diff options
Diffstat (limited to '2465/CH4')
-rw-r--r-- | 2465/CH4/EX4.1/Example_1.sce | 17 | ||||
-rw-r--r-- | 2465/CH4/EX4.10/Example_10.sce | 19 | ||||
-rw-r--r-- | 2465/CH4/EX4.11/Example_11.sce | 17 | ||||
-rw-r--r-- | 2465/CH4/EX4.12/Example_12.sce | 13 | ||||
-rw-r--r-- | 2465/CH4/EX4.13/Example_13.sce | 14 | ||||
-rw-r--r-- | 2465/CH4/EX4.15/Example_15.sce | 19 | ||||
-rw-r--r-- | 2465/CH4/EX4.16/Example_16.sce | 27 | ||||
-rw-r--r-- | 2465/CH4/EX4.17/Example_17.sce | 19 | ||||
-rw-r--r-- | 2465/CH4/EX4.18/Example_18.sce | 21 | ||||
-rw-r--r-- | 2465/CH4/EX4.19/Example_19.sce | 19 | ||||
-rw-r--r-- | 2465/CH4/EX4.2/Example_2.sce | 11 | ||||
-rw-r--r-- | 2465/CH4/EX4.3/Example_3.sce | 13 | ||||
-rw-r--r-- | 2465/CH4/EX4.4/Example_4.sce | 21 | ||||
-rw-r--r-- | 2465/CH4/EX4.5/Example_5.sce | 19 | ||||
-rw-r--r-- | 2465/CH4/EX4.6/Example_6.sce | 18 | ||||
-rw-r--r-- | 2465/CH4/EX4.7/Example_7.sce | 16 | ||||
-rw-r--r-- | 2465/CH4/EX4.8/Example_8.sce | 23 | ||||
-rw-r--r-- | 2465/CH4/EX4.9/Example_9.sce | 15 |
18 files changed, 321 insertions, 0 deletions
diff --git a/2465/CH4/EX4.1/Example_1.sce b/2465/CH4/EX4.1/Example_1.sce new file mode 100644 index 000000000..502811c64 --- /dev/null +++ b/2465/CH4/EX4.1/Example_1.sce @@ -0,0 +1,17 @@ +//Chapter-4,Example 1,Page 92
+clc;
+close;
+
+R= 2 // gas constant
+
+//as water temperature is 100 degree
+
+T = 273 + 100 // temperature in Kelvin
+
+w=R*T // work done
+
+q= 536*18 //heat in cal/mol
+
+delta_E= q-w
+
+printf('the amount of energy increased is %.1f cal/mol',delta_E)
diff --git a/2465/CH4/EX4.10/Example_10.sce b/2465/CH4/EX4.10/Example_10.sce new file mode 100644 index 000000000..b22df3756 --- /dev/null +++ b/2465/CH4/EX4.10/Example_10.sce @@ -0,0 +1,19 @@ +//Chapter-4,Example 10,Page 95
+clc;
+close;
+
+delta_H1 = 538 //latent heat of water at 100 degree
+
+T1= 273 + 100 //temperature in Kelvin
+
+T2= 273 +150 //temperature in Kelvin
+
+Cp_w = 1 // for water
+
+Cp_s = 8.1/18 //for steam
+
+delta_Cp = Cp_s - Cp_w
+
+delta_H2 = delta_H1 + delta_Cp*(T2-T1) //latent heat of water at 150 degree
+
+printf('the latent heat of water at 150 degree is %.2f cal/g',delta_H2)
diff --git a/2465/CH4/EX4.11/Example_11.sce b/2465/CH4/EX4.11/Example_11.sce new file mode 100644 index 000000000..da8cc6183 --- /dev/null +++ b/2465/CH4/EX4.11/Example_11.sce @@ -0,0 +1,17 @@ +//Chapter-4,Example 11,Page 96
+clc;
+close;
+
+R= 8.31 //gas constant
+
+T= 273+25 // temperature in Kelvin
+
+P1= 2 //pressure in atm
+
+P2= 1 //pressure in atm
+
+w= 2.303 *R*T*log10(P1/P2) //maximum work
+
+printf('maximum work done is %.f J', w)
+
+//mistake in textbook
diff --git a/2465/CH4/EX4.12/Example_12.sce b/2465/CH4/EX4.12/Example_12.sce new file mode 100644 index 000000000..e245a0a62 --- /dev/null +++ b/2465/CH4/EX4.12/Example_12.sce @@ -0,0 +1,13 @@ +//Chapter-4,Example 12,Page 96
+clc;
+close;
+
+q_rev= 19.14 //latent heat
+
+n= 18 //mols
+
+T= 273 //temperature in Kelvin
+
+dS= q_rev*n/T
+
+printf('the change of molar entropy is %.2f J/mol',dS)
diff --git a/2465/CH4/EX4.13/Example_13.sce b/2465/CH4/EX4.13/Example_13.sce new file mode 100644 index 000000000..f47b7d766 --- /dev/null +++ b/2465/CH4/EX4.13/Example_13.sce @@ -0,0 +1,14 @@ +//Chapter-4,Example 13,Page 96
+clc;
+close;
+
+
+q_rev= 12.19 //latent heat
+
+n= 32 //mols
+
+T= 273-182.9 //temperature in Kelvin
+
+dS= q_rev*n/T
+
+printf('the change of molar entropy is %.2f J/mol',dS)
diff --git a/2465/CH4/EX4.15/Example_15.sce b/2465/CH4/EX4.15/Example_15.sce new file mode 100644 index 000000000..95bcb4a40 --- /dev/null +++ b/2465/CH4/EX4.15/Example_15.sce @@ -0,0 +1,19 @@ +//Chapter-4,Example 15,Page 96
+clc;
+close;
+
+P1= 528 // pressure in mm of Hg
+
+P2= 760 // pressure in mm of Hg
+
+T2=100+273 //teperature in Kelvin
+
+delta_Hv= 545.5 *18 // latent heat of vapourisation of water in J/mol
+
+R= 1.987 //gas constant
+
+//from the integrated form of Clausius-Clapeyron equation
+
+T1= 1/((log10(P2/P1)*2.303*R/delta_Hv)+(1/T2))
+
+printf('the temperature of water is %.f K',T1)
diff --git a/2465/CH4/EX4.16/Example_16.sce b/2465/CH4/EX4.16/Example_16.sce new file mode 100644 index 000000000..0b2037a77 --- /dev/null +++ b/2465/CH4/EX4.16/Example_16.sce @@ -0,0 +1,27 @@ +//Chapter-4,Example 16,Page 97
+clc;
+close;
+
+//since the operation is isothermal & hte gas is ideal therefore..
+
+delta_E= 0 // from 1st law of thermodynamics
+
+P= 1 //pressure in atm
+
+V1= 10 // volume in cubic decimeter
+
+V2= 20 // volume in cubic decimeter
+
+W= P*(V2-V1)*(8.314/0.0821) // work done by system
+
+q=W //from 1st law of thermodynamics
+
+delta_H = delta_E + W
+
+printf(' q = %.2f J',q)
+
+printf('\n W = %.2f',W)
+
+printf('\n delta_E = %.f J',delta_E)
+
+printf('\n delta_H = %.2f J',delta_H)
diff --git a/2465/CH4/EX4.17/Example_17.sce b/2465/CH4/EX4.17/Example_17.sce new file mode 100644 index 000000000..483dd3afd --- /dev/null +++ b/2465/CH4/EX4.17/Example_17.sce @@ -0,0 +1,19 @@ +//Chapter-4,Example 17,Page 97
+clc();
+close();
+
+q= 300 //heat energy
+
+P= 2 // pressure in atm
+
+V1= 10 // volume in litre
+
+V2= 20 //volume in litre
+
+//since 1 lit.atm = 24.25 cal
+
+W=P*(V2-V1)*24.25 //work done
+
+delta_E= q-W //from the 1st law of thermodynamics
+
+printf('the change in internal energy is %.f cal',delta_E)
diff --git a/2465/CH4/EX4.18/Example_18.sce b/2465/CH4/EX4.18/Example_18.sce new file mode 100644 index 000000000..2c66dc887 --- /dev/null +++ b/2465/CH4/EX4.18/Example_18.sce @@ -0,0 +1,21 @@ +//Chapter-4,Example 18,Page 97
+clc();
+close();
+
+T1= 300 //temperature in Kelvin
+
+T2= 363 //temperature in Kelvin
+
+P1= 1 //pressure in atm
+
+P2=7 //pressure in atm
+
+Cv=5
+
+R=2 //gas constant
+
+Cp=Cv+R
+
+delta_S= Cp*log(T2/T1)+R*log(P1/P2) //entropy change
+
+printf('the entropy change is %.4f cal/deg ', delta_S)
diff --git a/2465/CH4/EX4.19/Example_19.sce b/2465/CH4/EX4.19/Example_19.sce new file mode 100644 index 000000000..ee777211a --- /dev/null +++ b/2465/CH4/EX4.19/Example_19.sce @@ -0,0 +1,19 @@ +//Chapter-4,Example 19,Page 97
+clc();
+close();
+
+T1= 300 //temperature in Kelvin
+
+T2= 310 //temperature in Kelvin
+
+Kp1=3.49*10^-2 //equilibrium constant
+
+delta_H=-11200
+
+R= 1.987 //gas constant
+
+//from Van't Hoff's Equation
+
+Kp2=Kp1*10^(delta_H*((1/T1)-(1/T2))/(2.303*R))
+
+printf('the value of Kp2 = %.6f/atm ', Kp2)
diff --git a/2465/CH4/EX4.2/Example_2.sce b/2465/CH4/EX4.2/Example_2.sce new file mode 100644 index 000000000..d2dbc03d9 --- /dev/null +++ b/2465/CH4/EX4.2/Example_2.sce @@ -0,0 +1,11 @@ +//Chapter-4,Example 2,Page 93
+clc;
+close;
+
+q= 990*4.2/10^3 //heat in kiloJoule
+
+w= 8.36*10^9/((10^3)*(10^7)) //work in kiloJoule
+
+delta_E = q-w
+
+printf('the internal energy change of system is %.3f kJ',delta_E)
diff --git a/2465/CH4/EX4.3/Example_3.sce b/2465/CH4/EX4.3/Example_3.sce new file mode 100644 index 000000000..d3037835a --- /dev/null +++ b/2465/CH4/EX4.3/Example_3.sce @@ -0,0 +1,13 @@ +//Chapter-4,Example 3,Page 93
+clc;
+close;
+
+n=1 // number of mol
+
+R= 8.314 // gas constant
+
+T = 273 + 27 // temperature in Kelvin
+
+w=n*R*T/1000 // work done in kiloJoule
+
+printf('work done by reaction ai 27 degree is %.4f kJ',w)
diff --git a/2465/CH4/EX4.4/Example_4.sce b/2465/CH4/EX4.4/Example_4.sce new file mode 100644 index 000000000..9afb4eb21 --- /dev/null +++ b/2465/CH4/EX4.4/Example_4.sce @@ -0,0 +1,21 @@ +//Chapter-4,Example 4,Page 93
+clc;
+close;
+
+q_v=-97000 //in cal
+
+R= 8.314 // gas constant
+
+T = 273 + 200 // temperature in Kelvin
+
+n_1= 1 //mols of gaseous reactant
+
+n_2= 1 // mols of gaseous product
+
+delta_n= n_2-n_1
+
+//q_p= q_v + delta_n*R*T
+
+q_p= q_v + delta_n*R*T
+
+printf('the heat combustion of carbon is %.f cals',q_p)
diff --git a/2465/CH4/EX4.5/Example_5.sce b/2465/CH4/EX4.5/Example_5.sce new file mode 100644 index 000000000..ce36439e7 --- /dev/null +++ b/2465/CH4/EX4.5/Example_5.sce @@ -0,0 +1,19 @@ +//Chapter-4,Example 5,Page 93
+clc;
+close;
+
+delta_H= -109 // heat change in Kcal
+
+n_1= 2 //mols of gaseous reactant
+
+n_2= 1 // mols of gaseous product
+
+delta_n= n_2-n_1
+
+T=500
+
+R= 2*10^-3
+
+delta_E = (delta_H) - (delta_n*R*T)
+
+printf('the value of delta_E is %.f Kcal',delta_E)
diff --git a/2465/CH4/EX4.6/Example_6.sce b/2465/CH4/EX4.6/Example_6.sce new file mode 100644 index 000000000..7509cd20a --- /dev/null +++ b/2465/CH4/EX4.6/Example_6.sce @@ -0,0 +1,18 @@ +//Chapter-4,Example 6,Page 93
+clc;
+close;
+
+delta_H1= -337.2 // Heat combustion for ethylene
+
+delta_H2=-68.3 // Heat combustion for hudrogen
+
+delta_H3= 372.8 // Heat combustion for ethane
+
+//Given reaction is...
+// C2H4(g) +H2(g) ---> C2H6(g)
+
+delta_H= delta_H1 + delta_H2 +delta_H3
+
+printf('the heat combustion for given reaction is %.2f Kcal',delta_H)
+
+
diff --git a/2465/CH4/EX4.7/Example_7.sce b/2465/CH4/EX4.7/Example_7.sce new file mode 100644 index 000000000..bb3bd5922 --- /dev/null +++ b/2465/CH4/EX4.7/Example_7.sce @@ -0,0 +1,16 @@ +//Chapter-4,Example 7,Page 94
+clc;
+close;
+
+delta_H1= 104 //for reaction.. H2(g)---> 2H(g)
+
+delta_H2= 120/2 //for reaction.. (1/2)O2(g)---> O(g)
+
+delta_H3= -58 //for reaction.. H2(g) + (1/2)O2(g)---> H2O(g)
+
+delta_H=delta_H1 + delta_H2 - delta_H3
+
+//there are two O-H bonds
+//therefore for one bond required heat energy is half of delta_H
+
+printf('the O-H bond energy is %.f Kcal',delta_H/2)
diff --git a/2465/CH4/EX4.8/Example_8.sce b/2465/CH4/EX4.8/Example_8.sce new file mode 100644 index 000000000..2a642f92e --- /dev/null +++ b/2465/CH4/EX4.8/Example_8.sce @@ -0,0 +1,23 @@ +//Chapter-4,Example 8,Page 94
+clc;
+close;
+
+delta_H_C= -393 // enthalpy for carbon
+
+delta_H_H2= -286 //enthalpy for hydrogen
+
+delta_H_C3H8=-2220 //enthalpy for propane
+
+// According to Hess's Law... delta_H1 = delta_H2 - delta_H3
+
+//delta_H2 for reaction... 3C +4H2 +5O2 ----> 3CO2 +4H2O
+
+delta_H2= 3*delta_H_C +4*delta_H_H2
+
+//delta_H2 for reaction... C3H8 + 5O2 ----> 3CO2 +4H2O
+
+delta_H3= delta_H_C3H8
+
+delta_Hf= delta_H2 - delta_H3 //enthalpy for propane at 298 K
+
+printf('the enthalpy of formation of propane at 298K is %.f Kcal', delta_Hf)
diff --git a/2465/CH4/EX4.9/Example_9.sce b/2465/CH4/EX4.9/Example_9.sce new file mode 100644 index 000000000..39c08702c --- /dev/null +++ b/2465/CH4/EX4.9/Example_9.sce @@ -0,0 +1,15 @@ +//Chapter-4,Example 9,Page 95
+clc;
+close;
+
+delta_H2= 2386 //enthalpy for.. yellow P---> H3PO4
+
+delta_H3= 2113 //enthalpy for.. red P---> H3PO4
+
+delta_HT = delta_H2- delta_H3 //enthalpy for...yellow P ---> red P
+
+// According to Hess's Law... delta_H1 = delta_H2 - delta_H3
+
+delta_HT = delta_H2 - delta_H3 // delta_H1 = delta_HT
+
+printf('the enthalpy change of transition from yellow P to red P is %.f cals',delta_HT)
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