001;//latent heat of vapourization in kJ
+mprintf('The latent heat of vapourization delHv of ethylene = %f kJ mol^-1',delHv);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.23/Example_5_23.sce b/1019/CH5/EX5.23/Example_5_23.sce new file mode 100644 index 000000000..b47aa79b1 --- /dev/null +++ b/1019/CH5/EX5.23/Example_5_23.sce @@ -0,0 +1,13 @@ +//Example 5.23
+clear;
+clc;
+
+//Given
+//log10(P)=-(1246.038/(t+221.354))+6.95926 vapour pressure in mm Hg as a function of temperature in oC
+T=298;//temperature in K
+
+//To determine the delHv
+//delHv=R*(T^2)*(dlogP/dT) by clausius clapeyron equation
+delHv=((2.303*1246.038*R*(T^2))/((T-51.796)^2))*0.001;//latent heat of vapourization in kJ mol^-1
+mprintf('The latent heat of vapourization delHv of thiophene = %f kJ mol^-1',delHv);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.24/Example_5_24.sce b/1019/CH5/EX5.24/Example_5_24.sce new file mode 100644 index 000000000..4f8c46e1d --- /dev/null +++ b/1019/CH5/EX5.24/Example_5_24.sce @@ -0,0 +1,20 @@ +//Example 5.24
+clear;
+clc;
+
+//Given
+p1=10;//vapour pressure of decane in Torr at temperature T1 K
+p2=400;//vapour pressure of decane in Torr at temperature T2 K
+T1=328.85;//initial temperature in K
+T2=423.75;//final temperature in K
+T=373;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine the delHv,Tb,delSv
+delHv=(T1*T2*R*log(p2/p1))/94.9;//integrated form of clausius clapeyron equation
+delSv=delHv/T;//entrpoy change during vapourization in J K^-1 mol^-1
+Tb=((((R*2.303*log10(76))/delHv)+(1/T1))^(-1))+186;//boiling temperature in K
+mprintf('delHv = %f J mol^-1',delHv);
+mprintf('\n delSv = %f J K^-1 mol^-1',delSv);
+mprintf('\n Tb = %f K',Tb);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.25/Example_5_25.sce b/1019/CH5/EX5.25/Example_5_25.sce new file mode 100644 index 000000000..a3c473480 --- /dev/null +++ b/1019/CH5/EX5.25/Example_5_25.sce @@ -0,0 +1,20 @@ +//Example 5.25
+clear;
+clc;
+
+//Given
+p1=10;//vapour pressure in mm Hg at temperature T1 K
+p2=40;//vapour pressure in mm Hg at temperature T2 K
+T1=358.95;//initial temperature in K
+T2=392.45;//final temperature in K
+Ts=325.75;//surrounding temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+ps=1;//
+//To determine the delHv,Tb,delSv
+delHv=(T1*T2*R*log(p2/p1))/33.5;//integrated form of clausius clapeyron equation
+Tb=((1/T1)-(19.147*log10(76)/delHv))^(-1);//boiling temperature in K
+delSv=delHv/Tb;//entropy in vapourization in J K^-1 mol^-1
+mprintf('(i) delHv = %f J mol^-1',delHv);
+mprintf('\n (ii) Tb = %f K',Tb);
+mprintf('\n (iii) delSv = %f J K^-1 mol^-1',delSv);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.3/Example_5_3.sce b/1019/CH5/EX5.3/Example_5_3.sce new file mode 100644 index 000000000..d1d639158 --- /dev/null +++ b/1019/CH5/EX5.3/Example_5_3.sce @@ -0,0 +1,19 @@ +//Example 5.3
+clear;
+clc;
+
+//Given
+n=2;//number of electrons transferred
+E=1.01463;//cell potential in V
+F=96500;//Farady charge in C
+T=298;//temperature in K
+p=-5*(10^(-5));//p=(delE/delT)p in V K^-1
+
+//To determine the free energy change delG
+delG=-n*F*E;//the free energy change in J
+mprintf('delG for Westron Cell = %f J',delG);
+delS=n*F*(p);//entropy change in J mol^-1
+mprintf('\n delS for Westron Cell = %f J K^-1',delS);
+delH=delG+(T*delS);//enthalpy change in J
+mprintf('\n delH for Westron Cell = %f J',delH);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.4/Example_5_4.sce b/1019/CH5/EX5.4/Example_5_4.sce new file mode 100644 index 000000000..fe910cd37 --- /dev/null +++ b/1019/CH5/EX5.4/Example_5_4.sce @@ -0,0 +1,27 @@ +//Example 5.4
+clear;
+clc;
+
+//Given
+n=1;//moles of gas
+V1=2;//initial volume in dm^3
+V2=20;//final volume in dm^3
+R=8.314;//gas constant in J K^-1 mol^-1
+T=300//temperature in K
+
+//To determine q,w,delE,delA,delG and delS
+w=-R*T*log(V2/V1);//work done in J
+delE=0;//isothermal expansion of ideal gas
+q=delE-w;//by 1st Law of thermodynamics
+delH=0;//delH=delE+del(n*R*T) and both are 0
+delA=-n*R*T*log(V2/V1);//helmoltz free energy in J
+delG=n*R*T*log(V1/V2);//Gibbs free energy in J
+delS=q/T;//entropy change in J K^-1
+mprintf('(i) w = %f J mol^-1',w);
+mprintf('\n (ii) delE = %f J since it is isothermal expansion of an ideal gas',delE);
+mprintf('\n (iii) q = %f J mol^-1',q);
+mprintf('\n (iv) delH = %f J mol^-1',delH);
+mprintf('\n (v) delA = %f J mol^-1',delA);
+mprintf('\n (vi) delG = %f J mol^-1',delG);
+mprintf('\n (vii) delS = %f J K^-1 mol^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.5/Example_5_5.sce b/1019/CH5/EX5.5/Example_5_5.sce new file mode 100644 index 000000000..9dee08ec8 --- /dev/null +++ b/1019/CH5/EX5.5/Example_5_5.sce @@ -0,0 +1,29 @@ +//Example 5.5
+clear;
+clc;
+
+//Given
+n=1;//moles of gas
+P1=10.1;//initial pressure in 10^5 N m^-2
+P2=1.01;//final pressure in 10^5 N m^-2
+R=8.314;//gas constant in J K^-1 mol^-1
+T=300//temperature in K
+
+//To determine delE,delH,delA;delG and delS
+w=0;//since the gas expands against zero pressure
+delG=n*R*T*log(P2/P1);//gibbs free energy in J
+delE=0;//isothermal expansion of ideal gas
+q=delE-w;//by 1st Law of thermodynamics
+delH=0;//delH=delE+del(n*R*T) and both are 0
+delA=n*R*T*log(P2/P1);//Helmoltz free energy in J
+delS=R*log(P2/P1);//entropy change in J K^-1
+delSsurr=0;//entropy of the surrounding
+delSuniv=delS+delSsurr;//entropy of the universe in J K^-1
+mprintf('(i) delE = %f J mol^-1 since it is isothermal expansion of an ideal gas',delE);
+mprintf('\n (ii) q = %f J mol^-1',q);
+mprintf('\n (iii) delH = %f J mol^-1',delH);
+mprintf('\n (iv) delA = %f J mol^-1',delA);
+mprintf('\n (v) delG = %f J mol^-1',delG);
+mprintf('\n (vi) delS of system = %f J K^-1 mol^-1',delS);
+mprintf('\n (vii) delS of universe = %f J K^-1 mol^-1',delSuniv);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.6/Example_5_6.sce b/1019/CH5/EX5.6/Example_5_6.sce new file mode 100644 index 000000000..e1f1a8b55 --- /dev/null +++ b/1019/CH5/EX5.6/Example_5_6.sce @@ -0,0 +1,24 @@ +//Example 5.6
+clear;
+clc;
+
+//Given
+n=1;//moles of toluene
+P=1;//pressure in atm
+R=8.314;//gas constant in J K^-1 mol^-1
+T=384//temperature in K
+delHv=363.3;//latent heat of vapourization in J g^-1
+
+//To determine q,delH,delG,delE and delS
+w=R*T;// work done in J
+qp=delHv*(n*92);
+delE=qp-w;//internal energy change in J
+delH=qp;//enthalpy change in J
+delG=0;//gibbs free energy in J
+delS=delH/T;//entropy change in J K^-1
+mprintf('(i) delE = %f J mol^-1 ',delE);
+mprintf('\n (ii) q = %f J mol^-1',qp);
+mprintf('\n (iii) delH = %f J mol^-1',delH);
+mprintf('\n (iv) delG = %f J mol^-1',delG);
+mprintf('\n (v) delS of system = %f J K^-1 mol^-1',delS);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.7/Example_5_7.sce b/1019/CH5/EX5.7/Example_5_7.sce new file mode 100644 index 000000000..4a82bb5ca --- /dev/null +++ b/1019/CH5/EX5.7/Example_5_7.sce @@ -0,0 +1,16 @@ +//Example 5.7
+clear;
+clc;
+
+//Given
+P1=2.15;//vapour pressure of water in mm of Hg
+P2=1.95;//vapour pressure of ice in mm of Hg
+R=8.314;//gas constant in J K^-1 mol^-1
+T=263//temperature in K
+
+//To determine the free energy change delG
+delG=R*T*log(P2/P1);//gibbs free energy in J mol^-1
+mprintf('(i) Free energy change = 0 J mol^-1');
+mprintf('\n (ii) Free energy change = %f J mol^-1',delG);
+mprintf('\n (iii) Total Free energy change = %f J mol^-1',delG);
+//end
\ No newline at end of file diff --git a/1019/CH5/EX5.9/Example_5_9.sce b/1019/CH5/EX5.9/Example_5_9.sce new file mode 100644 index 000000000..f4923f89c --- /dev/null +++ b/1019/CH5/EX5.9/Example_5_9.sce @@ -0,0 +1,28 @@ +//Example 5.9
+clear;
+clc;
+
+//Given
+Cpl=75.4;//heat capacity of water in J K^-1 mol^-1
+Cpv=33.2;//heat capacity of water vapour in J K^-1 mol^-1
+R=8.314;//gas constant in J K^-1 mol^-1
+T=300//temperature in K
+delHv=40850;//latent heat of vapourization in J mol^-1
+Tb=373;//boiling point of water in K
+P1=101325;//initial pressure in Pa
+P2=10132.5;//final pressure in Pa
+
+//To determine the free energy change delG
+delH1=Cpl*(Tb-T);//enthalpy change during 1st step in J
+delS1=Cpl*log(Tb/T);//entropy change during 1st step in J K^-1
+delH2=delHv;//enthalpy change during 2nd step in J
+delS2=delHv/Tb;//entropy change during 2nd step in J K^-1
+delH3=Cpv*(T-Tb);//enthalpy change during 3rd step in J
+delS3=Cpv*log(T/Tb);//entropy change during 3rd step in J K^-1
+delH4=0;//enthalpy change during final step in J
+delS4=R*log(P1/P2);//entropy change during final step in J K^-1
+delH=delH1+delH2+delH3+delH4;//total enthalpy in J
+delS=delS1+delS2+delS3+delS4;//total entropy change in J
+delG=delH-(T*delS);//gibbs free energy change in J
+mprintf('Free energy change = %f J mol^-1',delG);
+//end
\ No newline at end of file diff --git a/1019/CH6/EX6.1/Example_6_1.sce b/1019/CH6/EX6.1/Example_6_1.sce new file mode 100644 index 000000000..a69813562 --- /dev/null +++ b/1019/CH6/EX6.1/Example_6_1.sce @@ -0,0 +1,8 @@ +//Example 6.1 +clear; +clc; + +//To calculate the number of ways of distributing distinguishable molecules a,b,c between 3 energy levels +w=(3*2*1)/(1*1*1);//ways of distributing distinguishable molecules a,b,c between 3 energy levels +mprintf('ways of distributing distinguishable molecules a,b,c between 3 energy levels = %i',w); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.10/Example_6_10.sce b/1019/CH6/EX6.10/Example_6_10.sce new file mode 100644 index 000000000..88fd3e22a --- /dev/null +++ b/1019/CH6/EX6.10/Example_6_10.sce @@ -0,0 +1,23 @@ +//Example 6.10 +clear; +clc; + +//Given +Et=3.716;//transitioal energy in J +T=298;//temperature in K +Ht=6.193;//transitional enthalpy in J +R=8.314;//gas constant in J mol^-1 K^-1 +Cp=20.785;//transitional heat capacity in J K^-1 mol^-1 +h=6.626*10^(34);//plancks constant in Js +NA=6.023*10^(23);//Avogadro number +k=1.38*(10^(-23));//in J K^-1 +P=101325;//Pressure in N m^-2 +m=5.313*10^-26;//mass of one molecule of O2 in kg + +//To calculate the transitional entropy,work function,and free energy for oxygen gas +qt=7.906*10^6;//the transitional partion function +St=R*(2.5+log(qt));//transitional entropy in J K^-1 mol^-1 +Gt=-R*T*log(qt);//transitional free energy in J mol^-1 +mprintf('the transitional entropy = %f J K^-1 mol^-1',St); +mprintf('\n the transitional free energy = %f J mol^-1',Gt); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.12/Example_6_12.sce b/1019/CH6/EX6.12/Example_6_12.sce new file mode 100644 index 000000000..ad2f829fb --- /dev/null +++ b/1019/CH6/EX6.12/Example_6_12.sce @@ -0,0 +1,15 @@ +//Example 6.12
+clear;
+clc;
+
+//Given
+T=300;//temperature in K
+s=2;//symmetry number
+I=4.59;//moment of inertia in 10^(-47) kg m^2
+h=6.626;//plancks constant in 10^(34) Js
+k=1.38;//in (10^(-23)) J K^-1
+
+//To determine the rotational partition function
+Qr=((8*(%pi^2)*I*k*T)/(s*(h^2)))*0.001;//rotational partition function
+mprintf('rotational partition function,qr = %f',Qr);
+//end
\ No newline at end of file diff --git a/1019/CH6/EX6.13/Example_6_13.sce b/1019/CH6/EX6.13/Example_6_13.sce new file mode 100644 index 000000000..4f3c9c54b --- /dev/null +++ b/1019/CH6/EX6.13/Example_6_13.sce @@ -0,0 +1,19 @@ +//Example 6.13
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+P=1;//pressure in atm
+I=1.9373;//moment of inertia in 10^(-46) kg m^2
+R=8.314;//gas constant in J K^-1 mol^-1
+h=6.626;//plancks constant in 10^(34) Js
+k=1.38;//in (10^(-23)) J K^-1
+
+//To determine the rotational contributions to entropy and free energy
+Qr=(8*(%pi^2)*I*k)/(20*(h^2));//rotational transition function
+Sr=R*(1+log(Qr*T));//rotational contributions to entropy in J K^-1 mol^-1
+Gr=((R*T)-(T*Sr))*0.001;//rotational contributions to free energy in J mol^-1
+mprintf('rotational contributions to entropy = %f J K^-1 mol^-1',Sr);
+mprintf('\n rotational contributions to free energy = %f kJ mol^-1',Gr);
+//end
\ No newline at end of file diff --git a/1019/CH6/EX6.14/Example_6_14.sce b/1019/CH6/EX6.14/Example_6_14.sce new file mode 100644 index 000000000..caffc36b0 --- /dev/null +++ b/1019/CH6/EX6.14/Example_6_14.sce @@ -0,0 +1,15 @@ +//Example 6.14
+clear;
+clc;
+
+//Given
+T=300;//temperature in K
+w=4405;//vibrational frequency in cm^-1
+h=6.626*10^(34);//plancks constant in Js
+k=1.38*(10^(-23));//in J K^-1
+c=3*10^8;//speed of light in m s^-2
+
+//To determine the vibrational partition function for hydrogen molecule
+qv=(1-exp(-(h*c*w)/(k*T)));//the vibrational partition function for hydrogen molecule
+mprintf('the vibrational partition function for hydrogen molecule = %f',qv);
+//end
\ No newline at end of file diff --git a/1019/CH6/EX6.15/Example_6_15.sce b/1019/CH6/EX6.15/Example_6_15.sce new file mode 100644 index 000000000..c1519de93 --- /dev/null +++ b/1019/CH6/EX6.15/Example_6_15.sce @@ -0,0 +1,21 @@ +//Example 6.15
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+h=6.626;//plancks constant in 10^(34) Js
+k=1.3806;//in (10^(-23)) J K^-1
+c=2.997925;//speed of light in 10^8 m s^-2
+w=158020;//vibrational frequency in m^-1
+R=8.314;//gas constant in J mol^-1 K^-1
+
+//To determine the vibrational contributions
+x=(h*c*w)/(1000*k);//temperature in K
+Hv=R*T*((x/T)/(exp(x/298)-1));//the vibrational contributions to enthalpy in J
+Sv=R*(((x/T)/(exp(x/298)-1))-log(1)-(exp(-x/T)));//the vibrational contributions to entropy in J K^-1
+Gv=Hv-(T*Sv);////the vibrational contributions to free energy in J mol^-1
+mprintf('the vibrational contributions to enthalpy = %f J',Hv);
+mprintf('\n the vibrational contributions to entropy = %f J K^-1',Sv);
+mprintf('\n the vibrational contributions to free energy = %f J mol^-1',-Gv);
+//end
\ No newline at end of file diff --git a/1019/CH6/EX6.2/Example_6_2.sce b/1019/CH6/EX6.2/Example_6_2.sce new file mode 100644 index 000000000..2f9b60e46 --- /dev/null +++ b/1019/CH6/EX6.2/Example_6_2.sce @@ -0,0 +1,8 @@ +//Example 6.2 +clear; +clc; + +//To calculate the number of ways of distributing 4 between 4 energy levels so that there are 2 molecules in e1,1in e2 and 0 in e3 energy level +w=(4*3*2*1)/(2*1*1*1);//number of ways of distributing 4 between 4 energy levels so that there are 2 molecules in e1,1in e2 and 0 in e3 energy level +mprintf('number of ways of distributing 4 between 4 energy levels so that there are 2 molecules in e1,1in e2 and 0 in e3 energy level = %i',w); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.3/Example_6_3.sce b/1019/CH6/EX6.3/Example_6_3.sce new file mode 100644 index 000000000..d2a364dc0 --- /dev/null +++ b/1019/CH6/EX6.3/Example_6_3.sce @@ -0,0 +1,28 @@ +//Example 6.3 +clear; +clc; + +//To calculate the number of ways of distribute +//(i)2 distinguishable objects in two boxes +N=2;//number of objects +l=2;//number of boxes +w=l^N;//number of configurations +mprintf('the number of ways of distribute (i)2 distinguishable objects in 2 boxes = %i',w); + +//(ii)2 distinguishable objects in 3 boxes +N=2;//number of objects +l=3;//number of boxes +w=l^N;//number of configurations +mprintf('\n the number of ways of distribute (ii)2 distinguishable objects in 3 boxes = %i',w); + +//(iii)2 indistinguishable objects in 2 boxes +N=2;//number of objects +l=2;//number of boxes +w=(3*2*1)/(2*1*1);//number of configurations +mprintf('\n the number of ways of distribute (iii)2 indistinguishable objects in 2 boxes = %i',w); +//(iv)2 indistinguishable objects in 3 boxes +N=2;//number of objects +l=3;//number of boxes +w=(4*3*2*1)/(2*1*2*1);//number of configurations +mprintf('\n the number of ways of distribute (iv)2 indistinguishable objects in 3 boxes = %i',w); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.4/Example_6_4.sce b/1019/CH6/EX6.4/Example_6_4.sce new file mode 100644 index 000000000..cfbb530b4 --- /dev/null +++ b/1019/CH6/EX6.4/Example_6_4.sce @@ -0,0 +1,16 @@ +//Example 6.4 +clear; +clc; + +//Given +delHfus=6.0;//heatoffusion of water in kJ mol^-1 +T=273;//temperature in K +k=1.38*(10^(-23));//in J K^-1 +NA=6.023*(10^23);//avogadros number + +//To calculate the number of distinguishable states of water at 273 K +delS=(1000*delHfus)/(NA*T);//entropy change in J mol^-1 K^-1 +w=delS/(2.303*k);//w=log10(W) +W=10^(w);//number of distinguishable states +mprintf('number of distinguishable states = %i',W); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.5/Example_6_5.sce b/1019/CH6/EX6.5/Example_6_5.sce new file mode 100644 index 000000000..6e467af67 --- /dev/null +++ b/1019/CH6/EX6.5/Example_6_5.sce @@ -0,0 +1,18 @@ +//Example 6.5 +clear; +clc; + +//Given +T1=323;//temperature in K +T2=322;//temperature in K +T21=324;//temperature in K +n=2;//moles of water +NA=6.023*(10^23);//avogadros number +Cp=75;//specific heat at constant pressure of water in J K^-1 mol^-1 +k=1.38*(10^(-23));//in J K^-1 + +//To calculate the probability +delS=Cp*log(T2*T21/T1^2);//entropy change in J mol^-1 K^-1 +p=exp(delS/k);//probability +mprintf('Probability that 2 moles of water at 323 K will break down into two drops at 322 and 324 K each 1 mole is negligibly small = %f',p); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.6/Example_6_6.sce b/1019/CH6/EX6.6/Example_6_6.sce new file mode 100644 index 000000000..19903ccaa --- /dev/null +++ b/1019/CH6/EX6.6/Example_6_6.sce @@ -0,0 +1,16 @@ +//Example 6.6 +clear; +clc; + +//Given +delHfus=6.0;//heatoffusion of water in kJ mol^-1 +T=298;//temperature in K +k=1.38*(10^(-23));//in J K^-1 +N=10;//number of molecules + +//To calculate the probabilty that 10 molecules will be found in half of the container +delS=N*k*log(0.5);//entropy change in J mol^-1 K^-1 +w=delS/(2.303*k);//w=log10(W) +W=10^(w);//probabilty that 10 molecules will be found in half of the container +mprintf('probabilty that 10 molecules will be found in half of the container = %f',W); +//end
\ No newline at end of file diff --git a/1019/CH6/EX6.7/Example_6_7.sce b/1019/CH6/EX6.7/Example_6_7.sce new file mode 100644 index 000000000..ed4c8fe09 --- /dev/null +++ b/1019/CH6/EX6.7/Example_6_7.sce @@ -0,0 +1,15 @@ +//Example 6.7
+clear;
+clc;
+
+//Given
+NA=6.023*(10^23);//avogadros number
+W=6;//number of orientations
+n=1;//moles present
+N=NA;//number of particles
+R=8.314;//gas constant in J mol^-1 K^-1
+
+//To determine the residual entropy of a crystal in which the molecules can adapt 6 orientations of equal energy at 0 K
+S=R*log(W);//residual entropy in J K^-1 mol^-1
+mprintf('the residual entropy of a crystal in which the molecules can adapt 6 orientations of equal energy at 0 K = %f J K^-1 mol^-1',S);
+//end
\ No newline at end of file diff --git a/1019/CH6/EX6.8/Example_6_8.sce b/1019/CH6/EX6.8/Example_6_8.sce new file mode 100644 index 000000000..ae5a3b400 --- /dev/null +++ b/1019/CH6/EX6.8/Example_6_8.sce @@ -0,0 +1,17 @@ +//Example 6.8 +clear; +clc; + +//Given +V=24.4;//volume in dm^3 +T=298;//temperature in K +P=1;//pressure in atm +R=8.314;//gas constant in J mol^-1 K^-1 +h=6.62*10^(-26);//plancks constant in J s +m=5.313*10^(-26);//mass of 1 O2 molecule in kg +k=1.38*(10^(-23));//in J K^-1 + +//To calculate the transitional partion function +qt=(((2*%pi*m*k*T)^(3/2))*V)/((h^3)*10^9);//the transitional partion function +mprintf('the transitional partion function = %f * 10^30 ',qt); +//end
\ No newline at end of file diff --git a/1019/CH7/EX7.1/Example_7_1.sce b/1019/CH7/EX7.1/Example_7_1.sce new file mode 100644 index 000000000..c4a82673e --- /dev/null +++ b/1019/CH7/EX7.1/Example_7_1.sce @@ -0,0 +1,27 @@ +//Example 7.1 +clear; +clc; + +//Given +w2=4.450;//weight of solute in g +m2=98;//molecular mass of solute in g mol^-1 +W1=0.0822;//weight of solvent in kg +w1=82.2;//weight of solvent in g +m1=18;//molecular mass of solvent in g mol^-1 +p=1.029;//density of solution in g cm^-3 + +//To calculate mass percent,mole fraction,mole percent,molarity,molality,normality +P2=(w2/(w1+w2))*100;//Mass percent +mprintf('(a) Mass percent = %f',P2); +x2=(w2/m2)/((w1/m1)+(w2/m2));//Mole fraction +mprintf('\n (b) Mole fraction = %f',x2); +M=x2*100;//Mole percent +mprintf('\n (c) Mole percent = %f',M); +V=(w1+w2)/p;//volume in cm^3 +c2=(w2/m2)*(1000/V);//Molarity in mol dm^-3 +mprintf('\n (d) Molarity = %f mol dm^-3',c2); +M2=w2/(m2*W1);//Molality in mol kg^-1 +mprintf('\n (e) Molality = %f mol kg^-1',M2); +N=(w2/(m2/2))*((1000/V));//normality +mprintf('\n (f) normality = %f',N); +//end
\ No newline at end of file diff --git a/1019/CH7/EX7.10/Example_7_10.sce b/1019/CH7/EX7.10/Example_7_10.sce new file mode 100644 index 000000000..359cd58f0 --- /dev/null +++ b/1019/CH7/EX7.10/Example_7_10.sce @@ -0,0 +1,15 @@ +//Example 7.10
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+T=298;//temperature in K
+To=353;//Equillibrium temperature in K
+delHfus=19290;//Latent heat of fusion in J mol^-1
+
+//To determine the ideal solubility of napthlene at 298 K
+X=(delHfus/R)*((1/To)-(1/T));//X=log(x)
+x=exp(X);//x is the solubility
+mprintf('The ideal solubility of napthlene at 298 K = %f',x);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.11/Example_7_11.sce b/1019/CH7/EX7.11/Example_7_11.sce new file mode 100644 index 000000000..68f1bac5f --- /dev/null +++ b/1019/CH7/EX7.11/Example_7_11.sce @@ -0,0 +1,17 @@ +//Example 7.11
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+T=293;//temperature in K
+w2=2;//weight of the solute in g
+w1=100;//weight of solvent(benzene) in g
+M1=78;//molecular mass of solvent
+p1=74.66;//vapour pressure of pure benzene in mm Hg
+P1=74.01;//vapour pressure of benzene in the mixture in mm Hg
+
+//To determine the molecular weight of the hydrocarbon
+M2=(w2*M1*p1)/(w1*(p1-P1));//molecular weight of the hydrocarbon in g mol^-1
+mprintf('The molecular weight of the hydrocarbon is = %f g mol^-1',M2);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.12/Example_7_12.sce b/1019/CH7/EX7.12/Example_7_12.sce new file mode 100644 index 000000000..d6270351e --- /dev/null +++ b/1019/CH7/EX7.12/Example_7_12.sce @@ -0,0 +1,14 @@ +//Example 7.12
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+Tb=353.1;//Boiling temperature in K
+delHvap=30.67;//heat of vapourization of benzene in kJ mol^-1
+M1=78;//molecular mass of benzene in gm
+
+//To determine the molal boiling point elevation constant of benzene
+Kb=(R*(Tb^2)*M1)/(10^6*delHvap);//molal boiling point elevation constant of benzene in K kg mol^-1
+mprintf('The molal boiling point elevation constant of benzene is = %f K kg mol^-1',Kb);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.13/Example_7_13.sce b/1019/CH7/EX7.13/Example_7_13.sce new file mode 100644 index 000000000..95ccd468c --- /dev/null +++ b/1019/CH7/EX7.13/Example_7_13.sce @@ -0,0 +1,20 @@ +//Example 7.13
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+Tb=353.1;//temperature in K
+w2=13.86;//weight of the solute in g
+w1=100;//weight of solvent in g
+M1=78;//molecular mass of solvent in g
+M2=154;//molecular mass of solute in g
+delTb=2.3;//elevation in boiling point in K
+
+//To determine the Kb and delHvap
+m=(w2/M2)*(1000/w1);//molality in mol kg^-1
+Kb=delTb/m;//boiling point elevation constant in K mol^-1 kg
+delHvap=(R*(Tb^2)*M1)/(1000*Kb);//heat of vapourization in J mol^-1
+mprintf('The heat of vapourization, delHvap = %f J mol^-1',delHvap);
+mprintf('\n The molal boiling point elevation constant of benzene is = %f K kg mol^-1',Kb);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.14/Example_7_14.sce b/1019/CH7/EX7.14/Example_7_14.sce new file mode 100644 index 000000000..b71b955b5 --- /dev/null +++ b/1019/CH7/EX7.14/Example_7_14.sce @@ -0,0 +1,18 @@ +//Example 7.14
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+Tb=353.1;//temperature in K
+w2x=0.5;//weight of the solute x in g
+w2y=0.6;//weight of the solute y in g
+Mx=128;//molecular mass of solute x in g
+w1=50;//weight of solvent in g
+delTx=0.4;//elevation in boiling point due to x in K
+delTy=0.6;//elevation in boiling point due to y in K
+
+//To determine My
+My=w2y*Mx*delTx/(delTy*w2x);//molecular mass of y in g mol^-1
+mprintf('The molecular mass of y is %f g mol^-1',My);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.15/Example_7_15.sce b/1019/CH7/EX7.15/Example_7_15.sce new file mode 100644 index 000000000..8d88989cb --- /dev/null +++ b/1019/CH7/EX7.15/Example_7_15.sce @@ -0,0 +1,14 @@ +//Example 7.15
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+To=278.15;//Freezing temperature in K
+delHfus=9830;//heat of fusion of benzene in J mol^-1
+M1=78;//molecular mass of benzene in g
+
+//To determine the molal freezing point depression constant of benzene
+Kf=(R*(To^2)*M1)/(1000*delHfus);//molal freezing point depression constant of benzene in K kg mol^-1
+mprintf('The molal freezing point depression constant of benzene = %f K kg mol^-1',Kf);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.16/Example_7_16.sce b/1019/CH7/EX7.16/Example_7_16.sce new file mode 100644 index 000000000..907cdd227 --- /dev/null +++ b/1019/CH7/EX7.16/Example_7_16.sce @@ -0,0 +1,15 @@ +//Example 7.16
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+delTf=10;//Freezing temperature depression in K
+Kf=1.86;//molal freezing point depression constant of water K mol^-1 kg
+M2=32;//molecular mass of methyl alcohol in g
+w1=100;//mass of water in g
+
+//To determine the mass of methyl alcohol required
+w2=(delTf*M2*w1)/(Kf*1000);//mass of methyl alcohol required in g
+mprintf('The mass of methyl alcohol required = %f g',w2);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.17/Example_7_17.sce b/1019/CH7/EX7.17/Example_7_17.sce new file mode 100644 index 000000000..883722a4f --- /dev/null +++ b/1019/CH7/EX7.17/Example_7_17.sce @@ -0,0 +1,18 @@ +//Example 7.17
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+delTf=0.52;//Freezing temperature depression in K
+Kf=12;//molal freezing point depression constant of the solvent K mol^-1 kg
+w2=0.9;//mass of solute in g
+w1=180;//mass of solvent in g
+To=282;//freezing point of the solvent in K
+
+//To determine the molecular formula of solute,H2(CH2)n
+M2=(Kf*1000*w2)/(w1*delTf);//molecular mass of solute in g
+n=(M2-2)/14;
+mprintf('The molecular mass of the hydrocarbon = %f',M2);
+mprintf('\n The molecular formula of solute is H2(CH2)%i',n);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.19/Example_7_19.sce b/1019/CH7/EX7.19/Example_7_19.sce new file mode 100644 index 000000000..e9610fc02 --- /dev/null +++ b/1019/CH7/EX7.19/Example_7_19.sce @@ -0,0 +1,18 @@ +//Example 7.19
+clear;
+clc;
+
+//Given
+R=82;//gas constant in atm ml K^-1 mol^-1
+w2=2;//mass of solute in g
+M2=69000;//molecular mass of solvent in g mol^-1
+T=300.15;//temperature in K
+V=100;//volume of solution in ml
+
+//To determine the osmotic pressure in cms of (i)water (ii)mercury
+pi=(T*R*w2)/(M2*V);//the osmotic pressure in atm
+h1=(pi*1013250)/(980.67);//the osmotic pressure in cms of (i)water
+h2=pi*76;//the osmotic pressure in cms of (ii)mercury
+mprintf('the osmotic pressure in cms of (i)water = %f cm',h1);
+mprintf('\n the osmotic pressure in cms of (ii)mercury = %f cm',h2);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.2/Example_7_2.sce b/1019/CH7/EX7.2/Example_7_2.sce new file mode 100644 index 000000000..911610274 --- /dev/null +++ b/1019/CH7/EX7.2/Example_7_2.sce @@ -0,0 +1,15 @@ +//Example 7.2
+clear;
+clc;
+
+//Given
+R=0.08205;//gas constant in dm^3 atm K^-1 mol^-1
+b=0.0391;//Van der Waals constant in dm^3 mol^-1
+T=1273;//Temperature in K
+P=1000;//pressure in atm
+
+//To calculate the fugacity coefficient
+k=(b*P)/(R*T);//k=log(f/P)
+f=P*exp(k);//fugacity coefficient in atm
+mprintf('Fugacity coefficient = %f atm',f);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.20/Example_7_20.sce b/1019/CH7/EX7.20/Example_7_20.sce new file mode 100644 index 000000000..95e12c607 --- /dev/null +++ b/1019/CH7/EX7.20/Example_7_20.sce @@ -0,0 +1,16 @@ +//Example 7.20
+clear;
+clc;
+
+//Given
+R=82;//gas constant in atm cm^3 K^-1 mol^-1
+w2=1.35;//mass of solute in g
+h1=9.9;//the osmotic pressure in cm of water
+T=300;//temperature in K
+V=100;//volume of solution in ml
+
+//To determine the molecular mass of the polymer
+pi=(980.67*h1)/(1013250);//the osmotic pressure in atm
+M2=(w2*R*T)/(pi*V);//molecular mass of the polymer in g
+mprintf('The molecular mass of the polymer = %f g mol^-1',M2);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.21/Example_7_21.sce b/1019/CH7/EX7.21/Example_7_21.sce new file mode 100644 index 000000000..a058425bf --- /dev/null +++ b/1019/CH7/EX7.21/Example_7_21.sce @@ -0,0 +1,16 @@ +//Example 7.21
+clear;
+clc;
+
+//Given
+R=82;//gas constant in atm cm^3 K^-1 mol^-1
+w2=0.45;//mass of solute in g
+M2=180;//molecular mass of the solute in g mol^-1
+T=300;//temperature in K
+
+//To determine the height attained by water inside the tube and the osmotic pressure
+h=sqrt((w2*R*T*1013250)/(M2*980.67));//height attained by water inside the tube
+pi=(980.67*h)/(1013250);//the osmotic pressure in atm
+mprintf('The height attained by water inside the tube = %f cm',h);
+mprintf('\n The osmotic pressure = %f atm',pi);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.22/Example_7_22.sce b/1019/CH7/EX7.22/Example_7_22.sce new file mode 100644 index 000000000..9a26d352a --- /dev/null +++ b/1019/CH7/EX7.22/Example_7_22.sce @@ -0,0 +1,30 @@ +//Example 7.22
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+w2=17.1;//mass of sucrose in g
+w3=9;//mass of urea in g
+w4=6;//mass of urea in g
+M2=342;//molecular mass of sucrose in g mol^-1
+M3=180;//molecular mass of glucose in g mol^-1
+M4=60;//molecular mass of urea in g mol^-1
+T=300;//temperature in K
+V=3;//volume in dm^3
+
+//To determine the osmotic pressure and the weight average and number average molar mass
+n2=w2/M2;//moles of sucrose
+n3=w3/M3;//moles of glucose
+n4=w4/M4;//moles of urea
+x2=n2/(n2+n3+n4);//mole fraction of sucrose
+x2=n3/(n2+n3+n4);//mole fraction of glucose
+x2=n4/(n2+n3+n4);//mole fraction of urea
+Mw=((w2*M2)+(w3*M3)+(w4*M4))/(w2+w3+w4);//mass average molar mass in g mol^-1
+n1=n2+n3+n4;//moles of all solutes
+pi=(n1*R*T)/V;//the osmotic pressure in atm
+Mn=((w2+w3+w4)*R*T)/(pi*V);//number average molar mass in g mol^-1
+mprintf('The mass average molar mass = %f gm mol^-1',Mw);
+mprintf('\n The osmotic pressure = %f atm',pi);
+mprintf('\n The number average molar mass = %f gm mol^-1',Mn);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.23/Example_7_23.sce b/1019/CH7/EX7.23/Example_7_23.sce new file mode 100644 index 000000000..ada5640a0 --- /dev/null +++ b/1019/CH7/EX7.23/Example_7_23.sce @@ -0,0 +1,26 @@ +//Example 7.23
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+w2=2;//mass of solute in g
+M2=12400;//molecular mass of the solute in g mol^-1
+T=298;//temperature in K
+Kf=1.86;//freezing point depression constant for water
+w1=100;//weight of solvent in g
+Kb=0.52;//boiling point elevation constant for water
+p=24;//vapour pressure of water in mm Hg
+
+//To determine the height attained by water inside the tube and the osmotic pressure
+m=(w2/M2)*(1000/w1);//molality in mol kg^-1
+delTf=Kf*m;//depression in freezing point in oC
+delTb=Kb*m;//elevation in boiling point in oC
+pi=m*R*T*760;//osmotic pressure in mm Hg
+delp=(0.0016*18*p)/1000;//lowering of vapour pressure in mm Hg
+mprintf('The depression in freezing point = %f oC',delTf);
+mprintf('\n The elevation in boiling point = %f oC',delTb);
+mprintf('\n The osmotic pressure = %f mm Hg',pi);
+mprintf('\n The lowering of vapour pressure = %f mm Hg',delp);
+
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.24/Example_7_24.sce b/1019/CH7/EX7.24/Example_7_24.sce new file mode 100644 index 000000000..4ad348270 --- /dev/null +++ b/1019/CH7/EX7.24/Example_7_24.sce @@ -0,0 +1,14 @@ +//Example 7.24
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+T=310;//temperature in K
+delTf=0.402;//freezing temperature depression in K
+Kf=1.86;//freezing point depression constant of waater
+
+//To determine the osmotic pressure
+pi=(R*T*delTf)/(Kf);//the osmotic pressure in atm
+mprintf('The osmotic pressure = %f atm',pi);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.25/Example_7_25.sce b/1019/CH7/EX7.25/Example_7_25.sce new file mode 100644 index 000000000..fa9da87f8 --- /dev/null +++ b/1019/CH7/EX7.25/Example_7_25.sce @@ -0,0 +1,14 @@ +//Example 7.25
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+delTf=0.3;//freezing temperature depression in K
+Kf=1.86;//freezing point depression constant of waater
+m=0.1;//molality of acid solution in mol kg^-1
+
+//To determine the degree of dissociation
+a=(delTf/(Kf*m))-1;//degree of dissociation
+mprintf('The degree of dissociation = %f',a);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.26/Example_7_26.sce b/1019/CH7/EX7.26/Example_7_26.sce new file mode 100644 index 000000000..6a9cc7f29 --- /dev/null +++ b/1019/CH7/EX7.26/Example_7_26.sce @@ -0,0 +1,18 @@ +//Example 7.26
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+delTf=5.50-0.38;//freezing temperature depression for napthalene in K
+delTfo=5.50-1.66;//freezing temperature depression for benzoic acid in K
+m=1;//molality of acid solution in mol kg^-1
+
+//To determine the degree of dimerization ad the equillibrium constant
+Kf=delTf/m;//freezing point depression constant of benzene
+delTfc=Kf*m;//freezing temperature depression for benzoic acid (ideal) in K
+a=(1-(delTfo/delTfc))*2;//degree of dimerization
+K=((1-a)/(1-(a/2)))/((a/2)/(1-(a/2)))^2;//equillibrium constant
+mprintf('The degree of dimerization = %f',a);
+mprintf('\n The equillibrium constant = %i',K);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.27/Example_7_27.sce b/1019/CH7/EX7.27/Example_7_27.sce new file mode 100644 index 000000000..bd6eac733 --- /dev/null +++ b/1019/CH7/EX7.27/Example_7_27.sce @@ -0,0 +1,19 @@ +//Example 7.27
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+m=0.1;//molality of acid solution in mol kg^-1
+T=298;//temperature in K
+w1=1000;//mass of water in g
+
+//To determine the partial molar volume and the density
+V2=16.62+(1.5*1.77*sqrt(m))+(2*0.12*m);//partial molar volume in cm^3 mol^-1
+V=1003+(16.62*m)+(1.77*m^(3/2))+(0.12*m^2);//total volume in cm^3
+V1=(V-(m*V2))/55.55;//partial molar volume of water in cm^3 mol^-1
+p1=(w1+5.85)/V;//density of te solution in g cm^-3
+mprintf('The partial molar volume of water = %f cm^3 mol^-1',V1);
+mprintf('\n The partial molar volume of sodium chloride = %f cm^3 mol^-1',V2);
+mprintf('\n The density = %f g cm^-3',p1);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.28/Example_7_28.sce b/1019/CH7/EX7.28/Example_7_28.sce new file mode 100644 index 000000000..90e047d4e --- /dev/null +++ b/1019/CH7/EX7.28/Example_7_28.sce @@ -0,0 +1,20 @@ +//Example 7.28
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+T=298;//temperature in K
+w=1;//mass of solution in kg
+x1=0.531;//mole fraction of acetone
+x2=0.469;//mole fraction of chloroform
+M1=58;//molar mass of acetone in g mol^-1
+M2=119.5;//molar mass of chloroform in g mol^-1
+V1=74.166;//partial molar volume of acetone in cm^3 mol^-1
+V2=80.235;//partial molar volume of chloroform in cm^3 mol^-1
+//To determine the volume of the solution
+V=((w*1000)*((x1*V1)+(x2*V2)))/((x1*M1)+(x2*M2));//the volume of the solution in cm^3
+n=(w*1000)/((x1*M1)+(x2*M2));//total number of moles
+Vm=V/n;//mean molar volume of the solution in cm^3 mol^-1
+mprintf('The mean molar volume of the solution = %f cm^3 mol^-1',Vm);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.30/Example_7_30.sce b/1019/CH7/EX7.30/Example_7_30.sce new file mode 100644 index 000000000..32a747023 --- /dev/null +++ b/1019/CH7/EX7.30/Example_7_30.sce @@ -0,0 +1,18 @@ +//Example 7.30
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+T=298;//temperature in K
+x1=0.5;//mole fraction of chloroform
+x2=0.5;//mole fraction of p-xylene
+
+//To determine the volume of the solution
+V=x1*x2*(0.585+(0.085*(x1-x2))-(0.165*((x1-x2)^2)));;//the volume of the solution in cm^3 mol^-1
+delV2=0.585+0.085-0.165;//in cm^3 mol^-1
+delV1=0.585-0.085-0.165;//in cm^3 mol^-1
+mprintf('The mean molar volume of the solution = %f cm^3 mol^-1',V);
+mprintf('\n delV1 = %f cm^3 mol^-1',delV1);
+mprintf('\n delV2 = %f cm^3 mol^-1',delV2);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.32/Example_7_32.sce b/1019/CH7/EX7.32/Example_7_32.sce new file mode 100644 index 000000000..ffc8c2389 --- /dev/null +++ b/1019/CH7/EX7.32/Example_7_32.sce @@ -0,0 +1,17 @@ +//Example 7.30
+clear;
+clc;
+
+//Given
+R=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+T=363;//temperature in K
+P=734;//pressure in mm Hg
+ww=27;//mass percent of water
+wA=73;//mass percent of A
+Pw=526;//vapour pressure of water in mm Hg
+
+//To determine the volume of the solution
+PA=P-Pw;//partial pressure of A in mm Hg
+MA=(Pw*18*wA)/(ww*PA);//molar mass of A in g mol^-1
+mprintf('The molar mass of A = %i g mol^-1',MA);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.4/Example_7_4.sce b/1019/CH7/EX7.4/Example_7_4.sce new file mode 100644 index 000000000..e7bec8fad --- /dev/null +++ b/1019/CH7/EX7.4/Example_7_4.sce @@ -0,0 +1,33 @@ +//Example 7.1
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+T=298;//temperature in K
+
+//To calculate delGmix,delHmix and delSmix
+//(i) 10 moles of H + 10 moles of Ne
+n1=10;//moles of H
+n2=10;//moles of Ne
+x1=n1/(n1+n2);//mole fraction of H
+x2=n2/(n1+n2);//mole fraction of Ne
+delGmix1=R*T*((n1*log(x1))+(n2*log(x2)));//free energy change in J
+delSmix1=-delGmix1/T;//entropy change in J K^-1
+delHmix1=0;//since all gases are ideal
+mprintf('(i) delGmix = %f J \n delHmix = %f J \n delSmix = %f J K^-1',delGmix1,delHmix1,delSmix1);
+//(ii) 10 moles of H + 20 moles of Ne
+n21=10;//moles of H
+n22=20;//moles of Ne
+x21=n21/(n21+n22);//mole fraction of H
+x22=n22/(n21+n22);//mole fraction of Ne
+delGmix2=R*T*((n21*log(x21))+(n22*log(x22)));//free energy change in J
+delSmix2=-delGmix2/T;//entropy change in J K^-1
+delHmix2=0;//since all gases are ideal
+mprintf('\n (ii) delGmix = %f J \n delHmix = %f J \n delSmix = %f J K^-1',delGmix2,delHmix2,delSmix2);
+//(iii) 10 moles of Ne + 20 moles of equimolar mixture of Ne and He
+delGmix3=delGmix2-delGmix1//free energy change in J
+delSmix3=-delGmix3/T;//entropy change in J K^-1
+delHmix3=0;//since all gases are ideal
+mprintf('\n (iii) delGmix = %f J \n delHmix = %f J \n delSmix = %f J K^-1',delGmix3,delHmix3,delSmix3);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.6/Example_7_6.sce b/1019/CH7/EX7.6/Example_7_6.sce new file mode 100644 index 000000000..f95405386 --- /dev/null +++ b/1019/CH7/EX7.6/Example_7_6.sce @@ -0,0 +1,24 @@ +//Example 7.6
+clear;
+clc;
+
+//Given
+R=0.08205;//gas constant in atm dm^3 K^-1 mol^-1
+R1=8.314;//gas constant in J K^-1 mol^-1
+T=300;//temperature in K
+VCH4=4;//initial volume of methane in dm^3
+VAr=1;//initial volume of argon in dm^3
+Vf=3;//final volume
+P=1;//Initial Pressure in atm
+
+//To calculate delGmix,delHmix and delSmix
+nCH4=(P*VCH4)/(R*T);//moles of methane taken
+nAr=(P*VAr)/(R*T);//moles of Argon taken
+xCH4=nCH4/(nCH4+nAr);//mole fraction of methane in the mixture
+xAr=nAr/(nCH4+nAr);//mole fraction of Argon in the mixture
+pf=(R*T*(nCH4+nAr))/Vf;//final pressure in atm
+delGmix=R1*T*((nCH4*log(xCH4*pf))+(nAr*log(xAr*pf)));//free energy change in J
+delHmix=0;//since the gases are ideal
+delSmix=-delGmix/T;//entropy change in J K^-1
+mprintf('delGmix = %f J \n delHmix = %f J \n delSmix = %f J K^-1',delGmix,delHmix,delSmix);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.7/Example_7_7.sce b/1019/CH7/EX7.7/Example_7_7.sce new file mode 100644 index 000000000..0e6bc6000 --- /dev/null +++ b/1019/CH7/EX7.7/Example_7_7.sce @@ -0,0 +1,26 @@ +//Example 7.7
+clear;
+clc;
+
+//Given
+R=0.08205;//gas constant in atm dm^3 K^-1 mol^-1
+R1=8.314;//gas constant in J K^-1 mol^-1
+T=293;//temperature in K
+w1=100;//weight of ethanol taken in g
+w2=100;//weight of methanol taken in g
+p1=44.5;//vapour preaaure of pure ethanol in mm Hg
+p2=88.7;//vapour preaaure of pure methanol in mm Hg
+
+//To determine the vapour pressure of solution,partial vapour pressures and vapour phase composition
+n1=100/46;//moles of ethanol
+n2=100/32;//moles of methanol
+x1=n1/(n1+n2);//mole fraction of ethanol
+x2=n2/(n1+n2);//mole fraction of methanol
+P1=p1*x1;//partial pressure of ethanol in mm Hg
+P2=p2*x2;//partial pressure of methanol in mm Hg
+P=P1+P2;//vapour pressure of the solution in mm Hg
+y1=P1/P;//mole fraction of ethanol in the vapour phase
+y2=1-y1;//mole fraction of methanol in the vapour phase
+mprintf('(i) Vapour pressure of the solution = %f',P);
+mprintf('\n (ii) Partial vapour pressure of ethanol is %f mm Hg and that of methanol is %f mm Hg',P1,P2);
+mprintf('\n (iii) mole fraction of ethanol in vapour phase is %f and that of methanol is %f',y1,y2);
diff --git a/1019/CH7/EX7.8/Example_7_8.sce b/1019/CH7/EX7.8/Example_7_8.sce new file mode 100644 index 000000000..4e6193622 --- /dev/null +++ b/1019/CH7/EX7.8/Example_7_8.sce @@ -0,0 +1,18 @@ +//Example 7.8
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+T=298;//temperature in K
+n1=1;//moles of toluene
+n2=2;//moles of benzene
+
+//to determine the free energy of mixing
+x1=n1/(n1+n2);//mole fraction of toluene
+x2=n2/(n1+n2);//mole fraction of benzene
+delGmix=R*T*((n1*log(x1))+(n2*log(x2)));//free energy of mixing in J
+delSmix=-delGmix/T;//entropy change in J K^-1
+mprintf('The free energy of mixing = %f J',delGmix);
+mprintf('\n The entropy of mixing = %f J K^-1',delSmix);
+//end
\ No newline at end of file diff --git a/1019/CH7/EX7.9/Example_7_9.sce b/1019/CH7/EX7.9/Example_7_9.sce new file mode 100644 index 000000000..180e09f35 --- /dev/null +++ b/1019/CH7/EX7.9/Example_7_9.sce @@ -0,0 +1,19 @@ +//Example 7.9
+clear;
+clc;
+
+//Given
+R=8.314;//gas constant in J K^-1 mol^-1
+T=298;//temperature in K
+P=1;//pressure in atm
+kO2=4.34*(10^4);//Henrys constant for O2 in atm
+kN2=8.57*(10^4);//Henrys constant for N2 in atm
+
+//To determine the molality of O2 and N2 dissolved in water
+xO2=P/kO2;//mole fraction of O2
+xN2=P/kN2;//mole fraction of N2
+mO2=55.5*xO2;//molality of O2 in mol kg^-1
+mN2=55.5*xN2;//molality of N2 in mol kg^-1
+mprintf('Molality of O2 dissolved in water = %f mol kg^-1',mO2);
+mprintf('\n Molality of N2 dissolved in water = %f mol kg^-1',mN2);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.1/Example_8_1.sce b/1019/CH8/EX8.1/Example_8_1.sce new file mode 100644 index 000000000..a1e003f90 --- /dev/null +++ b/1019/CH8/EX8.1/Example_8_1.sce @@ -0,0 +1,18 @@ +//Example 8.1
+clear;
+clc;
+
+//Given
+Kp=0.10;//equillibrium constant at 300K
+Pa=20;// Partial pressure of A in atm
+Pm=1.0;///partial pressure of M in atm
+T=300;//Temperature in K
+R=8.314;// gas constant in J K^-1 mol^-1
+//To determine the free energy
+Qp=Pm/Pa;//reaction quotient
+delG=R*T*log(Qp/Kp);//free energy change
+mprintf('(a) delG = %f J mol^-1',delG);
+delG0=-R*T*log(Kp);//standard free energy in J mol^-1
+mprintf('\n (b) standard free energy = %f J mol^-1',delG0);
+mprintf('\n (c) Since delG is negetive,the reaction proceeds spontaneously in forward direction')
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.10/Example_8_10.sce b/1019/CH8/EX8.10/Example_8_10.sce new file mode 100644 index 000000000..721a4940b --- /dev/null +++ b/1019/CH8/EX8.10/Example_8_10.sce @@ -0,0 +1,13 @@ +//Example 8.10
+clear;
+clc;
+
+//Given
+T1=298;//initial temperature in K
+T2=308;//final temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine the value of delHo
+delHo=((R*T1*T2*log(2))/(T2-T1))*0.001;//delHo in kJ mol^-1
+mprintf('Enthalpy of reaction,delHo = %f kJ mol^-1',delHo);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.12/Example_8_12.sce b/1019/CH8/EX8.12/Example_8_12.sce new file mode 100644 index 000000000..20219bac1 --- /dev/null +++ b/1019/CH8/EX8.12/Example_8_12.sce @@ -0,0 +1,20 @@ +//Example 8.12
+clear;
+clc;
+
+//Given
+T1=1225;//initial temperature in K
+T2=1200;//final temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delHo=216.7;//standard enthalpy of the reaction in kJ
+K1=0.00328;//equillibrium constant at temperature T1
+
+//To determine equillibrium constant,delSo and delGo at temperature T2
+k=(log(K1)-((1000*delHo/R)*((1/T2)-(1/T1))));//k=log(K2)
+K2=exp(k);//equillibrium constant at T2
+delGo=R*T2*k/1000;//delGo in kJ mol^-1
+delSo=1000*((delHo+delGo)/T2);//delSo in J mol^-1 K^-1
+mprintf('equillibrium constant at 1200 K = %f',K2);
+mprintf('\n delGo at 1200 K = %f kJ mol^-1',delGo);
+mprintf('\n delSo at 1200 K = %f J K^-1 mol^-1',delSo);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.13/Example_8_13.sce b/1019/CH8/EX8.13/Example_8_13.sce new file mode 100644 index 000000000..3c2781282 --- /dev/null +++ b/1019/CH8/EX8.13/Example_8_13.sce @@ -0,0 +1,16 @@ +//Example 8.13
+clear;
+clc;
+
+//Given
+T=1225;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delHo=216.7;//standard enthalpy of the reaction in kJ
+K=0.00328;//equillibrium constant at temperature T1
+
+//To determine delSo and delGo at temperature T
+delGo=R*T*log(K)/1000;//delGo in kJ mol^-1
+delSo=1000*((delHo+delGo)/T);//delSo in J mol^-1 K^-1
+mprintf('delGo at 1225 K = %f kJ mol^-1',delGo);
+mprintf('\n delSo at 1225 K = %f J K^-1 mol^-1',delSo);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.15/Example_8_15.sce b/1019/CH8/EX8.15/Example_8_15.sce new file mode 100644 index 000000000..41336e333 --- /dev/null +++ b/1019/CH8/EX8.15/Example_8_15.sce @@ -0,0 +1,17 @@ +//Example 8.15
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delGfoH2Ol=-237.2;//standard enthalpy of formation of water in kJ mol^-1
+pH2O=23.7;//vapour pressure of water in mm Hg
+P=760;//standard pressure in mm Hg
+
+//To determine delGfoH2Og
+Kp=pH2O/P;//equillibrium constant for given reaction
+delGo=(-1)*R*T*log(Kp)/1000;//delGo in kJ mol^-1
+delGfoH2Og=delGo+delGfoH2Ol;//free energy of formation of water vapour in kJ mol^-1
+mprintf('Free energy of formation of water vapour,delGfoH2Og = %f kJ mol^-1',delGfoH2Og);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.16/Example_8_16.sce b/1019/CH8/EX8.16/Example_8_16.sce new file mode 100644 index 000000000..075293b98 --- /dev/null +++ b/1019/CH8/EX8.16/Example_8_16.sce @@ -0,0 +1,19 @@ +//Example 8.16
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delGfoCuO=-127.2;//standard enthalpy of formation of CuO in kJ mol^-1
+pH2O=23.7;//vapour pressure of water in mm Hg
+P=760;//standard pressure in mm Hg
+
+//To determine delGfoH2Og
+Kp=pH2O/P;//equillibrium constant for given reaction
+delGo=(-2)*delGfoCuO;//delGo in kJ mol^-1
+k=(-1000*delGo)/(R*T);//k=log(Kp)
+Kp=exp(k);//equillibrium constant Kp
+pO2=Kp*1;//partial pressure of O2 in atm
+mprintf('Partial pressure of O2 over CuO and Cu at 298 K = %f atm',pO2);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.18/Example_8_18.sce b/1019/CH8/EX8.18/Example_8_18.sce new file mode 100644 index 000000000..bafe51699 --- /dev/null +++ b/1019/CH8/EX8.18/Example_8_18.sce @@ -0,0 +1,14 @@ +//Example 8.18
+clear;
+clc;
+
+//Given
+delHo=241.82;//Enthalpy of reaction in kJ mol^-1
+delSo=44.4;//Entropy of the reaction in J K^-1 mol^-1
+K=1;//equillibrium constant for the reaction
+
+//To determine the temperature
+delGo=0;//since delGo=RTlog(k) and log(1)=0
+T=(delHo*1000)/delSo;//temperature in K
+mprintf('Temperature at which K=1 is %f K',T);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.19/Example_8_19.sce b/1019/CH8/EX8.19/Example_8_19.sce new file mode 100644 index 000000000..6bb038016 --- /dev/null +++ b/1019/CH8/EX8.19/Example_8_19.sce @@ -0,0 +1,17 @@ +//Example 8.19
+clear;
+clc;
+
+//Given
+delGo2=-36.7;//standard free energy change in conversion of fumarate to asparate in kJ mol^-1
+delGo3=-2.9;//standard free energy change in conversion of fumarate to malate in kJ mol^-1
+T=310;//Temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine the standard free energy change in conversion of malate to asparate and the equillibrium constant
+delGo1=delGo2-delGo3;//the standard free energy change in conversion of malate to asparate in kJ mol^-1
+k=(-1000*delGo1)/(R*T);//k=log(K)
+K=exp(k);//K is the equillibrium constant
+mprintf('Standard free energy change in conversion of malate to asparate = %f kJ mol^-1',delGo1);
+mprintf('\n The equillibrium constant at 310 K = %f',K);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.2/Example_8_2.sce b/1019/CH8/EX8.2/Example_8_2.sce new file mode 100644 index 000000000..9bd8c3d3f --- /dev/null +++ b/1019/CH8/EX8.2/Example_8_2.sce @@ -0,0 +1,27 @@ +//Example 8.2
+clear;
+clc;
+
+//Given
+p=0.35;
+Kp=p*10^-24;//equillibrium constant at 300K
+P0=1.0;//standard pressure in atm
+T=300;//Temperature in K
+R=0.082;// gas constant in atm dm^3 mol^-1K^-1
+C0=1;//in mol/dm^3
+Kp2=0.157;//Kp for reaction in (b)
+P=1;//pressure in atm
+
+// (a) To determine Kc1
+delv=(2+1)-2;
+c=p*((P0/(C0*R*T)))^delv
+Kc1=c*10^-24;//equillibrium constant
+mprintf('(a) Kc = %f *(10^-24)',c);
+
+//(b) To determine Kc2
+delv2=2-1;
+Kc2=Kp2*((P0/(C0*R*T)))^delv2;
+mprintf('\n (b) Kc = %f',Kc2);
+Kx=Kp2*(P0/P);//equillibrium constant
+mprintf('\n Kx = %f',Kx)
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.20/Example_8_20.sce b/1019/CH8/EX8.20/Example_8_20.sce new file mode 100644 index 000000000..60766ffa3 --- /dev/null +++ b/1019/CH8/EX8.20/Example_8_20.sce @@ -0,0 +1,24 @@ +//Example 8.20
+clear;
+clc;
+
+//Given
+T1=313;//1st temperature in K
+T2=333;//2nd temperature in K
+KT1=0.86;//Value of equillibrium constant at temperature T1
+KT2=0.35;//Value of equillibrium constant at temperature T2
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine the value of delGo,delHo and delSo
+delHo=(R*T1*T2*log(KT2/KT1))/(T2-T1);//delHo in J mol^-1
+delGo313=-R*T1*log(KT1);//value of delGo at T1 in J mol^-1
+delGo333=-R*T2*log(KT2);//value of delGo at T2 in J mol^-1
+delSo313=(delHo-delGo313)/T1;//value of delSo at T1 in J K^-1 mol^-1
+delSo333=(delHo-delGo333)/T2;//value of delSo at T2 in J K^-1 mol^-1
+mprintf('delHo = %f J mol^-1',delHo);
+mprintf('\n delGo at 313 K = %f J mol^-1',delGo313);
+mprintf('\n delGo at 333 K = %f J mol^-1',delGo333);
+mprintf('\n delSo at 313 K = %f J K^-1 mol^-1',delSo313);
+mprintf('\n delSo at 333 K = %f J K^-1 mol^-1',delSo333);
+
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.22/Example_8_22.sce b/1019/CH8/EX8.22/Example_8_22.sce new file mode 100644 index 000000000..9f9c28b29 --- /dev/null +++ b/1019/CH8/EX8.22/Example_8_22.sce @@ -0,0 +1,17 @@ +//Example 8.22
+clear;
+clc;
+
+//Given
+T=298;//Temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+k=4.814-(2059/T);//k=log(K),where K is the equillibrium constant
+
+//To determine the values of delGo,delHo and delSo
+delSo=4.814*R;//entropy change in J K^-1 mol^-1
+delGo=-R*T*k;//free energy change in J mol^-1
+delHo=delGo+(T*delSo);//enthalpy change in J mol^-1
+mprintf('delHo = %f J mol^-1',delHo);
+mprintf('\n delGo = %f J mol^-1',delGo);
+mprintf('\n delSo = %f J K^-1 mol^-1',delSo);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.23/Example_8_23.sce b/1019/CH8/EX8.23/Example_8_23.sce new file mode 100644 index 000000000..5778b4614 --- /dev/null +++ b/1019/CH8/EX8.23/Example_8_23.sce @@ -0,0 +1,21 @@ +//Example 8.23
+clear;
+clc;
+
+//Given
+T=298;//Temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+R1=0.082;//gas constant in atm dm^3 K^-1 mol^-1
+P=1;//pressure in atm
+a=0.167;//degree of dissociation
+
+//To determine Kp,Kc,delGoP and delGoC
+Kp=(4*(a^2)*P)/(1-(a^2));//Equillibrium constant in terms of pressure
+Kc=Kp*((R1*T)^(-1));//Equillibrium constant in terms of concentration
+delGoP=-0.001*R*T*log(Kp);//standard free energy in kJ
+delGoC=-0.001*R*T*log(Kc);//standard free energy in kJ
+mprintf('Kc = %f',Kc);
+mprintf('\n Kp = %f',Kp);
+mprintf('\n delGoP = %f kJ',delGoP);
+mprintf('\n delGoC = %f kJ',delGoC);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.27/Example_8_27.sce b/1019/CH8/EX8.27/Example_8_27.sce new file mode 100644 index 000000000..a042973be --- /dev/null +++ b/1019/CH8/EX8.27/Example_8_27.sce @@ -0,0 +1,18 @@ +//Example 8.27
+clear;
+clc;
+
+//Given
+T=1300;//Temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+p1=1.067*(10^5);//ratio of pressure of CO/CO2 for 1st reaction
+p2=1.835*(10^5);//ratio of pressure of CO/CO2 for 2nd reaction
+
+//To determine the values of delGo for required reaction
+Kp1=p1^2;//equillibrium constant for the 1st reaction
+Kp2=p2^2;//equillibrium constant for 2nd reaction
+delGo1=-0.001*R*T*log(Kp1);//free energy change in kJ
+delGo2=-0.001*R*T*log(Kp2);//free energy change in kJ
+delGoA=delGo2-delGo1;//delGo for required reaction in kJ
+mprintf('delGo for the formation of cobaltous silicate = %f kJ',delGoA);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.28/Example_8_28.sce b/1019/CH8/EX8.28/Example_8_28.sce new file mode 100644 index 000000000..cd928ebb3 --- /dev/null +++ b/1019/CH8/EX8.28/Example_8_28.sce @@ -0,0 +1,15 @@ +//Example 8.28
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delGfoO3=163.43;//free energy of formation of O3 in kJ
+delGfoO2=0;//free energy of formation of O2 in kJ
+
+//To determine the value of equillibrium constant
+delGo=(2*delGfoO3)-(3*delGfoO2);//delGo in kJ
+k=(-1000*delGo)/(2.303*R*T);//k=log10(K)
+mprintf('Equillibrium constant,K = 10^%f',k);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.3/Example_8_3.sce b/1019/CH8/EX8.3/Example_8_3.sce new file mode 100644 index 000000000..347396e7f --- /dev/null +++ b/1019/CH8/EX8.3/Example_8_3.sce @@ -0,0 +1,21 @@ +//Example 8.3
+clear;
+clc;
+
+//Given
+p=1.7;
+Kp=p*10^12;//equillibrium constant at 300K
+
+// (i) To determine Kp1
+p1=1/p;
+Kp1=1/Kp;//equillibrium constant
+mprintf('(i) Kp = %f * 10^-12',p1);
+
+//(ii) To determine Kc2
+p2=p1^2;
+Kp2=Kp1^2;//equillibrium constant
+mprintf('\n (ii) Kp = %f * 10^-24',p2);
+p3=1/p2;
+Kp3=1/Kp2;//equillibrium constant
+mprintf('\n (iii) Kp = %f * 10^24',p3)
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.32/Example_8_32.sce b/1019/CH8/EX8.32/Example_8_32.sce new file mode 100644 index 000000000..f75a04df2 --- /dev/null +++ b/1019/CH8/EX8.32/Example_8_32.sce @@ -0,0 +1,14 @@ +//Example 8.32
+clear;
+clc;
+
+//Given
+n1=1;//moles of acetic acid and ethanol initially mixed
+n2=0.667;//moles of easter and water produced
+
+//To determine the equillibrium constant
+n3=1-0.667;//moles of acid and ethanol remaining
+N=2;//total number of moles of reactants taken
+Ka=((n2/N)*(n2/N))/((n3/N)*(n3/N));
+mprintf('Equillibrium constant for the reaction = %f',Ka);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.4/Example_8_4.sce b/1019/CH8/EX8.4/Example_8_4.sce new file mode 100644 index 000000000..59a7c2273 --- /dev/null +++ b/1019/CH8/EX8.4/Example_8_4.sce @@ -0,0 +1,18 @@ +//Example 8.4
+clear;
+clc;
+
+//Given
+v=2;
+p=1;//pressure in atm
+V=1;//volume in L
+R=0.082;// gas constant in L atm K^-1 mol^-1
+T=298.15;// temperature in K
+w=3.176;// weight of N2O4 taken in g
+
+// To determine degree of dissociation a
+m1=(2*14)+(4*16);//molecular mass of N2O4 in g mol^-1
+m2=(w*R*T)/(p*V);//in g mol^-1
+a=(m1-m2)/m2((v-1));//degree of dissociation
+mprintf('Degree of dissociation = %f ',a);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.5/Example_8_5.sce b/1019/CH8/EX8.5/Example_8_5.sce new file mode 100644 index 000000000..93d3f4adc --- /dev/null +++ b/1019/CH8/EX8.5/Example_8_5.sce @@ -0,0 +1,23 @@ +//Example 8.5
+clear;
+clc;
+
+//Given
+delHfoC=0;//enthalpy of formation of graphite in kJ mol^-1
+delHfoH2=0;//enthalpy of formation of Hydrogen molecule in kJ mol^-1
+delHfoCH4=-74.83;//enthalpy of formation of methane in kJ mol^-1
+delSoC=5.68//standard entropy of graphite in J K^-1 mol^-1
+delSoH2=130.59//standard entropy of Hydrogen in J K^-1 mol^-1
+delSoCH4=186.19//standard entropy of methane in J K^-1 mol^-1
+T=298;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+
+//To determine the change in free energy delGo and Kp,the equillibrium constant
+delHo=1000*(delHfoCH4-(delHfoC+(2*delHfoH2)));//Net change in enthalpy in J mol^-1
+delSo=delSoCH4-(delSoC+(2*delSoH2));//Net change in entropy in J K^-1 mol^-1
+delGo=delHo-(T*delSo);//delGo in J mol^-1
+k=-1*delGo/(R*T);//k=log(Kp)
+Kp=exp(k);//equillibrium constant Kp
+mprintf('Change in free energy,delGo=%f J mol^-1',delGo);
+mprintf('\n Equillibrium constant,Kp=%f',Kp);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.6/Example_8_6.sce b/1019/CH8/EX8.6/Example_8_6.sce new file mode 100644 index 000000000..5afde5050 --- /dev/null +++ b/1019/CH8/EX8.6/Example_8_6.sce @@ -0,0 +1,19 @@ +//Example 8.6
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+p=101325;//pressure in N m^-2
+MoNH3=-16.6;//standard chemical potential of amonia at 298 K in kJ mol^-1
+MoN2=0;//standard chemical potential of nitrogen at 298 K in kJ mol^-1
+MoH2=0;//standard chemical potential of hydrogen at 298 K in kJ mol^-1
+
+//To determine the value of equillibrium constant Kp
+delGo=MoN2+(3*MoH2)-(2*MoNH3);//delGo in kJ
+k=(-1000*delGo)/(R*T);//k=log(Kp)
+Kp=exp(k);//equillibrium constant Kp
+mprintf('Change in free energy,delGo=%f kJ',delGo);
+mprintf('\n Equillibrium constant,Kp=%f',Kp);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.7/Example_8_7.sce b/1019/CH8/EX8.7/Example_8_7.sce new file mode 100644 index 000000000..b1b2d8572 --- /dev/null +++ b/1019/CH8/EX8.7/Example_8_7.sce @@ -0,0 +1,14 @@ +//Example 8.6
+clear;
+clc;
+
+//Given
+T=673;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+p=101325;//pressure in N m^-2
+Kp=1.64*10^(-4);//Equillibrium constant for the synthesis of amonia at 673 K
+
+//To determine the value of delGo
+delGo=(-1)*R*T*log(Kp);//delGo in J mol^-1
+mprintf('Change in free energy,delGo = %f J mol^-1',delGo);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.8/Example_8_8.sce b/1019/CH8/EX8.8/Example_8_8.sce new file mode 100644 index 000000000..2038deeac --- /dev/null +++ b/1019/CH8/EX8.8/Example_8_8.sce @@ -0,0 +1,14 @@ +//Example 8.8
+clear;
+clc;
+
+//Given
+T=298;//temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delGfoC2H4=68.12;//standard free energy change in the formation of ethylene in kJ mol^-1
+delGfoC2H6=-32.89;//standard free energy change in the formation of ethane in kJ mol^-1
+
+//To determine the value of delGo i.e. heat of hydrogenation of ethylene
+delGo=delGfoC2H6-delGfoC2H4;//heat of hydrogenation of ethylene in kJ mol^-1
+mprintf('Heat of hydrogenation of ethylene,delGo = %f kJ mol^-1',delGo);
+//end
\ No newline at end of file diff --git a/1019/CH8/EX8.9/Example_8_9.sce b/1019/CH8/EX8.9/Example_8_9.sce new file mode 100644 index 000000000..a5ef94c8e --- /dev/null +++ b/1019/CH8/EX8.9/Example_8_9.sce @@ -0,0 +1,27 @@ +//Example 8.9
+clear;
+clc;
+
+//Given
+T1=298;//initial temperature in K
+T2=1073;//final temperature in K
+R=8.314;//gas constant in J K^-1 mol^-1
+delGfoH2=0;//standard free energy change in the formation of hydrogen in kJ mol^-1
+delGfoC0=-137.27;//standard free energy change in the formation of CO in kJ mol^-1
+delGfoH2O=-228.59;//standard free energy change in the formation of water in kJ mol^-1
+delGfoC02=-394.38;//standard free energy change in the formation of CO2 in kJ mol^-1
+delHfoH2=0;//standard enthalpy in the formation of hydrogen in kJ mol^-1
+delHfoC0=-110.52;//standard enthalpy in the formation of CO in kJ mol^-1
+delHfoH2O=-241.83;//standard enthalpy in the formation of water in kJ mol^-1
+delHfoC02=-392.51;//standard enthalpy in the formation of CO2 in kJ mol^-1
+
+//To determine the value of Kp at T1 and T2
+delGo=delGfoH2+delGfoC02-(delGfoC0+delGfoH2O);//free energy change in kJ mol^-1
+delHo=delHfoH2+delHfoC02-(delHfoC0+delHfoH2O);//standard enthalpy change in kJ mol^-1
+k1=(-1000*delGo)/(R*T1);//k=log(Kp)
+Kp1=exp(k1);//equillibrium constant Kp at 298 K
+mprintf('Equillibrium constant,Kp at 298 K = %f ',Kp1);
+k2=((-1000*delHo/R)*((1/1073)-(1/298)))+k1;//equillibrium constant at 1073 K
+Kp2=exp(k2);//equillibrium constant Kp at 1073 K
+mprintf('\n Equillibrium constant,Kp at 1073 K = %f ',Kp2);
+//end
\ No newline at end of file |