diff options
Diffstat (limited to '821')
161 files changed, 1558 insertions, 0 deletions
diff --git a/821/CH1/EX1.1/1_1.sce b/821/CH1/EX1.1/1_1.sce new file mode 100755 index 000000000..96715ee4d --- /dev/null +++ b/821/CH1/EX1.1/1_1.sce @@ -0,0 +1,7 @@ +printf('The Atomic weight of 12.0111 for Natural Carbon shows that the 12C nuclide must be present to a larger extent.');
+printf('\nLet 100 atoms of natural carbon contain x atoms of 12C nuclide.\n');
+X=(13.0034-12.0111)*100/(13.0034-12.0000);//percentage of 12C in natural carbon//
+printf('Percentage of 12C in Natural carbon=X=%f',X);
+Y=100-X;//percentage of 13C in natural carbon//
+printf('\nPercentage of 13C in Natural carbon=%f',Y);
+
diff --git a/821/CH1/EX1.10/1_10.sce b/821/CH1/EX1.10/1_10.sce new file mode 100755 index 000000000..1caaac830 --- /dev/null +++ b/821/CH1/EX1.10/1_10.sce @@ -0,0 +1,12 @@ +h=6.625*10^-27;//plank's constant//
+g=10^3;//particle mass in grams//
+l1=1;//length of one dimensional box in cm//
+n1=1;
+n2=2;
+dE1=((n2^2-n1^2)*h^2)/(8*g*l1^2);//Energy difference between two energy levels of particle in eV//
+printf('Energy difference between two energy levels of particle=dE1=1*10^-44eV');
+l2=2*10^-8;//length of one dimensional box in cm//
+m=9.11*10^-28;//electron mass in grams//
+dE2=((n2^2-n1^2)*h^2)/(8*m*l2^2*1.6*10^-11);//Energy difference between two energy levels of electron in eV//
+printf('\nEnergy difference between two energy levels of electron=dE2=%feV',dE2);
+
diff --git a/821/CH1/EX1.2/1_2.sce b/821/CH1/EX1.2/1_2.sce new file mode 100755 index 000000000..822fc3484 --- /dev/null +++ b/821/CH1/EX1.2/1_2.sce @@ -0,0 +1,7 @@ +C=3*10^2;//velocity of light in megametre/sec//
+L=300*10^-9;//wavelength of radiation in metres//
+v=C*10^-6/L;//frequency of radiation in Teracycles per sec//
+printf('Frequency of radiation=v=%fTeracycles per sec=1.0*10^15Hz',v);
+v1=1/L;//wave number of radiation in per meter//
+printf('\nWavenumber of radiation=v1=3.3*10^4 per centimeter');
+
diff --git a/821/CH1/EX1.3/1_3.sce b/821/CH1/EX1.3/1_3.sce new file mode 100755 index 000000000..a104889bd --- /dev/null +++ b/821/CH1/EX1.3/1_3.sce @@ -0,0 +1,9 @@ +V=0.85;//external voltage in volts//
+e=1.6*10^-19;//electron charge in coloumbs//
+m=9.1*10^-28;//electron mass in grams//
+v=sqrt(2*V*e*10/m);//velocity of electron in motion in Kilocm per sec//
+printf('velocity of electron in motion=v=%f Kilocm per sec=5.47*10^7cm per sec',v);
+W=(3.198*10^-12)/(1.6*10^-12);//Threshold energy in eV//
+printf('\nThreshold energy of electron=W=%feV',W);
+v0=(3.198*10^-12)/(6.625*10^-15);//Threshold frequency in tera per sec//
+printf('\nThreshold frequency=v0=%fTera per sec=4.83*10^14per sec',v0);
diff --git a/821/CH1/EX1.4/1_4.sce b/821/CH1/EX1.4/1_4.sce new file mode 100755 index 000000000..79bbb992d --- /dev/null +++ b/821/CH1/EX1.4/1_4.sce @@ -0,0 +1,8 @@ +E=118.5*10^3*4.2*10^7;//energy of ions in ergs//
+C=3*10^10;//velocity of light in cm/sec//
+L=6.023*10^23;//Avagadro number//
+h=6.625*10^-27;//plank's constant//
+l=(L*h*C*10^8)/E;
+printf('wavelength required to cause ionization=l=%fAngstrums',l);
+
+
diff --git a/821/CH1/EX1.5/1_5.sce b/821/CH1/EX1.5/1_5.sce new file mode 100755 index 000000000..7328fd261 --- /dev/null +++ b/821/CH1/EX1.5/1_5.sce @@ -0,0 +1,7 @@ +n1=2;
+n2=4;
+dE=21.7*(10^-12)*((1/n1^2)-(1/n2^2));
+h=6.625*10^-27;//plank's constant//
+C=3*10^10;//velocity of light in cm/sec//
+l=h*C*10^8/dE;//Wavelength of second line in balmer series in Angstrums//
+printf('wavelength of the second line in balmer series=l=%fAngstrums',l);//here the answer given in textbook is slightly wrong the original answer should be the one comes through execution//
diff --git a/821/CH1/EX1.6/1_6.sce b/821/CH1/EX1.6/1_6.sce new file mode 100755 index 000000000..129ae2f33 --- /dev/null +++ b/821/CH1/EX1.6/1_6.sce @@ -0,0 +1,3 @@ +n1=1;
+dE=21.7*(10^-12)/(1.6*10^-12*n1^2);//energy required to promote an electron from ground to infinity in eV//
+printf('Ionisation potential for an electron=dE=%feV',dE);
diff --git a/821/CH1/EX1.7/1_7.sce b/821/CH1/EX1.7/1_7.sce new file mode 100755 index 000000000..92ce6604b --- /dev/null +++ b/821/CH1/EX1.7/1_7.sce @@ -0,0 +1,5 @@ +h=6.625*10^-27;//plank's constant//
+V=2*10^3;//velocity of Cricket Ball in cm/sec//
+m=170;//weight of Cricket Ball in grams//
+l=h/(m*V);//DeBroglie Wavelength of CricketBall in Angstrums//
+printf('DeBroglie Wavelength of CricketBall=l=%f=1.95*10^-24Angstrums',l);
diff --git a/821/CH1/EX1.8/1_8.sce b/821/CH1/EX1.8/1_8.sce new file mode 100755 index 000000000..9627caa6b --- /dev/null +++ b/821/CH1/EX1.8/1_8.sce @@ -0,0 +1,14 @@ +r1=0.53*10^-8;//Bohr radius in cm//
+r2=4*r1;//Bohr radius in second state in cm//
+printf('Bohr radius in second state=r2=2.12*10^-8cm');
+h=6.625*10^-27;//plank's constant//
+m=9.11*10^-28;//electron mass in grams//
+v2=h/(%pi*m*r2);//electron velocity in second state in cm per sec//
+printf('\nElectron velocity in second state=v2=%fcm per sec',v2);
+l=(h*10^8)/(m*v2);//De Broglie wavelength of electron in second state in Angstrums//
+printf('\nDe Broglie wavelength of electron in second state=l=%fAngstrums',l);
+e=1.6*10^-12;//electron charge in ergs//
+v=sqrt((2*(10^4)*e)/m);//velocity of the moving electron in second state in cm/sec//
+printf('\nVelocity of moving electron in second state=v=%fcm per sec',v);
+l1=(h*10^8)/(v*m);//De Broglie wavelength of moving elctron in Angstrums//
+printf('\nDe Broglie wavelength of moving electron in secondstate=l1=%fAngstrums',l1);
diff --git a/821/CH1/EX1.9/1_9.sce b/821/CH1/EX1.9/1_9.sce new file mode 100755 index 000000000..d5270b11b --- /dev/null +++ b/821/CH1/EX1.9/1_9.sce @@ -0,0 +1,11 @@ +m=9.11*10^-28;//electron mass in grams//
+v=1.1*10^8;//velocity of electron in cm per sec//
+p=m*v;//momentum of electron in gram cm per sec//
+printf('momentum of electron=p=10.01*10^-20gram cm per sec');
+dp=p*10^-2;//Uncertainity in momentum in gram cm per sec//
+printf('\nUncertainity in momentum=10.01*10^-22gram cm per sec');
+h=6.625*10^-27;//plank's constant//
+dx=(h*10^8)/(4*%pi*dp);//Uncertainity in position in Angstrum//
+printf('\nUncertainity in position=dx=%fAngstrum',dx);
+
+
diff --git a/821/CH2/EX2.1/2_1.sce b/821/CH2/EX2.1/2_1.sce new file mode 100755 index 000000000..6b3ea0125 --- /dev/null +++ b/821/CH2/EX2.1/2_1.sce @@ -0,0 +1,8 @@ +N0=3396;//no. of counts per minute given by radioactive nuclide at a given time//
+N=1000;//no. of counts per minute given by radioactive nuclide one hour later//
+thalf=0.693*60/(2.303*log(N0/N));//half life of nuclide in minutes//
+printf('Half life of radioactive nuclide=t1/2=%fminutes',thalf);
+t1=2.303*log(100/25)*thalf/0.693;//time required for the activity to decrease to 25% of the initial activity in minutes//
+printf('\nTime required for the activity to decrease to 25percent of the initial activity=t1=%fminutes',t1);
+t2=2.303*log(100/10)*thalf/0.693;//time required for the activity to decrease to 10% of the initial activity in minutes//
+printf('\nTime required for the activity to decrease to 10percent of the initial activity=t2=%fminutes',t2);
diff --git a/821/CH2/EX2.2/2_2.sce b/821/CH2/EX2.2/2_2.sce new file mode 100755 index 000000000..7a4e883e4 --- /dev/null +++ b/821/CH2/EX2.2/2_2.sce @@ -0,0 +1,5 @@ +R=3.7*10^10;//no. of alpha particles per second emitted by 1g of 226Ra//
+N=(6.023*10^23)/226;//no. of atoms of 226Ra//
+yr=3.15*10^7;//no of seconds in a year//
+thalf=0.693*N/(R*yr);//half life of 226Ra in years//
+printf('Half life of 226Ra molecule=t1/2=%fyears',thalf);//here the answer written in textbook is wrongly printed actual answer will be the one we are getting here//
diff --git a/821/CH2/EX2.3/2_3.sce b/821/CH2/EX2.3/2_3.sce new file mode 100755 index 000000000..b36183844 --- /dev/null +++ b/821/CH2/EX2.3/2_3.sce @@ -0,0 +1,6 @@ +thalf=14.8*60*60;//half life of 24Na atom in seconds//
+L=6.023*10^23;//Avagadro number//
+v=3.7*10^10;//1 Ci of radioactivity in disintegrations per second//
+w=(24*10^6*v*thalf)/(0.693*L);//weight of 1 Ci of 24Na in grams//
+printf('Weight of 1 Ci of 24Na=w=%fmicrograms=1.13*10^-7grams',w);
+
diff --git a/821/CH2/EX2.4/2_4.sce b/821/CH2/EX2.4/2_4.sce new file mode 100755 index 000000000..1199f62d4 --- /dev/null +++ b/821/CH2/EX2.4/2_4.sce @@ -0,0 +1,7 @@ +Mp=1.00728;//mass of proton in amu//
+Mn=1.00866;//mass of neutronin amu//
+MH=2.01355;//isotopic mass of H atom in amu//
+dM=((1*Mp)+(1*Mn)-MH);//dM value of H atom in amu//
+printf('dM value of H atom=dM=%famu',dM);
+BE=dM*931;//binding energy of H atom in MeV//
+printf('\nBinding energy of H atom=BE=%fMeV',BE);
diff --git a/821/CH2/EX2.5/2_5.sce b/821/CH2/EX2.5/2_5.sce new file mode 100755 index 000000000..39cdc8be5 --- /dev/null +++ b/821/CH2/EX2.5/2_5.sce @@ -0,0 +1,5 @@ +N0=15.3;//decay rate of Contemporary Carbon in disintegrations/min/gram//
+N=2.25;//decay rate of 14C specimen in disintegrtions/min/gram//
+thalf=5670;//half life of nuclide in years//
+t=2.303*log(N0/N)*thalf/0.693;//Age of the specimen in years//
+printf('Age of the specimen=t=%fyears',t);//here the answer given in textbook is actually wrong we get twice that of the answer which is shown through execution//
diff --git a/821/CH2/EX2.6/2_6.sce b/821/CH2/EX2.6/2_6.sce new file mode 100755 index 000000000..70c7f5cb3 --- /dev/null +++ b/821/CH2/EX2.6/2_6.sce @@ -0,0 +1,7 @@ +thalf=4.5*10^9;//half life of Uranium in years//
+printf('Here N0 and N must be in terms of Uranium.N is proportional to 1gram og Uranium');
+printf('\nN0 can be calculated from the given data.0.0453grams of 206Pb corresponds to 238*0.0453/206=0.0523grams of 238U,i.e 0.0453 grams of 206Pb must have been formed by the decaying of 0.523grams of 238U.\nSince N is proportional to 1,N0 is proportional to 1.0523.');
+N0=1.0523;
+N=1;
+t=2.303*log(N0/N)*thalf/0.693;//Age of the mineral in years//
+printf('\nAge of the mineral=t=%fyears=7.62*10^8years',t);//here also the answer given in textbook is wrong the one resulted through execution is the right one//
diff --git a/821/CH4/EX4.1/4_1.sce b/821/CH4/EX4.1/4_1.sce new file mode 100755 index 000000000..841c0521c --- /dev/null +++ b/821/CH4/EX4.1/4_1.sce @@ -0,0 +1,8 @@ +P=760;//pressure of 14g of nitrogen in mm of Hg//
+V=22.4;//Volume occupied by 14g of Nitrogen in litres//
+P1=380;//changed pressure of 14g of nitrogen in mm of Hg//
+V1=(P*V)/P1;//changed volume of 14g of nitrogen at 380mm pressure in litres//
+printf('Volume of 14g of Nitrogen at 380 mm Hg pressure=V1=%flitres',V1);
+V2=5.6;//changed volume of 14g of nitrogen in litres//
+P2=(P*V)/V2;//Pressure of 14g of nitrogen of volume 5.6litres//
+printf('\nPressure of 14g of Nitrogen of volume 5.6litres=P2=%fmm Hg',P2);
diff --git a/821/CH4/EX4.10/4_10.sce b/821/CH4/EX4.10/4_10.sce new file mode 100755 index 000000000..d5ad3eadc --- /dev/null +++ b/821/CH4/EX4.10/4_10.sce @@ -0,0 +1,7 @@ +printf('Density is proportional to molecular weight');
+printf('\nRelative density of O2 with respect to H2 is 32/2=16.');
+t1=12;//time for certain volume of gas to stream through hole in mins//
+p2=16;//Relative density of O2 w.r.t H2//
+t2=15;//time for same volume(as gas) of oxygen to stream through hole in mins//
+p1=((t1/t2)^2)*p2;//according to graham's law//
+printf('\nRealtive density of gas to oxygen=p1=%f',p1);//here the answer given in textbook is wrongly printed the actual answer is the one we got here through execution//
diff --git a/821/CH4/EX4.11/4_11.sce b/821/CH4/EX4.11/4_11.sce new file mode 100755 index 000000000..61973a02f --- /dev/null +++ b/821/CH4/EX4.11/4_11.sce @@ -0,0 +1,10 @@ +T=320;//47C in kelvin//
+R=8.31*10^7;//Universal gas constant in erg per degree per mole//
+M=32;//molecular weight of O2 in gram per mole//
+C2=(3*R*T)/M;//mean square velocity of Oxygen in (cm/sec)^2//
+Crms=sqrt(C2);//Root mean square velocity of Oxygen in cm/sec//
+printf('Root mean square velocity of Oxygen=Crms=%fcm/sec',Crms);
+Cm=sqrt(8*R*T/(%pi*M));//mean velocity of Oxygen in cm/sec//
+printf('\nMean velocity of Oxygen=Cm=%fcm/sec',Cm);
+Cmpv=sqrt(2*R*T/M);//mean probable velocity in cm/sec//
+printf('\nMean probable velocity of Oxygen=Cmpv=%fcm/sec',Cmpv);//in textbook in Cmpv value it is misprinted as 10^9 but it is actually 10^4//
diff --git a/821/CH4/EX4.12/4_12.sce b/821/CH4/EX4.12/4_12.sce new file mode 100755 index 000000000..1d2bc1061 --- /dev/null +++ b/821/CH4/EX4.12/4_12.sce @@ -0,0 +1,9 @@ +P=10^6;//pressure of gas in dyn per cm^2//
+p=0.00333;//density of gas in gram per cm^3//
+C2=3*P/p;//mean square velocity of gas in (cm/sec)^2//
+Crms=sqrt(C2);//Root mean square velocity of Oxygen in cm/sec//
+printf('Root mean square velocity of Gas=Crms=%fcm/sec',Crms);
+Cm=Crms/1.085;//mean velocity of Gas in cm/sec//
+printf('\nMean velocity of Oxygen=Cm=%fcm/sec',Cm);
+Cmpv=Cm/1.128;//mean probable velocity in cm/sec//
+printf('\nMean probable velocity of Oxygen=Cmpv=%fcm/sec',Cmpv);
diff --git a/821/CH4/EX4.13/4_13.sce b/821/CH4/EX4.13/4_13.sce new file mode 100755 index 000000000..fd4cad0a4 --- /dev/null +++ b/821/CH4/EX4.13/4_13.sce @@ -0,0 +1,5 @@ +printf('Mean free path is inversely proportional to N which is proportional to pressure');
+L1=5.8*10^-6;//mean free path of N2 at NTP in cm//
+K=58;//preesure is raised 58 times so factor is 58//
+L2=L1/K;//mean free path of N2 at 58atm pressure in cm//
+printf('\nMean free path of N2 at 58atm pressure=1*10^-7cm');
diff --git a/821/CH4/EX4.14/4_14.sce b/821/CH4/EX4.14/4_14.sce new file mode 100755 index 000000000..56085b18f --- /dev/null +++ b/821/CH4/EX4.14/4_14.sce @@ -0,0 +1,9 @@ +T=273;//temperature 0C in kelvin//
+R=8.31*10^7;//Universal gas constant in erg per degree per mole//
+M=28;//molecular weight of N2 in gram per mole//
+printf('Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10^23Molecules.');
+N=2.69*10^19;//no. of molecules in molecules per cm^3//
+G=3.78*10^-8;//collision diameter in cm//
+K=sqrt(%pi*R*T/M);
+Z11=2*N^2*G^2*K;//number of collisions per second of Nitrogen at 0C and 1atm//
+printf('\nNumber of molecular collisions per second of Nitrogen at 0C and 1atm=%f=10.43*10^28molecular collisions per sec per cm^3',Z11);
diff --git a/821/CH4/EX4.15/4_15.sce b/821/CH4/EX4.15/4_15.sce new file mode 100755 index 000000000..8f8a03f63 --- /dev/null +++ b/821/CH4/EX4.15/4_15.sce @@ -0,0 +1,18 @@ +T=273;//temperature 0C in kelvin//
+R=8.31*10^7;//Universal gas constant in erg per degree per mole//
+M=28;//molecular weight of N2 in gram per mole//
+printf('Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10^23Molecules.');
+N=2.69*10^19;//no. of molecules in molecules per cm^3//
+Cm=sqrt(8*R*T/(%pi*M));//mean velocity of Nitrogen in cm/sec//
+printf('\nMean velocity of Nitrogen=Cm=%fcm/sec',Cm);
+V=22400;//volume of nitrogen in cm^3//
+p=M/V;//Density of nitrogen in gram per cm^3//
+printf('\nDensity of Nitrogen=p=%f=1.25*10^-3gram per cm^3',p);
+n=10.99*10^-5;//Viscosity of N2 in poise//
+L=(3*n)/(Crms*p);//mean free path of nitrogen in cm//
+printf('\nMean free path of Nitrogen=L=5.81*10^-6cm');
+G=sqrt(1/(1.414*%pi*L*N));//Collision diameter of Nitrogen in cm//
+printf('\nCollission diameter of Nitrogen=G=3.80*10^-8cm');
+K=sqrt(%pi*R*T/M);
+Z11=2*N^2*G^2*K;//number of collisions per second of Nitrogen at 0C and 1atm//
+printf('\nNumber of molecular collisions per second of Nitrogen at NTP=%f=10.52*10^28molecular collisions per sec per cm^3',Z11);
diff --git a/821/CH4/EX4.16/4_16.sce b/821/CH4/EX4.16/4_16.sce new file mode 100755 index 000000000..6fdd04dfd --- /dev/null +++ b/821/CH4/EX4.16/4_16.sce @@ -0,0 +1,8 @@ +T1=27;//initial temperature in C//
+T2=177;//final temperature in C//
+printf('NH3 being a nonlinear molecule has 3 translational,3 rotational and 6 vibrational degrees of freedom.');
+printf('\nCv=3*0.5*R+3*0.5*R+6*R=9*R');
+Cv=18;//Molar heat capacity in cal per deg per mol//
+dU=Cv*(T2-T1);//Change in internal energy of a mole in cal per mole//
+printf('\nChange in internal energy of a mole of NH3=dU=%fcal per mole',dU);
+printf('\nThe actual increase in energy may not be 2700cal per mol\nbecause at the given temperature,none or only some of the vibrational degrees of freedom may be contributing to the total energy.');//answer should come as 2700 not 2400,it was just misprinted//
diff --git a/821/CH4/EX4.17/4_17.sce b/821/CH4/EX4.17/4_17.sce new file mode 100755 index 000000000..629b244e6 --- /dev/null +++ b/821/CH4/EX4.17/4_17.sce @@ -0,0 +1,8 @@ +printf('CH4 being a nonlinear molecule has 3 translational,3 rotational and 9 vibrational degrees of freedom.');
+printf('\nCv=3*0.5*R+3*0.5*R+9*R=12*R');
+Cv=24;//Molar heat capacity in cal per degree per mol//
+printf('\nActual Cv is 16cal per degree per mol');
+R=2;
+Cvr=16-R;//Real molar heat capapcity in cal per degree per mole//
+printf('\nReal molar heat capapcity of a mole of CH4=Cvr=%fcal per degree per mole',Cvr);
+printf('\nThe Real molar heat capacity at constant volume is 10cal less than the theoretical value.\nSince each vibrational degree of freedom can contribute 2cal,this means that 5 vibrational degrees\nare not contributing at the given temperature.');
diff --git a/821/CH4/EX4.18/4_18.sce b/821/CH4/EX4.18/4_18.sce new file mode 100755 index 000000000..95fdf525e --- /dev/null +++ b/821/CH4/EX4.18/4_18.sce @@ -0,0 +1,10 @@ +a=3.60;//Van der Waals constant for CO2 in L^2 atm per mol^2//
+b=4.30*10^-2;//Van der Waals constant for CO2 in litre per mol//
+N=300;//No. of moles of CO2 the cylinder contains//
+V=100;//volume of cylinder in litres//
+P=150;//Maximum pressure the cylinder can withstand in atm//
+R=0.082;
+K1=P+(N^2*a/V^2);
+K2=V-(N*b);
+T=(K1*K2)/(N*R);//The lowest temperature at which cylinder can explode in Kelvin//
+printf('The lowest temperature at which cylinder can explode=T=%fK',T);
diff --git a/821/CH4/EX4.19/4_19.sce b/821/CH4/EX4.19/4_19.sce new file mode 100755 index 000000000..3e4441505 --- /dev/null +++ b/821/CH4/EX4.19/4_19.sce @@ -0,0 +1,15 @@ +tA=280;//time of flow for liquid A in seconds//
+tB=200;//time of flow for liquid B in seconds//
+pA=1;//density of liquid A in gram per cm^3//
+pB=1.1;//density of liquid B in gram per cm^3//
+h=10;//height of liquid responsible for the flow in cm//
+g=980;//gravity constant in dyns//
+V=1;//volume of liquid in ml//
+L=10;//length of the capillary in cm//
+r=0.1;//radius of the capillary in cm//
+PA=h*pA*g;//Pressure of liquid A//
+PB=h*pB*g;//Pressure of liquid B//
+nA=(%pi*PA*tA*r^4)/(8*L*V);//Viscosity of Liquid A in centipoise//
+printf('\nViscosity of Liquid A=nA=%fcentipoise',nA);
+nB=(%pi*PB*tB*r^4)/(8*L*V);//Viscosity of Liquid B in centipoise//
+printf('\nViscosity of Liquid B=nB=%fcentipoise',nB);
diff --git a/821/CH4/EX4.2/4_2.sce b/821/CH4/EX4.2/4_2.sce new file mode 100755 index 000000000..12181dacb --- /dev/null +++ b/821/CH4/EX4.2/4_2.sce @@ -0,0 +1,5 @@ +V=5.6;//Volume occupied by 8g of Oxygen at 0 C in litres//
+T=273;//Temperature at which 8g of Oxygen occupies 5.6litres in Kelvin//
+V1=11.2;//Changed volume of 8g of Oxygen in litres//
+T1=(V1*T)/V;//Temperature at which 8g of Oxygen occupies 11.6litres in kelvin//
+printf('Temperature of 8g of Oxygen Occupying 11.6litres=T1=%fKelvin=273degrees',T1);
diff --git a/821/CH4/EX4.20/4_20.sce b/821/CH4/EX4.20/4_20.sce new file mode 100755 index 000000000..2297adf5e --- /dev/null +++ b/821/CH4/EX4.20/4_20.sce @@ -0,0 +1,6 @@ +ST=520;//surface tension of mercury in dyn per cm//
+d=13.6;//density of mercury in gram per cm^3//
+g=980;//gravity constant in dyns//
+r=0.1;//radius of the capillary tube in cm//
+h=(ST*2)/(g*r*d);//Depression observed in capillary tube in cm//
+printf('Depression observed in capillary tube=h=%fcm',h);//the answer is given wrong in textbook,it should actually be double the one given//
diff --git a/821/CH4/EX4.21/4_21.sce b/821/CH4/EX4.21/4_21.sce new file mode 100755 index 000000000..d06773e5a --- /dev/null +++ b/821/CH4/EX4.21/4_21.sce @@ -0,0 +1,6 @@ +T=293;//temperature 20C in kelvin//
+R=8.31*10^7;//Universal gas constant in erg per degree per mole//
+dr=165;//Change of surface tension in erg per cm^2 per mol litre//
+C=0.06;//concentration we are considering in M//
+t=(C*dr)/(R*T);//surface excess concentration of phenol in mol per cm^2//
+printf('surface excess concentration of phenol=t=4.08*10^-10mol per cm^2');
diff --git a/821/CH4/EX4.22/4_22.sce b/821/CH4/EX4.22/4_22.sce new file mode 100755 index 000000000..b2553a036 --- /dev/null +++ b/821/CH4/EX4.22/4_22.sce @@ -0,0 +1,10 @@ +M=256;//Molecular weight of acid in grams//
+w=2.56*10^-5;//weight of palmitic acid in grams//
+N=w/M;//No. of molecules of acid//
+A=123;//Total area occupied in cm^2//
+AM=A/N;//Area per molecule in cm^2//
+printf('Area per molecule=20.4*10^-16cm^2.\nThis is the area of crosssection A for a vertical position.');
+V=w/0.81;//Volume of acid in cm^3//
+VM=V/N;//Volume of one molecule of acid in cm^3//
+L=VM/AM;//Length of molecule in cm//
+printf('\nLength of a molecule=25.7*10^-8cm');
diff --git a/821/CH4/EX4.23/4_23.sce b/821/CH4/EX4.23/4_23.sce new file mode 100755 index 000000000..51eef4328 --- /dev/null +++ b/821/CH4/EX4.23/4_23.sce @@ -0,0 +1,12 @@ +T=298;//Temperature in kelvin//
+ST=0.2;//Lowered surface tension in dyn per cm//
+K=1.37*10^-16;//Boltzman's constant//
+AM=(K*T)/ST;//Area per molecule on the surface in cm^2//
+A=300;//Total area occupied in cm^2//
+N=A/AM;//no. of molecules//
+W=3*10^-7;//weight of insoluble substance in grams//
+w=W/N;//weight of one molecule in grams//
+N1=6.023*10^23;
+MW=w*N1;//Molecular weight of substance in grams//
+printf('Molecular weight of substance=MW=%f=123',MW);
+
diff --git a/821/CH4/EX4.24/4_24.sce b/821/CH4/EX4.24/4_24.sce new file mode 100755 index 000000000..ba75ff75c --- /dev/null +++ b/821/CH4/EX4.24/4_24.sce @@ -0,0 +1,8 @@ +MW=461;//molecular weight of lead iodide in grams//
+CR1=3000;//initial count rate in count per minute//
+CR2=900;//final count rate in count per minute//
+CR=CR1-CR2;
+printf('The count for lead iodide as absorbed=CR=%fcount per minute',CR);
+printf('\nThe ratio of weights of lead iodide and radio lead iodide in solution is equal to that of the same ratio on surface');
+printf('\nWeight of lead iodide in solution=0.0014grams');
+printf('\nWeight of radio lead iodide is proportional to the count.\nWeight of lead iodide on the surface=0.0014*21/9.\nMolecular weight of lead iodide=461.\nArea of the surface per gram=4266cm^2 g^-1');
diff --git a/821/CH4/EX4.25/4_25.sce b/821/CH4/EX4.25/4_25.sce new file mode 100755 index 000000000..9bdd8994e --- /dev/null +++ b/821/CH4/EX4.25/4_25.sce @@ -0,0 +1,8 @@ +printf('From the linear plot of the langmuir isotherm the intercept Xm=1/0.19=5.26milligrams');
+Xm=5.26*10^-3;
+printf('\nThis is the weight of N2 that forms a unimolecular layer of 2g charcoal.');
+MW=28;//molecular weight of N2//
+N=Xm*6.023*10^23/MW;//No. of molecules of N2//
+TA=N*16*10^-16;//Total area in cm^2//
+A=TA/2;//Area per gram in cm^2//
+printf('\nArea of N2 per gram=%fcm^2',A);
diff --git a/821/CH4/EX4.26/4_26.sce b/821/CH4/EX4.26/4_26.sce new file mode 100755 index 000000000..c2c61ceae --- /dev/null +++ b/821/CH4/EX4.26/4_26.sce @@ -0,0 +1,8 @@ +printf('From the linear plot of the langmuir isotherm the intercept=0.35*10^-3 and slope=9.47*10^-2');
+printf('\nVolume is the inverse of summation of intercept and slope and that is 10.52cc');
+Vm=10.52;//volume in cc//
+m=Vm/22400;//No. of moles of N2//
+N=m*6.023*10^23;//No. of molecules of N2//
+TA=N*16*10^-16;//Total area in cm^2//
+A=TA/17.5;//Area per gram in cm^2//
+printf('\nArea of N2 per gram=%fcm^2',A);
diff --git a/821/CH4/EX4.28/4_28.sce b/821/CH4/EX4.28/4_28.sce new file mode 100755 index 000000000..a7c4bff41 --- /dev/null +++ b/821/CH4/EX4.28/4_28.sce @@ -0,0 +1,10 @@ +d=13.56;//density of mercury in gr per cm^3//
+V=1/d;//Volume of mercury in cm^3//
+D=0.07*10^-4;//Diameter of the globule in cm//
+r=D/2;//radius of globule in cm//
+printf('One globule of mercury occpies a volume of 1.33*3.14*r^3cm^3\nSurface area of one globule=4*3.14*r^2');
+Vg=1.33*3.14*r^3;//volume of one globule in cm^3//
+y=0.0738/Vg;//No of globules in 0.0738 volume//
+printf('\nNo. of globules in 0.0738cm^3=y=%f',y);
+SA=y*4*3.14*r^2;//Surface area of y globules in cm^2//
+printf('\nSurface area of y globules=SA=%fcm^2',SA);
diff --git a/821/CH4/EX4.29/4_29.sce b/821/CH4/EX4.29/4_29.sce new file mode 100755 index 000000000..a3e55ea3d --- /dev/null +++ b/821/CH4/EX4.29/4_29.sce @@ -0,0 +1,10 @@ +T=300;//temperature in kelvin//
+R=8.31*10^7;//Universal gas constant in ergs//
+r=2*10^-5;//average radius of particles in cm//
+h=4.15*10^-3;//Vertical seperation in cm//
+p=1.21;//density of latex in gram per cm^3//
+g=980;//gravity constant in dynes//
+printf('Since the dispersion medium is water its density is p1=1.\nwhen the system equilibrated average number of colloidal particles seen in the field is halved so N0/N=2.\nthe required expression is derived based on kinetic energy of particles');
+p1=1;//density of water//
+L=log(2)*0.75*(R*T)/(%pi*g*h*(p-p1)*r^3);
+printf('\nValue of Avagadro number is L=%f=6.037*10^23molecules per mol.',L);
diff --git a/821/CH4/EX4.3/4_3.sce b/821/CH4/EX4.3/4_3.sce new file mode 100755 index 000000000..81e455d02 --- /dev/null +++ b/821/CH4/EX4.3/4_3.sce @@ -0,0 +1,7 @@ +P=380;//pressure of 11g of CO2 at 273K in mm of Hg//
+T=273;//Initial temperature of 11g of CO2 in kelvin//
+V=11.2;//Volume occupied by 11g of CO2 in litres at 273K//
+P1=760;//changed pressure of 11g of CO2 at 546K in mm of Hg//
+T1=546;//Final temperature of 11g of CO2 in kelvin//
+V1=(P*V*T1)/(T*P1);//changed volume of 11g of CO2 at 760mm pressure in litres//
+printf('Volume of 11g of CO2 at 760 mm Hg pressure at 546K=V1=%flitres',V1);
diff --git a/821/CH4/EX4.30/4_30.sce b/821/CH4/EX4.30/4_30.sce new file mode 100755 index 000000000..7a90f8287 --- /dev/null +++ b/821/CH4/EX4.30/4_30.sce @@ -0,0 +1,12 @@ +Vf=117.857*10^-9;//Volume falling within field of view in ml//
+AN=9.66;//Average no. of particles per count//
+N=AN/Vf;//No. of particles per count//
+printf('No .of particles per ml=N=%f',N);
+W=1.5*10^-7;//Weight of iron oxide per ml//
+w=W/N;//weight of one particle in grams//
+printf('\nWeight of one particle=w=1.83*10^-15grams');
+d=5.2;//density of iron oxide in g per cm^3//
+V=w/d;//Volume of one particle in cm^3//
+printf('\nVolume of one particle=V=3.52*10^-16cm^3');
+R=2.032*10^-5;//radius of particle in cm//
+printf('Radius of particle=R=2.032*10^-5cm\nDiameter of particle=4.064*10^-5cm');
diff --git a/821/CH4/EX4.31/4_31.sce b/821/CH4/EX4.31/4_31.sce new file mode 100755 index 000000000..3a03ae3c8 --- /dev/null +++ b/821/CH4/EX4.31/4_31.sce @@ -0,0 +1,8 @@ +D=78;//dielectric constant//
+V=0.009;//viscosity of suspension in dyn sec per cm^2//
+P=3.2;//potential gradient in volt per cm//
+t=38.2;
+d=0.033;//standard distance between electrodes in cm//
+u=90000;
+EP=(4*%pi*V*u*d)/(D*t*P);//Electrokinetic potential of the catalyst in volts//
+printf('Electrokinetic potential of the catalyst=EP=%fvolts',EP);
diff --git a/821/CH4/EX4.32/4_32.sce b/821/CH4/EX4.32/4_32.sce new file mode 100755 index 000000000..6e399c1a6 --- /dev/null +++ b/821/CH4/EX4.32/4_32.sce @@ -0,0 +1,12 @@ +D=80.36;//dielectric constant//
+V=0.01;//viscosity of suspension in dyn sec per cm^2//
+P=10;//potential gradient in volt per cm//
+v=3*10^-3;//observed velocity in cm per sec//
+u=90000;
+EP=(4*%pi*V*u*v)/(D*P);//Zeta potential of the catalyst in volts//
+printf('Zeta potential of the catalyst=EP=%fvolts',EP);
+printf('\nThe effective thickness of the double layer can be taken to be 1cm');
+e=9*10^-4;
+r=0.5*10^-4;
+N=4*%pi*e*r^2;//total no. of charges carried by a particle//
+printf('\nTotal no. of charges carried by a particle=N=2.83*10^-11');
diff --git a/821/CH4/EX4.33/4_33.sce b/821/CH4/EX4.33/4_33.sce new file mode 100755 index 000000000..81689875d --- /dev/null +++ b/821/CH4/EX4.33/4_33.sce @@ -0,0 +1,7 @@ +D=80.36;//dielectric constant//
+V=0.01;//viscosity of suspension in dyn sec per cm^2//
+P=100;//potential gradient in volt per cm//
+u=90000;
+EP=0.03;//Electrokinetic potential of the catalyst in volts//
+FW=(P*EP*D*3600)/(4*%pi*V*u);//amount of flow of water in cm per hour//
+printf('Amount of flow of water through diaphragm=FW=%fcm per hour',FW);
diff --git a/821/CH4/EX4.34/4_34.sce b/821/CH4/EX4.34/4_34.sce new file mode 100755 index 000000000..4a1d7b14a --- /dev/null +++ b/821/CH4/EX4.34/4_34.sce @@ -0,0 +1,8 @@ +D=80.36;//dielectric constant//
+V=0.01;//viscosity of suspension in dyn sec per cm^2//
+P=40;//potential gradient in volt per cm//
+r=0.05;//radius of capillary in cm//
+u=90000;
+ZP=0.05;//Zeta potential of the catalyst in volts//
+FW=(P*ZP*D*3600*r^2)/(4*V*u);//amount of flow of water in cc per hour//
+printf('Amount of flow of water through diaphragm=FW=%fcc per hour',FW);
diff --git a/821/CH4/EX4.35/4_35.sce b/821/CH4/EX4.35/4_35.sce new file mode 100755 index 000000000..4a44386c8 --- /dev/null +++ b/821/CH4/EX4.35/4_35.sce @@ -0,0 +1,14 @@ +C1=5*10^-4;
+C2=2.5*10^-4;
+x=(C1^2)/(C2+(2*C1));//The amount of Na+ transported at equilibrium in M//
+printf('The amount of Na+ transported at equilibrium=x=2*10^-4M');
+NaR=C2+x;//Na+ on RHS//
+printf('\nNa+ on RHS=4.5*10^-4M.\nCl- on RHS=x=2*10^-4M.\nNa+ on LHS=Cl- on LHS=C1-x=3*10^-4M');
+printf('\nE is in volts and F in coulombs.for homogeneity of units either R should be expressed in coulombs or F in calories.F is 96500C or 23060cal.');
+F=23060;
+n=1;
+T=300;//temperature in kelvin//
+R=2;
+K=1.5;
+E=(R*T*log(K))/(n*F);//Emf across membrane in volts//
+printf('\nEmf across membrane=E=%fvolts',E);
diff --git a/821/CH4/EX4.36/4_36.sce b/821/CH4/EX4.36/4_36.sce new file mode 100755 index 000000000..00bb8b323 --- /dev/null +++ b/821/CH4/EX4.36/4_36.sce @@ -0,0 +1,8 @@ +C1=0.04;
+C2=0.02;
+x=(C1^2)/(C2+(2*C1));//The amount of Na+ transported at equilibrium in M//
+printf('The amount of Na+ transported at equilibrium=x=%fM',x);
+NaR=C2+x;//Na+ on RHS//
+printf('\nNa+ on RHS=NaR=%fM',NaR);
+NaL=C1-x;//Na+ on LHS//
+printf('\nNa+ on LHS=NaL=%fM',NaL);
diff --git a/821/CH4/EX4.37/4_37.sce b/821/CH4/EX4.37/4_37.sce new file mode 100755 index 000000000..3cacfab9c --- /dev/null +++ b/821/CH4/EX4.37/4_37.sce @@ -0,0 +1,13 @@ +MW=58.45;//Molar weight of NaCl in grams//
+d=2.17;//density of NaCl in g/cc//
+MV=MW/d;//Molar volume in cc//
+printf('Molar Volume of NaCl=MV=%fcc',MV);
+printf('\nThis must contain 6.023*10^23 NaCl units.\nIn one unit cell of NaCl there are 8 corner Na+ ions and 6 on the face centres.\nThe number of Na+ ions in one unit cell is 8*1/8+6*1/2=4.\nThere are 12Cl- ions on the edges,and one in the centre.\nThe number of Cl- ions are 12*1/4+1=4 ');
+printf('\nThe volume of the unit cell containing 4 NaCl units=179*10^-24cc');
+a=5.63*10^-8;//unit cell length in cm//
+printf('\nUnit cell length of NaCl crystal=a=5.63*10^-8cm');
+Id=a/2;//Interionic distance in cm//
+printf('\nInterionic distance in the crystal=Id=2.815*10^-8cm');
+Q=5.9;//glancing angle in degrees//
+L=2*Id*sin(Q*%pi/180)*10^8;//wavelenth in angstrums//
+printf('\nWavelength of the Xrays used=L=%fAngstrums',L);
diff --git a/821/CH4/EX4.39/4_39.sce b/821/CH4/EX4.39/4_39.sce new file mode 100755 index 000000000..c48971985 --- /dev/null +++ b/821/CH4/EX4.39/4_39.sce @@ -0,0 +1,12 @@ +N=2;//No. of atoms or unit cells in BCC structure//
+L=6.023*10^23;//Avagadro number//
+a=4.291*10^-8;//Unit cell edge length in cm//
+Na=23;//weight of Na//
+p=(N*Na)/(L*a^3);//density of Na at room temperature in g/cm^3//
+printf('Density of Na at room temperature=p=%fg/cm^3',p);
+P=p*1.0398;//density of Na at -195degrees temperature in g/cm^3//
+printf('\nDensity of Na at -195degrees temperature=P=%fg/cm^3',P);
+a1=5.35*10^-8;//unit cell edge length in cm//
+N1=(P*L*a1^3)/(Na);//No. of unitcells at -195degrees//
+printf('\nNumber of unitcells at -195degrees=N1=%f',N1);
+printf('\nAt -195degrees temperature Na have 4 unitcells which means it assumes an FCC structure.');
diff --git a/821/CH4/EX4.4/4_4.sce b/821/CH4/EX4.4/4_4.sce new file mode 100755 index 000000000..5119d8837 --- /dev/null +++ b/821/CH4/EX4.4/4_4.sce @@ -0,0 +1,6 @@ +P=570/760;//pressure of 10L of H2 at 300K in Atmospheres//
+T=300;//Temperature of 10L of H2 in kelvin//
+V=10;//Volume occupied by H2 at 300K in litres//
+R=0.082;//Value of R in litre-atmospheres//
+n=(P*V)/(R*T);//Number of moles of hydrogen present//
+printf('Number of moles of Hydrogen present=n=%fmol',n);
diff --git a/821/CH4/EX4.5/4_5.sce b/821/CH4/EX4.5/4_5.sce new file mode 100755 index 000000000..224dc4db9 --- /dev/null +++ b/821/CH4/EX4.5/4_5.sce @@ -0,0 +1,8 @@ +TP=746;//Total pressure of gas at 298K in mm of Hg//
+PP=24;//Partial pressure of Water vapour at 298K in mm of Hg//
+PPG=TP-PP;//partial pressure of gas at 298K in mm of Hg//
+printf('partial pressure of dry gas at 298k=%fmm of Hg',PPG);
+V=200;//Volume occupied by gas at 298K in millilitres//
+P1=760;
+V1=(PPG*V)/P1;//Volume of dry gas at a pressure of 760mm Hg//
+printf('\nVolume of dry gas at pressure 760mm of Hg=V1=%fml',V1);
diff --git a/821/CH4/EX4.6/4_6.sce b/821/CH4/EX4.6/4_6.sce new file mode 100755 index 000000000..0a48c2694 --- /dev/null +++ b/821/CH4/EX4.6/4_6.sce @@ -0,0 +1,16 @@ +mA=0.11;//amount of gas A in grams//
+MWA=44;//Molecular weight of gas A in grams//
+nA=mA/MWA;//No. of moles of gas A//
+printf('No. of moles of gas A=nA=%fmol',nA);
+mB=0.17;//amount of gas B in grams//
+MWB=34;//Molecular weight of gas B in grams//
+nB=mB/MWB;//No. of moles of gas B//
+printf('\nNo. of moles of gas B=nB=%fmol',nB);
+MFA=nA/(nA+nB);//mole fraction of gas A//
+printf('\nmole fraction of gas A=MFA=%f',MFA);
+MFB=nB/(nA+nB);//mole fraction of gas B//
+printf('\nmole fraction of gas B=MFB=%f',MFB);
+PPA=MFA*759;//Partial pressure of gas A//
+printf('\nPartial pressure of gas A=PPA=%fmm of Hg',PPA);
+PPB=MFB*759;//Partial pressure of gas B//
+printf('\nPartial pressure of gas B=PPB=%fmm of Hg',PPB);
diff --git a/821/CH4/EX4.7/4_7.sce b/821/CH4/EX4.7/4_7.sce new file mode 100755 index 000000000..c6d93deef --- /dev/null +++ b/821/CH4/EX4.7/4_7.sce @@ -0,0 +1,11 @@ +P1=0.3;//pressure of gas A in atm//
+VA=50;//volume of gas A in ml//
+V=500;//Volume of vessel in ml//
+PA=P1*VA/V;//Pressure of gas A in vessel//
+printf('Pressure of gas A in vessel=PA=%fatm',PA);
+P2=0.4;//pressure of gas B in atm//
+VB=250;//volume of gas B in ml//
+PB=P2*VB/V;//Pressure of gas A in vessel//
+printf('\nPressure of gas B in vessel=PB=%fatm',PB);
+P=PA+PB;//Total pressure in vessel in atm//
+printf('\nTotal pressure in vessel=P=%fatm',P);
diff --git a/821/CH4/EX4.8/4_8.sce b/821/CH4/EX4.8/4_8.sce new file mode 100755 index 000000000..20893de3c --- /dev/null +++ b/821/CH4/EX4.8/4_8.sce @@ -0,0 +1,10 @@ +printf('The volume of 1 mole of H2 at 1 atm and 273.15K=22.415Litres');
+printf('\nThe volume at 0.5atm is obtained by considering the constancy of PV');
+P1=1;
+V1=22.415;
+P2=0.5;
+V2=(P1*V1)/P2;//Volume at 0.5atm in litres//
+printf('\nVolume at 0.5atm of H2 at 273.15K=V2=%fLitres',V2);
+MH2=2;//molecular weight of H2//
+DH2=MH2/V2;//density of H2 in gram per litre//
+printf('\nDensity of H2=DH2=%fgram per litre',DH2);
diff --git a/821/CH4/EX4.9/4_9.sce b/821/CH4/EX4.9/4_9.sce new file mode 100755 index 000000000..e17a41d91 --- /dev/null +++ b/821/CH4/EX4.9/4_9.sce @@ -0,0 +1,10 @@ +printf('One mole of H2 Occupies 22.415Litres at 273.15K');
+printf('\nThe volume at 300K is obtained by considering the constancy of V/T');
+T1=273.15;
+V1=22.415;
+T2=300.15;
+V2=(V1/T1)*T2;//Volume at 300K in litres//
+printf('\nVolume of 1mole of H2 at 300.15K=V2=%fLitres',V2);
+MH2=2;//molecular weight of H2//
+DH2=MH2/V2;//density of H2 in gram per litre//
+printf('\nDensity of H2=DH2=%fgram per litre',DH2);
diff --git a/821/CH5/EX5.1/5_1.sce b/821/CH5/EX5.1/5_1.sce new file mode 100755 index 000000000..4cb87840f --- /dev/null +++ b/821/CH5/EX5.1/5_1.sce @@ -0,0 +1,5 @@ +P1=10;//initial pressure in atm//
+P2=1;//final pressure in atm//
+Wrev=0.0821*273*log(P1/P2);//maximum work required for the expansion of gas in litre-atm//
+printf('Maximum work required for the expansion of 1mol of ideal gas from 10atm to 1atm=Wrev=%flitre atm',Wrev);
+printf('\nIf the gas is now compressed back from 1atm to 10atm under isothermal conditions,Wrev=-51.608530litre atm');
diff --git a/821/CH5/EX5.10/5_10.sce b/821/CH5/EX5.10/5_10.sce new file mode 100755 index 000000000..b0e2ae978 --- /dev/null +++ b/821/CH5/EX5.10/5_10.sce @@ -0,0 +1,5 @@ +dHfH2O=-68.32;//dHf value of H2O in kcal per mol//
+dHfCO2=-94.05;//dHf value of CO2 in kcal per mol//
+dH298=-208.34;//enthalphy change at 298K in Kcal//
+dHfCH3COOH=(2*dHfH2O+2*dHfCO2-dH298);//dHf value of CH3COOH in Kcal per mol//
+printf('Standard enthalpy of formation of CH3COOH=%fKcal per mol',dHfCH3COOH);
diff --git a/821/CH5/EX5.11/5_11.sce b/821/CH5/EX5.11/5_11.sce new file mode 100755 index 000000000..f5b7bc29e --- /dev/null +++ b/821/CH5/EX5.11/5_11.sce @@ -0,0 +1,7 @@ +dHfZnSO4l=-36.43;//Standard heat of formation of ZnSO4 in kcal//
+dHfH2SO4l=-193.91;//Standard heat of formation of H2SO4 in kcal//
+dHfZnSO4aq=-19.45;//Heats of Dilution of ZnSO4 in kcal//
+dHfH2SO4aq=-22.99;//Heats of Dilution of H2SO4 in kcal//
+dH=dHfZnSO4l+dHfH2SO4l+dHfH2SO4aq;
+dHf298=dH-dHfZnSO4aq;//Standarad enthalphy of formation of ZnSO4 at 298K in Kcal//
+printf('Standard enthalpy of formation of ZnSO4 at 298K=%fKcal',dHf298);
diff --git a/821/CH5/EX5.12/5_12.sce b/821/CH5/EX5.12/5_12.sce new file mode 100755 index 000000000..bfd0cc0ed --- /dev/null +++ b/821/CH5/EX5.12/5_12.sce @@ -0,0 +1,8 @@ +CvH2=6.95*10^-3;//Mean molar capacity of H2 in Kcal per deg//
+CvO2=6.97*10^-3;//Mean molar capacity of O2 in Kcal per deg//
+CvH2O=8.7*10^-3;//Mean molar capacity of H2O in Kcal per deg//
+dCv=CvH2O-CvH2-0.5*CvO2;//Net mean molar capapcity in Kcal per deg//
+printf('Net mean molar capacity=dCv=%fKcal per deg',dCv);
+dHf=-57.8;//Standard heat of formation of water vapour in Kcal//
+dHf373=dHf+(dCv*(373-298));//Heat of formation of water vapour at 373K in Kcal//
+printf('\nHeat of formation of water vapour at 373K=%fKcal',dHf373);//here the answer is written wrong in textbook the correct answer is given here//
diff --git a/821/CH5/EX5.13/5_13.sce b/821/CH5/EX5.13/5_13.sce new file mode 100755 index 000000000..63d54e998 --- /dev/null +++ b/821/CH5/EX5.13/5_13.sce @@ -0,0 +1,8 @@ +CvCO2=1.331;//Mean molar capacity of CO2 in Kcal per deg//
+CvC=6.714*10^-3;//Mean molar capacity of C in Kcal per deg//
+CvO2=-77.09*10^-7;//Mean molar capacity of O2 in Kcal per deg//
+T1=300;//Initial Temperature in kelvin//
+T2=400;//final temperature in Kelvin//
+dHf400=-94.06;//Enthalpy change in the formation of CO2 at 400k in Kcal//
+dHf300=dHf400-((CvCO2*(T2-T1))-(0.5*CvC*(T2^2-T1^2))-(0.33*CvO2*(37*10^6)))*10^-3;//Enthropy of formation of CO2 at 300K in Kcal//
+printf('\nEnthropy of formation of CO2 at 300K=%fKcal',dHf300);//in textbook the answer is not having negative answer//
diff --git a/821/CH5/EX5.14/5_14.sce b/821/CH5/EX5.14/5_14.sce new file mode 100755 index 000000000..17b9aa0c3 --- /dev/null +++ b/821/CH5/EX5.14/5_14.sce @@ -0,0 +1,6 @@ +dHfNH3=-11.04;//Standard heat of formation of NH3 in kcal//
+dHfH2O=-57.80;//Standard heat of formation of H2O in kcal//
+dHfNO=21.60;//Standard heat of formation of NO in kcal//
+dHfO2=0;//Standard heat of formation of O2 in Kcal//
+dH=4*dHfNO+6*dHfH2O-4*dHfNH3-5*dHfO2;//Enthalpy change for the oxidation of NH3//
+printf('Enthalpy change for the oxidation of NH3 at 298K=%fKcal',dH);
diff --git a/821/CH5/EX5.15/5_15.sce b/821/CH5/EX5.15/5_15.sce new file mode 100755 index 000000000..15723090b --- /dev/null +++ b/821/CH5/EX5.15/5_15.sce @@ -0,0 +1,13 @@ +T=773;//temperature in kelvin//
+dH298=-216240;//Enthalpy change for the oxidation of NH3 in cal//
+CvO2=6.148+T*3.102*10^-3-(T^2)*9.23*10^-7;//heat capacity of O2 in cal deg^-1 mol^-1//
+CvH2O=7.219+T*2.374*10^-3+(T^2)*2.67*10^-7;//Heat capacity of H2O in cal deg^-1 mol^-1//
+CvNO=6.50+T*1*10^-3;//Heat capacity of NO in cal per deg per mol//
+CvNH3=6.189+T*7.887*10^-3-(T^2)*7.28*10^-7;//Heat capacity of H2O in cal deg^-1 mol^-1//
+dCv=4*CvNO+6*CvH2O-4*CvNH3-5*CvO2;//Heat capacity change for oxidation//
+da=13.818;
+db=-28.814*10^-3;
+dc=91.29*10^-7;
+T1=298;//temperature in kelvin//
+dH773=dH298+(T-T1)*da+(T^2-T1^2)*0.5*db+(T^3-T1^3)*0.33*dc;//Enthalpy change for the oxidation of NH3 at 773k in cal//
+printf('\nEnthalpy change for the oxidation of NH3 at 773K=%fcal=-215.69Kcal',dH773);//here the answer is slightly different//
diff --git a/821/CH5/EX5.16/5_16.sce b/821/CH5/EX5.16/5_16.sce new file mode 100755 index 000000000..fd4dd8365 --- /dev/null +++ b/821/CH5/EX5.16/5_16.sce @@ -0,0 +1,6 @@ +ECH=99;//bond energy of CH bond in Kcal//
+EClCl=58;//bond energy of ClCl bond in Kcal//
+ECCl=78;//bond energy of CCl bond in Kcal//
+EHCl=103;//bond energy of HCl bond in Kcal//
+dE=4*ECH+4*EClCl-4*ECCl-4*EHCl;//Net change in bond energy in kcal//
+printf('Enthalpy change of the reaction=dE=%fKcal',dE);
diff --git a/821/CH5/EX5.17/5_17.sce b/821/CH5/EX5.17/5_17.sce new file mode 100755 index 000000000..6631bf20d --- /dev/null +++ b/821/CH5/EX5.17/5_17.sce @@ -0,0 +1,8 @@ +T1=298;//initial temperature in kelvin//
+pO2=0.2;//percentage of oxygen in air//
+pN2=0.8;//percentage of nitrogen in air//
+dH298=-310;//Heat of combustion of acetylene in Kcal//
+printf('The combustion of acetelyne proceeds according to the equation C2H2+2.5 O2 = 2CO2+H2O,\nHeat of combustion of acetylene=dH298=-310KCal.');
+printf('\nThe scheme of calculation can be represented as Reactants at 298 = Products at T2K.\nThe gases present in the flame zone after combustion are CO2,H2O and unreacted N2 of the air.');
+printf('\nSince 2.5mol of O2 have been utilized for combustion,4*2.5mol i.e 10mol of N2 are present in the flame zone.\nThe heat generated at 298K,dH298 is utilized in raising the products at 298K to a temeprature of T2K.');
+printf('\nThe Heat capacity equations given in the question can be solved and after solving we will get a quadratic equation that is T2^2+4511*T2-2.273*10^7=0 \nMaximum temperature reached by an oxy-acetylene flame at constant pressure=T2=3015.0000K.');
diff --git a/821/CH5/EX5.18/5_18.sce b/821/CH5/EX5.18/5_18.sce new file mode 100755 index 000000000..980f897ee --- /dev/null +++ b/821/CH5/EX5.18/5_18.sce @@ -0,0 +1,19 @@ +dH=-57800;//Enthalpy change in cal//
+R=1.99;//universal gas constant//
+T=298;//temperture in kelvin//
+dn=-0.5;//change in no. of moles//
+dU=dH-(R*T*dn);//heat of combustion in cal//
+printf('The air used contains 2 mol of N2.\ndU=-57500cal.');
+CpN2=8.3;//Cp value of N2//
+CpH2O=11.3;//Cp value of H2O//
+CvN2=CpN2-R;//Cv value of N2//
+CvH2O=CpH2O-R;//Cv value of H2O//
+printf('\nCv value of N2=CvN2=%fKcal deg^-1 mol^-1',CvN2);
+printf('\nCv value of H2O=CvH2O=%fKcal deg^-1 mol^-1',CvH2O);
+printf('\nSolving integration we get T2=2940K.The maximum pressure can be calculated from PV=nRT equation.');
+P1=1;//initial pressure in atm//
+n1=3.5;
+n2=3;
+T2=2940;//final temperature in Kelvin//
+P2=(P1*n2*T2)/(n1*T);//maximum pressure in atm//
+printf('\nMaximum pressure of the reaction=P2=%fatm',P2);
diff --git a/821/CH5/EX5.2/5_2.sce b/821/CH5/EX5.2/5_2.sce new file mode 100755 index 000000000..e556c299e --- /dev/null +++ b/821/CH5/EX5.2/5_2.sce @@ -0,0 +1,7 @@ +n=5;//no. of moles of Benzene//
+R=8.314*10^7;//Universal gas constant//
+K=80;//Evaporated temperature of benzene in centigrades//
+T=270+K;//evaporated temperature of benzene in kelvins//
+Wrev=n*R*T;//Maximum work done in ergs//
+printf('Maximum work done=Wrev=%fergs=1.454*10^11ergs',Wrev);
+
diff --git a/821/CH5/EX5.20/5_20.sce b/821/CH5/EX5.20/5_20.sce new file mode 100755 index 000000000..0bf265fee --- /dev/null +++ b/821/CH5/EX5.20/5_20.sce @@ -0,0 +1,11 @@ +printf('The maximum possible efficiency is obtained only in a reversible carnots cycle according to which ME=dT/T.');
+T=300;//initial temperature of steam engine in kelvin//
+T1=373;//operating temperature in kelvin//
+ME1=(T1-T)/T1;//maximum possible efficiency for temperature range of 300k to 373k//
+printf('\nMaximum possible efficiency for operating temperature between 300K and 373K=ME1=%f',ME1);
+T2=630;//operating temperature in kelvin//
+ME2=(T2-T)/T2;//maximum possible efficiency for temperature range of 300k to 630k//
+printf('\nMaximum possible efficiency for operating temperature between 300K and 630K=ME1=%f',ME2);
+T3=510;//operating temperature in kelvin//
+ME3=(T3-T)/T3;//maximum possible efficiency for temperature range of 300k to 510k//
+printf('\nMaximum possible efficiency for operating temperature between 300K and 510K=ME3=%f',ME3);
diff --git a/821/CH5/EX5.22/5_22.sce b/821/CH5/EX5.22/5_22.sce new file mode 100755 index 000000000..85665be65 --- /dev/null +++ b/821/CH5/EX5.22/5_22.sce @@ -0,0 +1,9 @@ +T1=300;//initial temperature in kelvin//
+T2=400;//final temperature in kelvin//
+P1=2;//initial pressure in atm//
+P2=10;//final pressure in atm//
+R=1.99;
+n=5;//no.of moles//
+Cv=6.95;
+dS=n*Cv*2.303*log10(T2/T1)-n*2.303*R*log10(P2/P1);
+printf('Change in entropy=dS=%feu',dS);
diff --git a/821/CH5/EX5.23/5_23.sce b/821/CH5/EX5.23/5_23.sce new file mode 100755 index 000000000..bf75b163e --- /dev/null +++ b/821/CH5/EX5.23/5_23.sce @@ -0,0 +1,4 @@ +Lv=6896;//Latent heat of vapourization in cal per mol//
+Tb=341.7;//Boiling temperature in kelvin//
+dS=Lv/Tb;//entropy change accompanying the vapourization in eu//
+printf('Entropy change accompanying the vapourization=dS=%feu or cal per deg',dS);
diff --git a/821/CH5/EX5.24/5_24.sce b/821/CH5/EX5.24/5_24.sce new file mode 100755 index 000000000..8fc32bffb --- /dev/null +++ b/821/CH5/EX5.24/5_24.sce @@ -0,0 +1,9 @@ +T1=300;//initial temperature in kelvin//
+T2=600;//final temperature in kelvin//
+T3=373;//initial temperature in kelvin//
+T4=746;//final temperature in kelvin//
+Cv=6.09;//molar heat capacity in cal per deg//
+dS2=Cv*2.303*log10(T2/T1);//change in entropy for temperature change between 300k to 600k//
+printf('Change in entropy for temperature change between 300k to 600k=dS2=%fcal per deg',dS2);
+dS4=Cv*2.303*log10(T4/T3);//change in entropy for temperature change between 373k to 746k//
+printf('\nChange in entropy for temperature change between 373k to 746k=dS4=%fcal per deg',dS4);
diff --git a/821/CH5/EX5.25/5_25.sce b/821/CH5/EX5.25/5_25.sce new file mode 100755 index 000000000..9c4cb87c1 --- /dev/null +++ b/821/CH5/EX5.25/5_25.sce @@ -0,0 +1,23 @@ +T=300;//initial temperature in kelvin//
+T1=692;//temperature in kelvin//
+T2=1180;//temperature in kelvin//
+T3=1500;//final temperature in kelvin//
+Lf=1800;//Latent heat of fusion in cal per mol//
+Tf=692;//temperature of fusion in kelvin//
+Lv=27700;//Latent heat of vapourization in cal per mol//
+Tv=1180;//temperature of vapourization in kelvin//
+SH=0.092;//specific heat of Zn in cal per deg per gram//
+w=65.38;//atomic weight of Zn//
+Cv=w*SH;//molar heat capacity in cal per deg//
+dS1=Cv*2.303*log10(T1/T);//change in entropy for temperature change between 300k to 692k//
+printf('Change in entropy for temperature change between 300k to 692k=dS1=%feu',dS1);
+dS2=Lf/Tf;//change in entropy in the process of fusion in eu//
+printf('\nChange in entropy in the process of fusion=dS2=%feu',dS2);
+dS3=Cv*2.303*log10(T2/T1);//change in entropy for temperature change between 692k to 1180k//
+printf('\nChange in entropy for temperature change between 692k to 1180k=dS3=%feu',dS3);
+dS4=Lv/Tv;//change in entropy in the process of vapourization in eu//
+printf('\nChange in entropy in the process of vapourization=dS4=%feu',dS4);
+dS5=Cv*2.303*log10(T3/T2);//change in entropy for temperature change between 1180k to 1500k//
+printf('\nChange in entropy for temperature change between 1180k to 1500k=dS5=%feu',dS5);
+dStotal=dS1+dS2+dS3+dS4+dS5;//total entropy change accompanying the process in eu//
+printf('\nTotal entropy change accompanying the process=dStotal=%feu',dStotal);
diff --git a/821/CH5/EX5.26/5_26.sce b/821/CH5/EX5.26/5_26.sce new file mode 100755 index 000000000..82edab5e9 --- /dev/null +++ b/821/CH5/EX5.26/5_26.sce @@ -0,0 +1,11 @@ +printf('This is a spontaneous process.Since dS is independent of the manner in which the processes are conducted,\nthe above irreversible process can be considered to take place reversibly.')
+T1=300;//initial temperature in kelvin//
+T2=370;//final temperature in kelvin//
+Cv=18;//molar heat capacity in cal per deg//
+dSH2O=-Cv*2.303*log10(T2/T1);//change in entropy of H2O for temperature change between 300k to 370k//
+printf('\nChange in entropy of H2O for temperature change between 300k to 370k=dSH2O=%feu',dSH2O);
+dT=T2-T1;//change in temperature in kelvin//
+dSthermostat=Cv*dT/T1;//change in entropy of thermostat in eu//
+printf('\nChange in entropy of thermostat=dSthermostat=%feu',dSthermostat);
+dS=dSH2O+dSthermostat;//net entropy chane in eu//
+printf('\nNet entropy change=dS=%feu',dS);
diff --git a/821/CH5/EX5.27/5_27.sce b/821/CH5/EX5.27/5_27.sce new file mode 100755 index 000000000..f31cadde1 --- /dev/null +++ b/821/CH5/EX5.27/5_27.sce @@ -0,0 +1,6 @@ +dGC2H6O=-41.77;//Standard free energy change of C2H6O in Kcal per mol//
+dGCO2=-94.26;//Standard free energy change of CO2 in Kcal per mol//
+dGH2O=-56.69;//Standard free energy change of C2H6O in Kcal per mol//
+dGO2=0;//Standard free energy change of O2 in Kcal per mol//
+dG=2*dGCO2+3*dGH2O-dGC2H6O-3*dGO2;//Standard free energy change in Kcal//
+printf('Standard free energy change accompanying the combustion of ethanol at 298K=%fKcal',dG);
diff --git a/821/CH5/EX5.28/5_28.sce b/821/CH5/EX5.28/5_28.sce new file mode 100755 index 000000000..6cb614941 --- /dev/null +++ b/821/CH5/EX5.28/5_28.sce @@ -0,0 +1,7 @@ +dGCH3COOH=-93.8;//Standard free energy change of CH3COOH in Kcal per mol//
+dGCH3CH2OH=-41.77;//Standard free energy change of CH3CH2OH in Kcal per mol//
+dGH2O=-56.69;//Standard free energy change of C2H6O in Kcal per mol//
+dGO2=0;//Standard free energy change of O2 in Kcal per mol//
+dG=dGCH3COOH+dGH2O-dGCH3CH2OH-dGO2;//Standard free energy change in Kcal//
+printf('Standard free energy change accompanying the Oxidation of ethanol to acetic acid=%fKcal',dG);
+printf('\nSince dG is negative the reaction is feasible under these conditions.');
diff --git a/821/CH5/EX5.29/5_29.sce b/821/CH5/EX5.29/5_29.sce new file mode 100755 index 000000000..7f5f79c40 --- /dev/null +++ b/821/CH5/EX5.29/5_29.sce @@ -0,0 +1,7 @@ +dGAgCl=-26.22;//Standard free energy change of AgCl in Kcal per mol//
+dGAgI=-15.85;//Standard free energy change of AgI in Kcal per mol//
+dGKCl=-97.59;//Standard free energy change of KCl in Kcal per mol//
+dGKI=-77.03;//Standard free energy change of KI in Kcal per mol//
+dG=dGAgCl+dGKI-dGAgI-dGKCl;//Standard free energy change in Kcal//
+printf('Standard free energy change accompanying the given reaction=%fKcal',dG);
+printf('\nSince dG is positive the reaction is not feasible under these conditions but the reverse reation is possible.');
diff --git a/821/CH5/EX5.3/5_3.sce b/821/CH5/EX5.3/5_3.sce new file mode 100755 index 000000000..0aca07140 --- /dev/null +++ b/821/CH5/EX5.3/5_3.sce @@ -0,0 +1,11 @@ +w=0.454;//weight of TNT in grams//
+T=298;//temperature in kelvin//
+R=2*10^-3;
+dn=5;
+m=w/227;//mol of TNT exploded//
+printf('mol of TNT exploded=%fmol',m);
+H=1.64;//Heat liberated in kcal//
+dU=-H/m;
+printf('\ndU=%fKcal per mol',dU);
+dH=dU+(R*T*dn);
+printf('\ndH=%fKcal per mol',dH);
diff --git a/821/CH5/EX5.30/5_30.sce b/821/CH5/EX5.30/5_30.sce new file mode 100755 index 000000000..78da120f6 --- /dev/null +++ b/821/CH5/EX5.30/5_30.sce @@ -0,0 +1,8 @@ +SFe2O3=21.5;//Standard Entropy of Fe2O3 in cal per deg per mol//
+SCO=47.3;//Standard Entropy of CO in cal per deg per mol//
+SFe=6.5;//Standard Entropy of Fe in cal per deg per mol//
+SCO2=51.1;//Standard Entropy of CO2 in cal per deg per mol//
+dS=2*SFe+3*SCO2-SFe2O3-3*SCO;//Standard entropy change in cal per deg per mol//
+printf('Standard entropy change accompanying the reduction of Fe2O3 by CO=%fcal per deg per mol',dS);
+printf('\nHere it will be noticed that dS is very small but definitely positive.');
+printf('\nThe small value due to the fact that there is no change in number of moles of gases\nwhich are major contributors to the entropy change.');
diff --git a/821/CH5/EX5.31/5_31.sce b/821/CH5/EX5.31/5_31.sce new file mode 100755 index 000000000..cb65bd11a --- /dev/null +++ b/821/CH5/EX5.31/5_31.sce @@ -0,0 +1,5 @@ +SCaCO3=22.2;//Standard Entropy of CaCO3 in cal per deg per mol//
+SCaO=9.5;//Standard Entropy of CaO in cal per deg per mol//
+SCO2=51.1;//Standard Entropy of CO2 in cal per deg per mol//
+dS=SCaO+SCO2-SCaCO3;//Standard entropy change in cal per deg per mol//
+printf('Standard entropy change in the given reaction=%fcal per deg per mol',dS);
diff --git a/821/CH5/EX5.32/5_32.sce b/821/CH5/EX5.32/5_32.sce new file mode 100755 index 000000000..c2360db0f --- /dev/null +++ b/821/CH5/EX5.32/5_32.sce @@ -0,0 +1,6 @@ +SFe2O3=21.5;//Standard Entropy of Fe2O3 in cal per deg per mol//
+SH2=31.21;//Standard Entropy of H2 in cal per deg per mol//
+SFe=6.5;//Standard Entropy of Fe in cal per deg per mol//
+SH2O=16.75;//Standard Entropy of H2O in cal per deg per mol//
+dS=2*SFe+3*SH2O-SFe2O3-3*SH2;//Standard entropy change in cal per deg per mol//
+printf('Standard entropy change in the given reaction=%fcal per deg per mol',dS);
diff --git a/821/CH5/EX5.33/5_33.sce b/821/CH5/EX5.33/5_33.sce new file mode 100755 index 000000000..0c12f4977 --- /dev/null +++ b/821/CH5/EX5.33/5_33.sce @@ -0,0 +1,12 @@ +T=273;//temperature in kelvin//
+P1=1;//pressure in atm//
+M=28;//molecular weight of N2 in grams//
+printf('For an ideal gas density=M*P/R*T.\nWhere M is molecular weight=28.\nTemperature being constant,if p1 and p2 are the densities at 1atm and 50atm respectively');
+P2=1.25;//pressure in atm//
+p1=50;//density at 1atm pressure//
+p2=p1*P2;
+printf('\nIf N2 gas behaved ideally at 273k,the ideal pressure Pideal is given by p2*R*T/M');
+Pideal=p2*0.0821*T/M;
+printf('\nIdeal pressure=Pideal=%fatm',Pideal);
+f=p1^2/Pideal;
+printf('\nFugacity of the N2gas=f=%fatm',f);
diff --git a/821/CH5/EX5.34/5_34.sce b/821/CH5/EX5.34/5_34.sce new file mode 100755 index 000000000..b36e1d95e --- /dev/null +++ b/821/CH5/EX5.34/5_34.sce @@ -0,0 +1,12 @@ +M=18;//molecular weight of water in grams//
+Dl=0.958;//density of water in liquid form in gram per cm^3//
+Dv=5.98*10^-4;//density of water in vapour form in gram per cm^3//
+MVl=M/Dl;//molar volume of water in liquid form in cm^3//
+MVv=M/Dv;//molar volume of water in vapour form in cm^3//
+L=540;//Latent heat of vapourization in cal per gram//
+T=373;//temperature in kelvin//
+dV=30.11;
+dP=(L*M)/(T*dV);
+printf('dP/dT=0.357atm per deg=28.55mm of Hg per deg');
+dS=dP*dV;//change in entropy in cal per deg per mol//
+printf('\nChange in entropy=dS=%fcal per deg per mol',dS);
diff --git a/821/CH5/EX5.35/5_35.sce b/821/CH5/EX5.35/5_35.sce new file mode 100755 index 000000000..01a5f321c --- /dev/null +++ b/821/CH5/EX5.35/5_35.sce @@ -0,0 +1,10 @@ +dS=21.0;//standard entropy change of benzene in cal per deg per mol//
+Tb=353;//boiling point of benzene in kelvin//
+Lv=dS*Tb;//latent heat of vapourization of benzene in cal per mol//
+printf('Latent heat of vapourization of benzene=Lv=%fcal per mol',Lv);
+T=300;//temperature in kelvin//
+R=1.99;//universal gas constant//
+P=10^(Lv*((1/Tb)-(1/T))/(2.303*R));//vapour pressure of benzene in atm//
+PHg=P*760;//vapour pressure in mm of Hg//
+printf('\nvapour pressure of benzene=P=%fatm',P);
+printf('\nVapour pressure of benzene in mm of Hg=PHg=%fmm of Hg',PHg);
diff --git a/821/CH5/EX5.36/5_36.sce b/821/CH5/EX5.36/5_36.sce new file mode 100755 index 000000000..def390ed1 --- /dev/null +++ b/821/CH5/EX5.36/5_36.sce @@ -0,0 +1,8 @@ +T2=373;//final temperature in kelvin//
+Lv=540*18;//latent heat of vapourization of water in cal per mol//
+P1=1/20;//initial pressure in atm//
+P2=1;//final pressure in atm//
+R=1.99;//universal gas constant//
+T1=1/((1/T2)+(2.303*R*log10(P2/P1)/Lv));
+printf('\noperating temperature of the reactor=T1=%fK',T1);
+printf('\nHence the plant can be operated at a temperature of 303.500Kelvin or 30.500degrees \nsince at a temperature higher than this the liquid phase no longer exists.');
diff --git a/821/CH5/EX5.37/5_37.sce b/821/CH5/EX5.37/5_37.sce new file mode 100755 index 000000000..4b4e2906a --- /dev/null +++ b/821/CH5/EX5.37/5_37.sce @@ -0,0 +1,6 @@ +T1=373;//initial temperature in kelvin//
+Lv=540*18;//latent heat of vapourization of water in cal per mol//
+T2=423;//final temperature in kelvin//
+R=1.99;//universal gas constant//
+P=10^(-Lv*((1/T2)-(1/T1))/(2.303*R));//vapour pressure of water in atm//
+printf('\npressure of water at which we can produce superheated steam=P=%fatm',P);
diff --git a/821/CH5/EX5.38/5_38.sce b/821/CH5/EX5.38/5_38.sce new file mode 100755 index 000000000..0711ed77b --- /dev/null +++ b/821/CH5/EX5.38/5_38.sce @@ -0,0 +1,21 @@ +PPSO3=1;//partial pressure of SO3 in atm//
+PPSO2=0.2;//partial pressure of SO2 in atm//
+PPO2=0.05;//partial pressure of O2 in atm//
+Kp=3.5;
+R=0.0821;//universal gas constant//
+T=1000;//temperture in kelvin//
+n1=3;
+n2=2;
+dn=n2-n1;//change in no. of moles//
+Kc=Kp/((R*T)^dn);
+printf('Kc for the reaction=Kc=%flitre per mol',Kc);
+P=2;//pressure in atm//
+Kinfinite=Kp/(P^dn);
+printf('\nKinfinite of the reaction=Kinfinite=%fper atm',Kinfinite);
+Qp=(PPSO3^2)/(PPSO2^2*PPO2);//reation quotient involving pressures//
+printf('\nReaction quotient invloving pressures=Qp=%f',Qp);
+dGO=-2.303*1.99*10^-3*T*log10(Kp);
+printf('\ndGO=%f',dGO);
+dG=dGO+(2.303*1.99*10^-3*T*log10(Qp));
+printf('\nstandard free energy change in the reaction at 1000k=dG=%fKcal',dG);
+printf('\nIt must be noted that under these conditions dG is positive.\nso it is the dissociation of SO3 that is spontaneous.');//here the answer given in the textbook is wrong the right one is that we got here through execution//
diff --git a/821/CH5/EX5.39/5_39.sce b/821/CH5/EX5.39/5_39.sce new file mode 100755 index 000000000..6d103fa10 --- /dev/null +++ b/821/CH5/EX5.39/5_39.sce @@ -0,0 +1,5 @@ +Kp=1.60;//Kp for water at equilibrium//
+Kc=1.60;//at equilibrium Kp=Kc//
+x=0.56;
+printf('Kp=Kc=x^2/(1-x)^2=1.60\nUpon solving above equation we get x=0.56\nTotal no. of moles=1-x+1-x+x+x=2\nmole percent of CO=mole percent of H2O=0.56*50=28');
+printf('\nmole percent of CO2=mole percent of H2=0.44*50=22\nSo 100 volumes of the mixture will contain 28volumes of CO,28volumes of H2O\n22volumes of CO2 and 22volumes of H2.');
diff --git a/821/CH5/EX5.4/5_4.sce b/821/CH5/EX5.4/5_4.sce new file mode 100755 index 000000000..3b706f0c6 --- /dev/null +++ b/821/CH5/EX5.4/5_4.sce @@ -0,0 +1,8 @@ +dH=-817;
+dU=-820;
+P1=1;//initial pressure in atm//
+V=1;//volume in litre//
+K=(1.99*10^-3)/0.082;//multiplying factor to convert litre atm to calories//
+P2=P1+((dH-dU)/(V*K));//final pressure in atm//
+printf('Final pressure of the system=P2=%fatm',P2);
+printf('\nIn this problem P1 and P2 are in atm and V being in litres,The VdP term in litre atm.\ndH and dU are in Kcal and so the VdP term is converted into Kcal.\nIt is seen that the pressure developed is enormous.\nThis takes place in a confined space,an explosion occurs.\nThe final pressure P2 can also be calculated from the ideal gas equation PV=nRT which gives the same result.');//in textbook the answer given is bit wrong and it should be one we get through execution//
diff --git a/821/CH5/EX5.40/5_40.sce b/821/CH5/EX5.40/5_40.sce new file mode 100755 index 000000000..5b024951e --- /dev/null +++ b/821/CH5/EX5.40/5_40.sce @@ -0,0 +1,8 @@ +T1=1000;//initial temperature in kelvin//
+Kp1=0.72;//Kp value at T1 temperature//
+T2=1260;//final temperature in kelvin//
+Kp2=1.60;//Kp value at T2 temperature//
+R=1.99;//universal gas constant//
+printf('Since Kp2>Kp1 one would expect dH to be positive i.e Endothermic reaction.');
+dH=(2.303*R*T1*T2*log10(Kp2/Kp1))/(T2-T1);
+printf('\ndH for the reaction=dH=%fcal=7.702Kcal',dH);
diff --git a/821/CH5/EX5.41/5_41.sce b/821/CH5/EX5.41/5_41.sce new file mode 100755 index 000000000..86115a786 --- /dev/null +++ b/821/CH5/EX5.41/5_41.sce @@ -0,0 +1,10 @@ +P=10;//pressure in atm//
+printf('The total pressure P=PH2+PN2+PNH3.At all times PH2=3PN2\nSo PNH3=P-4*PN2 or PN2=0.25*(P-PNH3).');
+printf('\nLet x represent the mole fraction of NH3 at equilibrium.\nThen 1-x represents the sum of the mole fraction of N2 and H2.');
+x=0.0122;//yield of NH3 in moles//
+PNH3=x*P;//pressure of NH3 in atm//
+xNH3=x;
+PN2=0.25*(1-x)*P;//pressure of N2 in atm//
+PH2=0.75*(1-x)*P;//pressure of H2 in atm//
+Kp=(PNH3^2)/(PN2*PH2^3);
+printf('\nKp value for the reaction=Kp=%f=1.5*10^-5atm^-2',Kp);
diff --git a/821/CH5/EX5.42/5_42.sce b/821/CH5/EX5.42/5_42.sce new file mode 100755 index 000000000..c895266a7 --- /dev/null +++ b/821/CH5/EX5.42/5_42.sce @@ -0,0 +1,5 @@ +printf('If the yield of NH3 is 12.2mol percent,x=0.122');
+x=0.122;//yield of NH3 in moles//
+Kp=1.48*10^-5;//Kp value of the equation in atm^-2//
+P=sqrt((256*x^2)/(27*Kp*(1-x)^4));//pressure of the system in atm//
+printf('\nPressure of the system=P=%fatm',P);
diff --git a/821/CH5/EX5.44/5_44.sce b/821/CH5/EX5.44/5_44.sce new file mode 100755 index 000000000..cf6d02e6f --- /dev/null +++ b/821/CH5/EX5.44/5_44.sce @@ -0,0 +1,17 @@ +printf('PCl5=PCl3+Cl2');
+M0=208.5;
+P=1;//pressure in atm//
+w=4.59;//weight of PCl5 in grams//
+V=1.7;//volume of PCl5 in gm^3//
+T=523;//temperature in kelvin//
+R=0.0821;//universal gas constant//
+M=(w*R*T)/(P*V);
+printf('\nMolecular weight of PCl5=M=%f',M);
+a=(M0-M)/(M*(2-1));
+printf('\na=%f',a);
+Kp=(a^2*P)/(1-a^2);
+printf('\nKp for the reaction=Kp=%f',Kp);
+printf('\nIf total pressure is increased at the same temperature,Kp being constant,a should decrease.\nLet the degree of dissociation when P=2atm be a1 at the same temperature.');
+P1=2;
+a1=sqrt(Kp/(P1+Kp));
+printf('\na1=%f',a1);//the answers in textbook are slightly different from what we got here but it's nothing wrong//
diff --git a/821/CH5/EX5.45/5_45.sce b/821/CH5/EX5.45/5_45.sce new file mode 100755 index 000000000..32eac9873 --- /dev/null +++ b/821/CH5/EX5.45/5_45.sce @@ -0,0 +1,3 @@ +a=0.5;//dissociation constant//
+Kp=(a^2*P)/(1-a^2);
+printf('Total pressure required to bring 50percent dissociation=P=3*Kp');
diff --git a/821/CH5/EX5.46/5_46.sce b/821/CH5/EX5.46/5_46.sce new file mode 100755 index 000000000..20dd8bf7a --- /dev/null +++ b/821/CH5/EX5.46/5_46.sce @@ -0,0 +1,9 @@ +a=0.21;//dissociation constant//
+P=1;//pressure in atm//
+Kp=(a^2*P)/(1-a^2);
+printf('\nKp for the reaction=Kp=%f',Kp);
+printf('\nP=1atm=0.4+PCOCl2+PCO+PCl2\nsum of pressure reactants=0.6atm\nCOCl2=CO+Cl2\nHere P is the total pressure due to all the participants of the reaction and is equal to 0.6atm.');
+P1=0.6;
+a1=sqrt(Kp/(P1+Kp));
+printf('\na1=%f',a1);
+printf('\nfor Pressure of 0.6atm a has increased,thereby indicating that the forward reaction is favoured.');//here the value of a1 is slightly different from textbook//
diff --git a/821/CH5/EX5.47/5_47.sce b/821/CH5/EX5.47/5_47.sce new file mode 100755 index 000000000..1c714ec61 --- /dev/null +++ b/821/CH5/EX5.47/5_47.sce @@ -0,0 +1,18 @@ +acid=1/3;
+ester=2/3;
+alcohol=1/3;
+water=2/3;
+K=(ester*water)/(acid*alcohol);
+printf('K value of the reaction=K=%f',K);
+printf('\nIn finding no. of moles we end up with quadratic equation 3*x^2-24*x+20=0.\nupon solving the equation we get x=7.05 and 0.945.');
+printf('\nThe first solution is not admissible since the maximum yield of the ester cannot exceed one mol of acetic acid.\nHence x=0.945 i.e yield of the ester is 94.5percent.');
+printf('\nThis problem illustrates the influence of an increased concentration of the reactant\nSince using one mole of each reactant the yield of ester is only 66.66percent');
+acid=1-x;
+ester=x;
+alcohol=1-x;
+water=1+x;
+printf('\nIn finding no. of moles we end up with quadratic equation 3*x^2-9*x+4=0.\nupon solving the equation we get x=2.457 and 0.5425.');
+printf('\nThe first solution is not admissible,x=0.5425.\nyield of the ester in the presence of the product water has decreased from 66.67percent to 54.25percent.');
+printf('\nother homogeneous equilibria in the liquid phase such as ionization of weak acids\nionization of weak bases,hydrolysis of salts,etc.,can be treated likewise.');
+
+
diff --git a/821/CH5/EX5.48/5_48.sce b/821/CH5/EX5.48/5_48.sce new file mode 100755 index 000000000..80e21f4cf --- /dev/null +++ b/821/CH5/EX5.48/5_48.sce @@ -0,0 +1,8 @@ +Kp=29.64;//dissociation pressure of CaCO3 in mm of Hg//
+printf('CaCO3+C=CaO+2CO for the dissociation of CaCO3=CaO+CO2');
+printf('\nKp1=PCO2=29.64/760=0.039atm\nFor the reduction of CO2 by C,CO2+C=2CO.Kp2=PCO^2/PCO2\nVolume percent=mol percent=mole fraction*100\nPCO=xCO*total pressure=xCO since total pressure=1atm\nxCO=PCO=0.724atm and xCO2=PCO2=0.276atm');
+printf('\nKp2=PCO^2/PCO2=(0.724)^2/0.276=1.9atm');
+Kp1=0.039;//dissociation pressure of CaCO3 in atm//
+Kp2=1.9;
+Kp3=Kp1*Kp2;//equilibrium constant for overall reaction in atm^2//
+printf('\nEquilibrium constant for overall reaction=Kp3=%f=7.41*10^-2atm^2',Kp3);
diff --git a/821/CH5/EX5.49/5_49.sce b/821/CH5/EX5.49/5_49.sce new file mode 100755 index 000000000..441db138b --- /dev/null +++ b/821/CH5/EX5.49/5_49.sce @@ -0,0 +1,7 @@ +MH2O=18;//Molecular weight of H2O in grams//
+WH2O=100;//weight of H2O in grams//
+W=3.6;//weight of oraganic substance in grams//
+dP=0.0855;//Lowering in vapour pressure in mm of Hg//
+P=23.76;//Vapour pressure of Organic substance in mm of Hg//
+M=(W*MH2O*P)/(WH2O*dP);//Molecular weight of Organic substance in grams//
+printf('Molecular weight of Organic substance=M=%fgrams',M);
diff --git a/821/CH5/EX5.5/5_5.sce b/821/CH5/EX5.5/5_5.sce new file mode 100755 index 000000000..2ba08e477 --- /dev/null +++ b/821/CH5/EX5.5/5_5.sce @@ -0,0 +1,5 @@ +dHfFe2O3=-196.5;//dHf value of Fe2O3 in kcal per mol//
+dHfCO2=-94.05;//dHf value of CO2 in kcal per mol//
+dHfCO=-26.41;//dHf value of CO in kcal per mol//
+dH298=(3*dHfCO2-dHfFe2O3-3*dHfCO);//enthalphy change at 298K in Kcal//
+printf('Enthalpy change for the reduction of Fe2O3 by Co at 298K=%fKcal',dH298);
diff --git a/821/CH5/EX5.50/5_50.sce b/821/CH5/EX5.50/5_50.sce new file mode 100755 index 000000000..2aafc7e3e --- /dev/null +++ b/821/CH5/EX5.50/5_50.sce @@ -0,0 +1,11 @@ +T0=373;//temperature in kelvin//
+R=1.99;//value of R in cal per deg per mol//
+lv=540;//Latent heat of vaporization in cal per grams//
+Kb=(R*T0^2)/(1000*lv);//value of Kb in deg per mol//
+printf('Value of Kb=%fKbdeg per mol',Kb);
+T1=373.57;//temperature in kelvin//
+W2=5;//weight of urea in grams//
+W1=75;//weight of water boils in grams//
+dT=T1-T0;//change in temperature//
+M2=(Kb*1000*W2)/(dT*W1);//Molecular weight of urea in grams//
+printf('\nMolecular weight of urea=M2=%fgrams',M2);
diff --git a/821/CH5/EX5.51/5_51.sce b/821/CH5/EX5.51/5_51.sce new file mode 100755 index 000000000..5225eaf91 --- /dev/null +++ b/821/CH5/EX5.51/5_51.sce @@ -0,0 +1,8 @@ +lf=80;//Latent heat of fusion of Ice in cal per gram//
+lv=540;//Latent heat of evaporation of water in cal per gram//
+Tf=273;//fusion temperature in kelvin//
+Tb=373;//Boiling temperature in kelvin//
+dTb=Tb-Tf;//change in temperature//
+dTf=(dTb*Tf^2*lv)/(Tb^2*lf*1000);
+FP=0-dTf;//Freezing point of the solution//
+printf('The Freezing point of the solution=FP=%f',FP);
diff --git a/821/CH5/EX5.52/5_52.sce b/821/CH5/EX5.52/5_52.sce new file mode 100755 index 000000000..2a3c88c95 --- /dev/null +++ b/821/CH5/EX5.52/5_52.sce @@ -0,0 +1,10 @@ +Kf=1.86;//Kf for water//
+M2=32;//Molecular weight of Methanol in grams//
+W1=10000;//weight of H2O in grams//
+dTf=10;//lowering freezing point of water//
+W2=(dTf*W1*M2)/(1000*Kf);//weight of methanol in grams//
+printf('weight of Methanol=W2=%fgrams=1.72Kg',W2);
+M3=62;//Molecular weight of Ethylene in grams//
+W3=(dTf*W1*M3)/(1000*Kf);//weight of methanol in grams//
+printf('\nweight of Ethylene=W3=%fgrams=3.334Kg',W3);
+
diff --git a/821/CH5/EX5.53/5_53.sce b/821/CH5/EX5.53/5_53.sce new file mode 100755 index 000000000..2e253024b --- /dev/null +++ b/821/CH5/EX5.53/5_53.sce @@ -0,0 +1,12 @@ +Kf=1.86;//Kf for water//
+printf('At -18.6degrees temperature only Ice seperates from the Aqueous solution\nThe weight of Glycol remains constant through its concentration Actually increases since ice seperates out on cooling.');
+printf('\nInitial Concentration=3334grams per 10Kg of Water.');
+dT=18.6;
+m=dT/Kf;
+w=10;//initial content of water in litres//
+w1=5.376;
+W=w-w1;//weight of ice seperating in kilograms
+printf('\nIf dT=18.6 then m=10');
+printf('\nAt -18.6Degrees 10 mol of glycol are present in 1000grams of water,i.e 620grams in 1000grams of water,or 6200grams in 10Kg of water.');
+printf('\n3334grams of glycol would be present in 10*3334/6200,i.e 5.376Kg of Water.');
+printf('\nWt. of Ice seperating=W=%fKg',W);
diff --git a/821/CH5/EX5.54/5_54.sce b/821/CH5/EX5.54/5_54.sce new file mode 100755 index 000000000..a9db29afa --- /dev/null +++ b/821/CH5/EX5.54/5_54.sce @@ -0,0 +1,20 @@ +T=300;//temperature in kelvin//
+R=0.0821;//universal gas constant//
+Ws=0.171;//Weight of Sucrose in the solution in grams//
+Ms=342;//Molecular weight of Sucrose in grams//
+l=0.05;//volume of solution in litres//
+ns=(Ws)/(Ms*l);//no. of moles of sucrose in the solution//
+printf('Osmatic pressure being a colligative property depends only on the no. of mol of solute present and not on their nature.');
+printf('\nNo. of moles of Sucrose in a litre solution=ns=%fmol per litre',ns);
+Wg=0.18;//Weight of glucose in the solution in grams//
+Mg=180;//Molecular weight of glucose in grams//
+ng=(Wg)/(Mg*l);//no. of moles of glucose in the solution//
+printf('\nNo. of moles of Glucose in a litre solution=ns=%fmol per litre',ng);
+Wu=0.06;//Weight of Urea in the solution in grams//
+Mu=60;//Molecular weight of Urea in grams//
+nu=(Wu)/(Mu*l);//no. of moles of Urea in the solution//
+printf('\nNo. of moles of Urea in a litre solution=ns=%fmol per litre',nu);
+c=ns+ng+nu;//total no of moles in a litre solution//
+printf('\nTotal no. of moles in a litre solution=c=%fmol per litre',c);
+OP=c*R*T;//Osmatic pressure in atms//
+printf('\nOsmatic pressure of the solution at 300k=OP=%fatm',OP);
diff --git a/821/CH5/EX5.55/5_55.sce b/821/CH5/EX5.55/5_55.sce new file mode 100755 index 000000000..7ce9567ac --- /dev/null +++ b/821/CH5/EX5.55/5_55.sce @@ -0,0 +1,11 @@ +Hb=2.4;//rise of the benzene solution in mm//
+Db=0.88;//Density of Benzene solution in g per cm^3//
+Dm=13.6;//Density of mercury solution in g per cm^3//
+Hm=(Hb*Db)/Dm;//Rise of mercury solution in mm//
+printf('The Osmatic pressure given in terms of the height of a liquid column must be converted into an equivalent height of a mercury column.');
+printf('\nEquivalent Height of the mercury column=Hm=%fmm',Hm);
+printf('\nThe density of solution is equal to density of solvent since the solution is dilute.\nLet M2 be the molecular weight of polymer.');
+R=0.0821;//Universal gas constant//
+T=310;//temperature in kelvin//
+M2=(2.5*R*T*760)/Hm;//Molecular weight of polymer in grams//
+printf('\nMolecular weight of polymer=M2=%f=3.11*10^5grams',M2);
diff --git a/821/CH5/EX5.56/5_56.sce b/821/CH5/EX5.56/5_56.sce new file mode 100755 index 000000000..8422bba0c --- /dev/null +++ b/821/CH5/EX5.56/5_56.sce @@ -0,0 +1,7 @@ +T=310;//temperature in kelvin//
+R=0.0821;//universal gas constant//
+OP=7.65;//Osmatic pressure in atm//
+c=OP/(R*T);
+printf('Concentration of glucose=c=%fM',c);
+W=c*180;//weight in one litre in grams//
+printf('\nWeight in one litre=W=%fgrams',W);
diff --git a/821/CH5/EX5.57/5_57.sce b/821/CH5/EX5.57/5_57.sce new file mode 100755 index 000000000..e130c96f7 --- /dev/null +++ b/821/CH5/EX5.57/5_57.sce @@ -0,0 +1,14 @@ +T0=373;//temperature in kelvin//
+Kb=0.52;//value of Kb in deg per mol//
+T1=373.208;//temperature in kelvin//
+W2=3.40;//weight of BaCl2 in grams//
+W1=100;//weight of water boils in grams//
+dTb=T1-T0;//change in temperature//
+Mobs=(Kb*1000*W2)/(dTb*W1);//Molecular weight of BaCl2 observed in grams//
+printf('Molecular weight of BaCl2 observed=Mobs=%fgrams',Mobs);
+Mthr=208.4;//Theoritical Molecular weight of BaCl2 in grams//
+i=Mthr/Mobs;
+a=0.5*(i-1);//apparent degree of dissociation of BaCl2//
+printf('\nApparent degree of dissociation of BaCl2=a=%f',a);
+
+
diff --git a/821/CH5/EX5.58/5_58.sce b/821/CH5/EX5.58/5_58.sce new file mode 100755 index 000000000..fa19dad82 --- /dev/null +++ b/821/CH5/EX5.58/5_58.sce @@ -0,0 +1,14 @@ +Kf=1.86;//Kf for water//
+m=0.1;//no. of moles of organic mono carboxilic acid//
+dTf=Kf*m;//Theoritical change in temperature//
+printf('Theoritical change in temperature=dTf=%f',dTf);
+dTobs=0.220;//Observed change in temperature//
+i=dTobs/dTf;
+printf('\nSince dTobs is greater than dTf ionization must have occurred in aqueous solution.\nAn organic monobasic acid RCOOH ionizes as RCOOH = RCOO- + H+.');
+a=i-1;//degree of ionization//
+printf('\nDegree of ionization=a=%f',a);
+printf('\nIf the acid dissolved as such in its molecular form as species dTthr=5.12*0.1=0.512,dTobs=0.265.');
+printf('\nThis value is nearly half the expected value,suggesting that the molecule exists as associated molecules in solution.');
+i1=0.265/0.512;
+a1=2*(1-i1);
+printf('\nThe acid is demerized to the extent of a1=96.500000percent.')
diff --git a/821/CH5/EX5.6/5_6.sce b/821/CH5/EX5.6/5_6.sce new file mode 100755 index 000000000..b0e2ae978 --- /dev/null +++ b/821/CH5/EX5.6/5_6.sce @@ -0,0 +1,5 @@ +dHfH2O=-68.32;//dHf value of H2O in kcal per mol//
+dHfCO2=-94.05;//dHf value of CO2 in kcal per mol//
+dH298=-208.34;//enthalphy change at 298K in Kcal//
+dHfCH3COOH=(2*dHfH2O+2*dHfCO2-dH298);//dHf value of CH3COOH in Kcal per mol//
+printf('Standard enthalpy of formation of CH3COOH=%fKcal per mol',dHfCH3COOH);
diff --git a/821/CH5/EX5.7/5_7.sce b/821/CH5/EX5.7/5_7.sce new file mode 100755 index 000000000..ce9312cea --- /dev/null +++ b/821/CH5/EX5.7/5_7.sce @@ -0,0 +1,10 @@ +T=298;//temperature in kelvin//
+R=1.987*10^-3;
+dn=-4;
+dU=-1148.93;//Internal energy change of n-Heptane in kcal per mol//
+dHf=dU+(R*T*dn);//dHf value of C7H16 in Kcal per mol//
+printf('dHf value of C7H16=dHf=%fKcal per mol',dHf);
+dHfH2O=-68.32;//dHf value of Fe2O3 in kcal per mol//
+dHfCO2=-94.05;//dHf value of CO2 in kcal per mol//
+dHfC7H16=(8*dHfH2O+7*dHfCO2-dHf);//dHf value of CH3COOH in Kcal per mol//
+printf('\nStandard enthalpy of formation of C7H16=%fKcal per mol',dHfC7H16);
diff --git a/821/CH5/EX5.8/5_8.sce b/821/CH5/EX5.8/5_8.sce new file mode 100755 index 000000000..8c11ed573 --- /dev/null +++ b/821/CH5/EX5.8/5_8.sce @@ -0,0 +1,4 @@ +dHf=-67.63;//enthalpy change value in kcal//
+dHfCO2=-94.05;//Heat of formation value of CO2 in kcal//
+dHfCO=dHfCO2-dHf;//Heat of formation value of CO in Kcal//
+printf('Heat of formation of CO=%fKcal',dHfCO);
diff --git a/821/CH5/EX5.9/5_9.sce b/821/CH5/EX5.9/5_9.sce new file mode 100755 index 000000000..d40cacb02 --- /dev/null +++ b/821/CH5/EX5.9/5_9.sce @@ -0,0 +1,4 @@ +dHfDiam=-94.50;//heat of formation value of Diamond in kcal//
+dHfGrap=-94.05;//Heat of formation value of Graphite in kcal//
+dHf=dHfGrap-dHfDiam;//Enthalpy change when graphite converted to diamond in Kcal//
+printf('Enthalpy change when graphite converted to diamond=%fKcal',dHf);
diff --git a/821/CH6/EX6.1/6_1.sce b/821/CH6/EX6.1/6_1.sce new file mode 100755 index 000000000..a256603fe --- /dev/null +++ b/821/CH6/EX6.1/6_1.sce @@ -0,0 +1,12 @@ +MW=249.6;//molecular weight of CuSO4.5H2O in grams//
+w=0.3120;//weight of CuSO4.5H2O in grams//
+V=0.25;//volume of the solution in litres//
+printf('From Equation (a) 2 mol of CuSO4.5H2O liberates 1 mol of I2,i.e. 2 equivalents.\nHence the equlivalent weight of CuSO4.5H2O=mol.wt/1.');
+printf('\nFrom equation (b) the equivalent weight of CuSO4.5H2O is mol.wt/2 since 1mol of CuSO4.5H2O reacts with 2 mol of OH-,i.e 2 equivalents.');
+W=w/V;//weight of CuSO4.5H2O in one litre solution in grams//
+printf('\nWeight of CuSO4.5H2O in a litre of the solution=W=%fgrams.',W);
+Na=W/MW;//Normality of the solution for (a)//
+printf('\nNormality of the solution for (a)=Na=%f',Na);
+Nb=W*2/MW;//Normality of the solution for (b)//
+printf('\nNormality of the solution for (b)=Nb=%f',Nb);
+printf('\nIn the first case 1ml of the solution contains 5*10^-3equivalents or 5 equivalents of CuSO4.5H2O,\nand in the second case 1ml of the solution will contain 10m eq of CuSO4.5H2O.');
diff --git a/821/CH6/EX6.2/6_2.sce b/821/CH6/EX6.2/6_2.sce new file mode 100755 index 000000000..834793843 --- /dev/null +++ b/821/CH6/EX6.2/6_2.sce @@ -0,0 +1,23 @@ +v=180;//volume of conc. H2SO4 in ml//
+n=6.61;//Normality of the solution//
+N=1000*n/v;
+printf('The Noramality or Strength of the Conc. acid=N=%fN',N);
+printf('\n1 eq.per litre=0.5mol per litre in the case of H2SO4 since the eq.wt=0.5 the mol.wt.');
+printf('\n 6.6N soln=6.61 eq per litre=3.305mol per litre.\n Strength of the diluted solution=3.305M');
+SG=1.84;//super gravity of Conc. H2SO4//
+w=SG*v;//weight of 180ml of conc. H2SO4 in grams//
+printf('\nWt of 180ml of conc.H2SO4=w=%fgrams.',w);
+printf('\nThis actually contains 6.61*49grams of H2SO4.\n percentage of H2SO4 by weight=97.8');
+sg=1.198;//specific gravity of the diluted solution//
+V=1000;//volume of the diluted solution in ml//
+W=sg*V;//weight of one litre of the diluted solution in grams//
+printf('\nWt of 1 litre of the diluted solution=W=%fgrams ',W);
+WH2O=w+W;//weight of water in grams//
+printf('\ntherefore Weight of water=WH2O=%fgrams.',WH2O);
+printf('\nIf the percent of H2SO4 by wt in the diluted solution is y.\nWt of H2SO4 in 1litre of the diluted solution=49*6.61grams.so y value comes as 27.04percent');
+M=3.305*1000/WH2O;//molality of the solution//
+printf('\nMolality of the solution=M=%f',M);
+mf=0.064;//mole fraction of H2SO4//
+mfH2O=1-mf;//mole fraction of water//
+printf('\nMol of sulphuric acid is 329.9/98=3.305.\nMol of water=874.1/18=48.561.\nMol fraction of H2SO4=0.064.');
+printf('\nMole fraction of water=mfH2O=%f',mfH2O);
diff --git a/821/CH6/EX6.3/6_3.sce b/821/CH6/EX6.3/6_3.sce new file mode 100755 index 000000000..80429d81d --- /dev/null +++ b/821/CH6/EX6.3/6_3.sce @@ -0,0 +1,6 @@ +N2=79.2;//percentage of Nitrogen in air//
+O2=20.8;//percentage of Oxygen in air//
+b=76.93;//Weight percent of N2 in air//
+printf('Weight percent of N2 in air=b=%f',b);
+a=100-b;//weight percent of O2 in air//
+printf('\nWeight percent of O2 in air=a=%f',a);
diff --git a/821/CH6/EX6.4/6_4.sce b/821/CH6/EX6.4/6_4.sce new file mode 100755 index 000000000..ad601c15a --- /dev/null +++ b/821/CH6/EX6.4/6_4.sce @@ -0,0 +1,14 @@ +N2=0.79;//partial pressure of Nitrogen in air//
+O2=0.21;//partial pressure of Oxygen in air//
+AN2=0.015;//Absorption coefficient of N2//
+AO2=0.028;//Absorption coefficient of O2//
+l=22.4;
+printf('Absorption coefficient being the solubility of the gas at partial pressure of 1atm of the gas,\nThe solubilities in mol per litre of the two gases are');
+SN2=N2*AN2/l;//solubility of N2//
+SO2=O2*AO2/l;//solubility of O2//
+printf('\nSolubility of N2=SN2=%f=5.29*10^-4mol per litre.',SN2);
+printf('\nSolubility of O2=SO2=%f=2.625*10^-4mol per litre',SO2);
+VO2=(SO2*100)/(SN2+SO2);
+printf('\nThe mole or Volume percent of O2=%f',VO2);
+VN2=100-VO2;
+printf('\nThe mole or volume percent of N2=%f',VN2);
diff --git a/821/CH6/EX6.5/6_5.sce b/821/CH6/EX6.5/6_5.sce new file mode 100755 index 000000000..173db90d0 --- /dev/null +++ b/821/CH6/EX6.5/6_5.sce @@ -0,0 +1,11 @@ +printf('Upon solving the equations PA=0.9atm,PB=0.3atm');
+PA=0.9;//vapour pressure of A//
+PB=0.3;//Vapour pressure of B//
+xA=0.33;
+xB=0.66;
+yA=(xA*PA)/(xA*PA+xB*PB);
+printf('\nComposition of Vapour A in the mixture=yA=%f',yA);
+yB=1-yA;
+printf('\nComposition of Vapour B in the mixture=yB=%f',yB);
+VP=yA*PA+yB*PB;//total vapour pressure of the mixture//
+printf('\nTotal vapour pressure of the mixture=VP=%f',VP);
diff --git a/821/CH6/EX6.6/6_6.sce b/821/CH6/EX6.6/6_6.sce new file mode 100755 index 000000000..57d261bf9 --- /dev/null +++ b/821/CH6/EX6.6/6_6.sce @@ -0,0 +1,7 @@ +yA=0.60;
+xA1=0.40;
+xA=0.5*yA+0.5*xA1;
+printf('Let PA and PB represent the vapour pressures of pure A and pure B respectively.');
+printf('\nFrom 1 mol of solution after distillation,we get 0.5mol of distillate and 0.5mol of residue.');
+printf('\nVapour pressure of substance A=PA=900.00000mm of Hg');
+printf('\nVapour pressure of substance B=PB=400.00000mm of Hg');
diff --git a/821/CH6/EX6.7/6_7.sce b/821/CH6/EX6.7/6_7.sce new file mode 100755 index 000000000..24588153f --- /dev/null +++ b/821/CH6/EX6.7/6_7.sce @@ -0,0 +1,9 @@ +wA=162;
+wB=100;
+VPB=641;//vapour pressure of water//
+VPA=119;//vapour pressure of oraganic substance//
+MB=18;//Molecular weight of H2O//
+printf('Even though the boiling part of A might be higher,it distills out at a low temperature 95.3degrees.');
+printf('\nIf A were to distill at 95.3degrees,the distillation will have to be carried out at a reduced pressure of about 119mm of mercury');
+MA=(wA*MB*VPB)/(wB*VPA);
+printf('\nMolecular weight of A=MA=%fgrams',MA);
diff --git a/821/CH6/EX6.8/6_8.sce b/821/CH6/EX6.8/6_8.sce new file mode 100755 index 000000000..fea1083cb --- /dev/null +++ b/821/CH6/EX6.8/6_8.sce @@ -0,0 +1,11 @@ +w=50;//weight of acid A in grams//
+x=1;
+y=0.2;
+K=5;
+n=5;
+wn=w*(x/(x+K*y))^n;
+printf('wn=%fgrams',wn);
+y1=1;
+w0=w*(x/(x+K*y1));
+printf('\nw0=%fgrams',w0);
+printf('\nIt is seen that the first process leaves only 1.563grams of A with the aq. layer,\nwhereas the secondone using all available solvent in a single lot leaves 8.333grams in aqueous layer.\nIn the process (a)96.88percent of A is extracted,whereas in (b) only 83.67percent A is extracted.');
diff --git a/821/CH6/EX6.9/6_9.sce b/821/CH6/EX6.9/6_9.sce new file mode 100755 index 000000000..44797d1eb --- /dev/null +++ b/821/CH6/EX6.9/6_9.sce @@ -0,0 +1,24 @@ +AN=0.096;//normality of H2SO4 in aqua layer//
+ON=0.014;//normality of H2SO4 in org. layer//
+AV=13.3;//amount of H2SO4 required in aq. layer for neutralization//
+OV=7.15;//amount of H2SO4 requred in org. layer for neutralization//
+AS=AN*AV/10;//strength of NH3 in aq. layer//
+printf('Strength of NH3 in aq. layer=AS=%fN.',AS);
+OS=ON*OV/20;//strength of NH3 in org. layer//
+printf('\nStrength of NH3 in org. layer=OS=%fN.',OS);
+K=AS/OS;//equilibrium constant//
+printf('\nEquilibrium constant=K=%f',K);
+AV1=20.0;//amount of H2SO4 required in aq. layer at equilibrium//
+OV1=8.0;//amount of H2SO4 required in org. layer at equilibrium//
+AN1=AV1*AN/5;//Normality of NH3 in aq. layer//
+printf('\nNormality of NH3 in aq. layer=AN1=%fN.',AN1);
+ON1=OV1*ON/10;//Normality of NH3 in org. layer//
+printf('\nNormality of NH3 in org. layer=ON1=%fN.',ON1);
+printf('\nIn the aq.layer NH3 includes the free ammonia(uncombined).\nNH3t and that which has combined with Cu2+ to form the complex ion NH3.');
+printf('\nNH3aq=NH3t+NH3combined.\nThe value of NH3t can be obtained from the value of K.\nK=25.49=NH3t/NH3combined.');
+NH3t=K*ON1;
+printf('\nSince nernsts law holds good for the same species present in both phases.\nNH3t=%f=0.2855N.',NH3t);
+NH3c=AN1-NH3t
+printf('\nNH3c=%f=0.0985N',NH3c);
+printf('\n0.025mol per litre of Cu2+ combines with 0.0985mol per litre of NH3.\n1 mol per litre of Cu2+ combines with 0.0985/0.025=3.936mol per litre of NH3.');
+printf('\nor 1mol of Cu2+ combines with 4mol of NH3,i.e the value of x is 4.\nThe formula of the complex ion is thus (Cu(NH3)4)2+');
diff --git a/821/CH7/EX7.1/7_1.sce b/821/CH7/EX7.1/7_1.sce new file mode 100755 index 000000000..db902fe79 --- /dev/null +++ b/821/CH7/EX7.1/7_1.sce @@ -0,0 +1,4 @@ +q=4.0*10^-3;//quantity of electricity in coulombs//
+e=1.6*10^-19;//charge of an electron in coulombs//
+N=q/e;//no. of electrons per second//
+printf('No. of electrons per second=N=%f=2.5*10^16',N);
diff --git a/821/CH7/EX7.10/7_10.sce b/821/CH7/EX7.10/7_10.sce new file mode 100755 index 000000000..1dae4cead --- /dev/null +++ b/821/CH7/EX7.10/7_10.sce @@ -0,0 +1,14 @@ +SCsat=4.63*10^-6;//Specific conductance of saturated solution in ohm^-1cm^-1//
+SCused=1.12*10^-6;//specific conductance of the water used in the experiment//
+SC0Na2SO4=130.1;//specific conductance of Na2SO4 in ohm^-1cm^-1//
+SC0BaCl2=139.9;//specific conductance of 1/2BaCl2 in ohm^-1cm^-1//
+SC0NaCl=126.5;//specific conductance of NaCl in ohm^-1cm^-1//
+SC0=SC0Na2SO4-SC0NaCl+SC0BaCl2;//effective specific conductance in ohm^-1cm^2//
+printf('SC0=%fohm^-1cm^-1',SC0);
+SC=SCsat-SCused;
+printf('\nSpecific conductance of the experiment=SC=%f=3.51*10^-6ohm^-1cm^-1',SC);
+S=(SC*1000)/SC0;//Solubility of the solution//
+printf('\nSolubility of the solution=S=%f=2.437*10^-5gram equivalent per litre',S);
+printf('\n1mol of BaSO4=2equivalents')
+SBaSO4=S/2;//Solubility of the BaSO4 solution//
+printf('\nSolubility of the BaSO4 solution=SBaSO4=%f=1.218*10^-5mol litre^-1',SBaSO4);
diff --git a/821/CH7/EX7.11/7_11.sce b/821/CH7/EX7.11/7_11.sce new file mode 100755 index 000000000..38ca1a0ba --- /dev/null +++ b/821/CH7/EX7.11/7_11.sce @@ -0,0 +1,11 @@ +M1=0.1;//molarity of KCl//
+IKCl=0.5*(M1*1^2+M1*1^2);//Iconic strength of KCl//
+printf('Iconic strength of KCl=IKCl=%f',IKCl);
+M2=0.2;//molarity of K2SO4//
+IK2SO4=0.5*(2*M2*1^2+M2*2^2);//Iconic strength of K2SO4//
+printf('\nIconic strength of K2SO4=IK2SO4=%f',IK2SO4);
+M3=0.2;//molarity of MgCl2//
+IMgCl2=0.5*(M3*2^2+2*M3*1^2);//Iconic strength of MgCl2//
+printf('\nIconic strength of MgCl2=IMgCl2=%f',IMgCl2);
+I=IKCl+IK2SO4+IMgCl2;//total iconic strength of the mixture//
+printf('\nTotal Iconic strength of the mixture=I=%f',I);
diff --git a/821/CH7/EX7.12/7_12.sce b/821/CH7/EX7.12/7_12.sce new file mode 100755 index 000000000..55cba64b7 --- /dev/null +++ b/821/CH7/EX7.12/7_12.sce @@ -0,0 +1,5 @@ +r=0.96;
+I=(log10(r)/-0.51)^2;//Iconic strength of the HCl solution//
+printf('Iconic strength of the HCl solution=I=%f',I);
+printf('\nHCl being a 1,1 electrolyte I=c');
+printf('\nSo the maximum concentration of HCl to be used=I=c=1.20*10^-3');
diff --git a/821/CH7/EX7.13/7_13.sce b/821/CH7/EX7.13/7_13.sce new file mode 100755 index 000000000..aba40e4f8 --- /dev/null +++ b/821/CH7/EX7.13/7_13.sce @@ -0,0 +1,8 @@ +w=3.55;//weight of the salt in grams//
+MW=258.2;//Molecular weight of the salt//
+printf('KAl(SO4)2 = K+ + Al3+ + 2SO42-');
+c=w*4/MW;//Concentration of the salt//
+printf('\nConcentration of the salt=c=%fM',c);
+SO4=2*c;//Concentration of SO42- in the solution//
+printf('\nconcentration of K+ = Al3+ =%fM',c);
+printf('\nConcentration of SO42- =%fM',SO4);
diff --git a/821/CH7/EX7.14/7_14.sce b/821/CH7/EX7.14/7_14.sce new file mode 100755 index 000000000..bd02e99ae --- /dev/null +++ b/821/CH7/EX7.14/7_14.sce @@ -0,0 +1,5 @@ +c=0.1;//concentration of the solution//
+a=1.332*10^-2;//Ionization constant//
+Ka=(c*a^2)/(1-a);
+printf('CH3COOH = CH3COO- + H+');
+printf('\nKa value for the reaction=Ka=%f=1.799*10^-5',Ka);
diff --git a/821/CH7/EX7.15/7_15.sce b/821/CH7/EX7.15/7_15.sce new file mode 100755 index 000000000..4e6ea8ac4 --- /dev/null +++ b/821/CH7/EX7.15/7_15.sce @@ -0,0 +1,12 @@ +c=0.1;//concentration of the solution//
+Kb=1.8*10^-5;
+printf('The value of a should be calculated first using Kb=(c*a^2)/(1-a)\nThis gives rise to a quadratic equation which can be solved to obtain the value of a.');
+printf('\nUsually it is permissible to use approximation methods if K<10^-5\nOne can neglect a in comparison to 1 and solve for a.\nA better way is to use the method of succesive approximations.\nThis will be illustrated using the above equation');
+printf('\nFirst find the approximate value of a by neglecting the value of a in comparison with 1.\nLet the approximate value be a1');
+a1=1.342*10^-2;
+a2=1.332*10^-2;
+printf('\nWe repeat this procedure till 2 consecutive values of a do not differ significantly.');
+a3=1.332*10^-2;
+OH=a3*c;//concentration of OH- in the solution//
+printf('\nSince the values of a2 and a3 are the same the correct value of a=1.332*10^-2\nThe approximate value is greater than the correct value by about 1percent.');
+printf('\nThe concentration of OH- =%f=1.332*10^-3M',OH);
diff --git a/821/CH7/EX7.18/7_18.sce b/821/CH7/EX7.18/7_18.sce new file mode 100755 index 000000000..cdb60dc02 --- /dev/null +++ b/821/CH7/EX7.18/7_18.sce @@ -0,0 +1,8 @@ +Kw=1.0^10^-14;
+KH1=1.0*10^-5;//KH value of H+ ion in RCOOH//
+KH2=1.0^10^-10;//KH value of H+ ion in HCN//
+Kb1=Kw/KH1;//Kb value for RCOO- ion//
+printf('Kb value for RCOO- ion=10^-9 ');
+Kb2=Kw/KH2;//Kb value for CN- ion//
+printf('\nKb value for CN- ion=10^-4');
+printf('\nCN- is about 10^5 times stronger than RCOO- as a base.');
diff --git a/821/CH7/EX7.19/7_19.sce b/821/CH7/EX7.19/7_19.sce new file mode 100755 index 000000000..eab0189c6 --- /dev/null +++ b/821/CH7/EX7.19/7_19.sce @@ -0,0 +1,16 @@ +c1=1;//concentration of HCl//
+PH1=-log10(c1);
+printf('PH for the 1M HCl solution=PH1=%f',PH1);
+c2=5.2*10^-4;//concentration of H+ in the solution//
+PH2=-log10(c2);
+printf('\nPH for the solution=PH2=%f',PH2);
+c3=0.025;//concentration of 0.025M HClO4//
+PH3=-log10(c3);
+printf('\nPH for the 0.025M HClO4 solution=PH3=%f',PH3);
+PH4=4.45;
+c4=10^(-PH4);//concentration of the solution//
+printf('\nConcentration of the solution=c4=%f=3.548*10^-5',c4);
+POH5=1.30;
+PH5=14-POH5;
+c5=10^(-PH5);//concentration of the solution//
+printf('\nConcentration of the solution=c5=1.995*10^-13');
diff --git a/821/CH7/EX7.2/7_2.sce b/821/CH7/EX7.2/7_2.sce new file mode 100755 index 000000000..c87167862 --- /dev/null +++ b/821/CH7/EX7.2/7_2.sce @@ -0,0 +1,18 @@ +i=3;//current passed through the solution in amps//
+t=5;//amount of time current passed through in hours//
+q=(i*t)/26.8;//quantity of electricity passed in farads//
+printf('Quantity of electricity passed=q=%fFarads',q);
+printf('\nIf all the current is used in the deposition of Ni,i.e 100percent efficiency 0.56 equivalents of Ni should be deposited at the cathode.');
+N=0.56*0.60;//No. of equivalents of Ni deposited//
+printf('\nNo. of equivalents of Ni deposited=N=%f',N);
+w=58.71;//weight of Ni in grams//
+wd=N*w/2;//weight of Ni actually deposited in grams//
+printf('\nWeight of Ni actually deposited=wd=%fgrams',wd);
+TA=32;//total area of the cathode in cm^2 for 2faces//
+d=8.9;//density of Ni in gram per cm^3//
+V=wd/d;//volume of the Ni deposited in cm^3//
+printf('\nVolume of the Ni deposited=V=%fcm^3',V);
+T=V/TA;//thickness of the deposit in cm//
+printf('\nThickness of the deposit=T=%fcm',T);
+printf('\nOut of 0.56Farad, 0.336Farad is used for Ni deposition\nhence 0.224Farad is used for liberation of hydrogen.');
+printf('\n0.224 equivalent of hydrogen is=11.2*0.224=2.51litres.');
diff --git a/821/CH7/EX7.20/7_20.sce b/821/CH7/EX7.20/7_20.sce new file mode 100755 index 000000000..67a118894 --- /dev/null +++ b/821/CH7/EX7.20/7_20.sce @@ -0,0 +1,8 @@ +a=1.33*10^-2;//Ionization constant//
+c=0.1;//concentration of the solution//
+OH=a*c;
+printf('OH- =a*c=1.33*10^-3');
+POH=-log10(OH);//POH of the solution//
+printf('\nPOH of the solution=POH=%f',POH);
+PH=14-POH;//PH of the solution//
+printf('\nPH of the solution=PH=%f',PH);
diff --git a/821/CH7/EX7.21/7_21.sce b/821/CH7/EX7.21/7_21.sce new file mode 100755 index 000000000..24d7f1ff6 --- /dev/null +++ b/821/CH7/EX7.21/7_21.sce @@ -0,0 +1,8 @@ +c=0.01;//concentration of the solution//
+r=10^(-0.51*sqrt(c));
+printf('r=%f',r);
+a=r*c;//ionization constant//
+printf('\nIonization constant=a=%f',a);
+PH=-log10(a);//PH of the solution//
+printf('\nPH of the solution=PH=%f',PH);
+printf('\nBy assuming ideal behaviour PH=-log10(10^-2)=2.00');
diff --git a/821/CH7/EX7.22/7_22.sce b/821/CH7/EX7.22/7_22.sce new file mode 100755 index 000000000..cdc80b9d1 --- /dev/null +++ b/821/CH7/EX7.22/7_22.sce @@ -0,0 +1,15 @@ +CNH3=0.1;//concentration of NH3 solution//
+CNH4Cl=0.1;//concentration of NH4Cl solution//
+POH=4.74;
+PH=14-POH+log10(CNH3/CNH4Cl);
+printf('PH of the solution=PH=%f',PH);
+printf('\nOn adding 0.01mol of HCl,assuming that no volume change occurs,0.01mol of NH4Cl is produced.\nTherefore,the concentration of NH3 decreases by 0.01 and that of NH4Cl increases by 0.01 ');
+C1NH3=0.09;
+C1NH4Cl=0.11;
+PH1=14-POH+log10(C1NH3/C1NH4Cl);
+printf('\nPH of the solution=PH1=%f',PH1);
+printf('\nOn adding 0.01mol of NaOH,assuming that no volume change occurs,0.01mol of NH3 is produced.\nTherefore,the concentration of NH3 increases by 0.01 and that of NH4Cl decreases by 0.01 ');
+C2NH3=0.11;
+C2NH4Cl=0.09;
+PH2=14-POH+log10(C2NH3/C2NH4Cl);
+printf('\nPH of the solution=PH2=%f',PH2);
diff --git a/821/CH7/EX7.23/7_23.sce b/821/CH7/EX7.23/7_23.sce new file mode 100755 index 000000000..4f60dda86 --- /dev/null +++ b/821/CH7/EX7.23/7_23.sce @@ -0,0 +1,12 @@ +c=0.1;//concentration of the solution//
+Kw=1.0*10^-14;
+Ka=7.24*10^-10;//dissociation constant of HCN//
+printf('For a salt of this type the hydrolysis reaction is\nCN- +H2O = HCN + OH-');
+Kh=Kw/Ka;//hydrolysis constant//
+printf('\nHydrolysis constant of the solution=Kh=%f=1.381*10^-5',Kh);
+printf('\nIonization constant is generally calculated using Kh=(c*a^2)/(1-a)');
+printf('\nIonization constant=a=0.011680=1.168*10^-2');
+printf('\nThe degree of hydrolysis is 1.168percent');
+PKw=23.14;
+PH=0.5*(PKw+log10(c));//PH of the 0.1M NaCN solution//
+printf('\nPH of the 0.1M NaCN solution=PH=%f',PH);
diff --git a/821/CH7/EX7.25/7_25.sce b/821/CH7/EX7.25/7_25.sce new file mode 100755 index 000000000..e5e17c133 --- /dev/null +++ b/821/CH7/EX7.25/7_25.sce @@ -0,0 +1,14 @@ +MW=332;//molecular weight of Ag2CrO4 in grams//
+s=2.5*10^-2;//solubility of Ag2CrO4 in g per litre//
+S=s/MW;//Solubility of Ag2CrO4 in mol per litre//
+printf('Solubility of Ag2CrO4=S=%f=7.5*10^-5mol per litre',S);
+Ag=2*7.5*10^-5;//Solubility of Ag component in mol per litre//
+CrO4=7.5*10^-5;//Solubility of CrO4 component in mol per litre//
+Ksp=Ag*CrO4;//value of Ksp//
+printf('\nValue of Ksp for the reaction=Ksp=%f=1.7*10^-12',Ksp);
+MWAgCl=143.5;//Molecular weight of AgCl//
+Ksp1=1.1*10^-10;//Ksp value of AgCl//
+S1=sqrt(Ksp1);//Solubility of AgCl in mol per litre//
+printf('\nSolubility of AgCl=S1=%f=1.05*10^-5mol per litre',S1);
+s1=S1*MWAgCl;//solubility of AgCl in g per litre//
+printf('\nSolubility of AgCl=s1=%f=1.50*10^-3gram per litre',s1);
diff --git a/821/CH7/EX7.26/7_26.sce b/821/CH7/EX7.26/7_26.sce new file mode 100755 index 000000000..b2fb5cd16 --- /dev/null +++ b/821/CH7/EX7.26/7_26.sce @@ -0,0 +1,9 @@ +MW=372;//molecular weight of Li3Na3(AlF6)2 in grams//
+s=0.74;//solubility of Li3Na3(AlF6)2 in g per litre//
+S=s/MW;//Solubility of Li3Na3(AlF6)2 in mol per litre//
+printf('Solubility of Li3Na3(AlF6)2=S=%fmol per litre',S);
+Li=3*S;//Solubility of Li component in mol per litre//
+Na=3*S;//Solubility of Na component in mol per litre//
+AlF6=2*s;//Solubility of AlF6 component in mol per litre//
+Ksp=(Li^3)*(Na^3)*(AlF6^2);//value of Solubility product//
+printf('\nValue of Ksp for the reaction=Ksp=%f=7.47*10^-19',Ksp);
diff --git a/821/CH7/EX7.27/7_27.sce b/821/CH7/EX7.27/7_27.sce new file mode 100755 index 000000000..b134a51f9 --- /dev/null +++ b/821/CH7/EX7.27/7_27.sce @@ -0,0 +1,9 @@ +Ksp=2.2*10^-8;//Solubility product of PbSO4//
+Pb=0.01;//concentration of Pb in Pb(NO3)2//
+SO4=Ksp/Pb;//Concentration of SO4 in PbSO4 solution//
+printf('Let us first calculate the maximum concentration of SO4 that can remain in equilibrium with PbSO4 if the concentration of Pb is 0.01M');
+printf('\nConcentration of SO4 in PbSO4 solution=Ksp=2.2*10^-6M\nThe concentration of SO4 should be greater than 2.2*10^-4M in order to precipitate Pb from a 0.01M solution as PbSO4');
+Pb2=Ksp/0.001;
+printf('\nConcentration of Pb in PbSO4 solution=Pb2=2.2*10^-5mol per litre');
+printf('\nHence out of 0.01moles of Pb in a litre only 2.2*10^-5mol per litre remain in solution.the precipitation is almost 99.78percent complete.');
+printf('\nTherefore it can be said that Pb is quantitatively precipitated in these conditions.');
diff --git a/821/CH7/EX7.28/7_28.sce b/821/CH7/EX7.28/7_28.sce new file mode 100755 index 000000000..2a5e15975 --- /dev/null +++ b/821/CH7/EX7.28/7_28.sce @@ -0,0 +1,14 @@ +Cu=0.1;//concentration of Cu2+ ions in the solution//
+Mn=0.1;//concentration of Mn2+ ions in the solution//
+H=0.3;//concentration of H+ ions in the solution//
+KspCuS=1.0*10^-44;//Solubility product of CuS//
+KspMnS=1.4*10^-15;//Solubility product of MnS//
+K1=9.1*10^-8;//K1 value of the H2S solution//
+K2=1.2*10^-15;//K2 value of the H2S solution//
+K=K1*K2;//K value of H2S//
+printf('K value of H2S solution=K=1.1*10^-22');
+S2=K/H^2;//Concentration of S2- in the solution//
+printf('Concentration of S2 in the solution=1.22*10^-22\nThe Iconic product of CuS=1.22*10^-22*10^-1 is >> Ksp for CuS and so it precipitates.');
+printf('\nIf MnS were to precipitate the S2 should be greater than the equilibrium concentration of S2\ni.e Mn*S2 = 1.4*10^-15 so S2eq=1.4*10^-14');
+printf('\nThe S2 should be greater than 1.4*10^-14 so that MnS will precipitate.\nLet the S2 desired be 1.1*10^-11\nIn order to get this concentration of S2 the required H is 10^-6M');
+printf('\nThe solution should have a PH of greater than 6 i.e PH >> 6');
diff --git a/821/CH7/EX7.29/7_29.sce b/821/CH7/EX7.29/7_29.sce new file mode 100755 index 000000000..af638b991 --- /dev/null +++ b/821/CH7/EX7.29/7_29.sce @@ -0,0 +1,11 @@ +Al=0.01;//concentration of Al3+ ions in the solution//
+Mg=0.01;//concentration of Mg2+ ions in the solution//
+NH4Cl=2;//concentration of NH4Cl in the solution//
+printf('NH3 + H2O = NH4+ + OH-');
+KspMgOH2=3.4*10^-11;//Solubility product of Mg(OH)2//
+KspAlOH3=5.0*10^-33;//Solubility product of Al(OH)3//
+Kb=1.8*10^-5;//Kb value of the NH3//
+printf('\nNH4+ in solution = NH4+ from added NH4Cl that derived from the reaction between NH3 and H2O = NH4+ from NH4Cl(since the other quantity is too small)');
+printf('\nNH3 = Original concentration since amount dissociated is very low\nwe get OH- = 1.8*10^-5\nThe iconic product for Mg(OH)2 is (Mg2+)(OH-)^2 or (10^-2)(1.8*10^-5)^2 i.e 3.24*10^-12\nIt is less than 3.4*10^-11,Ksp,So it is not precipitated.');
+printf('\nHowever,in the case of Al(OH)3 the Iconic product=(10^-2)(1.8*10^-5)^3=5.832*10^-17>>The Ksp for Al(OH)3 i.e 5.0*10^-33.');
+printf('\nso Al(OH)3 gets precipitated.');
diff --git a/821/CH7/EX7.3/7_3.sce b/821/CH7/EX7.3/7_3.sce new file mode 100755 index 000000000..e168d8693 --- /dev/null +++ b/821/CH7/EX7.3/7_3.sce @@ -0,0 +1,10 @@ +i=5*10^-3;//steady current given by the cell in amps//
+w=1.7399;//amount of MnO2 used in grams//
+MW=86.95;//molecular weight of MnO2 in grams//
+F=96500;//farad value in coulombs//
+C=0.02*F;//charge value in coulombs//
+printf('From the cathode reaction 2mol of MnO2=2Farad.');
+printf('\n0.02Farad means the charge=C=%fCoulombs.',C);
+q=1930;//charge in coloumbs corresponding to 0.02Farad//
+t=q/i;//amount of time cell suplies the current in seconds//
+printf('\nAmount of time cell supplies the current=t=%f=3.86*10^5seconds',t);
diff --git a/821/CH7/EX7.30/7_30.sce b/821/CH7/EX7.30/7_30.sce new file mode 100755 index 000000000..4f322ea41 --- /dev/null +++ b/821/CH7/EX7.30/7_30.sce @@ -0,0 +1,7 @@ +Fe=0.01;//concentration of Fe3+ ions in the solution//
+Ksp=3.8*10^-38;//Solubility product of Fe(OH)3//
+OH=(Ksp/Fe)^0.333;//Concentration of OH- ions in the solution//
+printf('Concentration of OH- ions in the solution=OH=%f=1.561*10^-12',OH);
+printf('\nAt this PH,Fe(OH)3 starts precipitating and precipitation is complete (Fe3+)=10^-6M\n(10^-6)(OH-)^3 = 3.8*10^-38 ');
+printf('\nupon solving this we get (OH-)=3.362*10^-11\nPOH=10.48 or PH=3.52');
+printf('\nAt this PH the precipitation of Fe(OH)3 is almost complete.');
diff --git a/821/CH7/EX7.31/7_31.sce b/821/CH7/EX7.31/7_31.sce new file mode 100755 index 000000000..787b06b78 --- /dev/null +++ b/821/CH7/EX7.31/7_31.sce @@ -0,0 +1,11 @@ +C=0.01;//concentration of Ca(NO3)2 solution//
+Ksp=3.2*10^-11;//Solubility product of Fe(OH)3//
+printf('CaF2 = Ca2+ + 2F-\n(Ca2+)(F-)^2 = 4*S^3 = 3.2*10^-11.');
+printf('\nLet S1 be the solubility in 0.01M Ca(NO3)2\nCa(NO3)2 can be assumed to dissociate completely so that (Ca2+) from Ca(NO3)2 is 0.01M');
+S=(Ksp/4)^0.33;//solubility in mol per litre//
+printf('\nSolubility of CaF2 solution=S=%f=2.18*10^-4mol per litre',S);
+printf('\nThe solubility product relationship should be true,irrespective of the source Ca2+\nCompared to the concentration of Ca2+ ions obtained from Ca(NO3)2,that of Ca2+ ions from CaF2 is negligible');
+S1=sqrt(Ksp/0.04);//solubility in 0.01M ca(NO3)2//
+printf('\nBut the F- ions are obtained only from CaF2 and so (F-)=2*S1\nKsp = 3.2*10^-11=(S1+0.01)*(2*S1)^2=(0.01)*(2*S1)^2 since S1 is negligible compared to 0.01');
+printf('\nSolubility in 0.01M Ca(NO3)2 solution=S1=%f=2.83*10^-5',S1);
+printf('\nThus the value of S1 can be seen to be less than that of S');
diff --git a/821/CH7/EX7.32/7_32.sce b/821/CH7/EX7.32/7_32.sce new file mode 100755 index 000000000..3842a5603 --- /dev/null +++ b/821/CH7/EX7.32/7_32.sce @@ -0,0 +1,13 @@ +T=298;//temperature in kelvin//
+E=0.22;//emf of the cell in volts//
+dE=-0.00065;//Temperature coefficient of the emf in volt per degree//
+c=4.184;//1 cal =4.184 joules//
+n=1;
+F=96500;//1Farad value//
+printf('The positive electrode is the cathode and the negative electrode is the anode in a galvanic cell\nAnode reaction 1/2H2 = H+ + e-\nCathode reaction AgCl + e- = Ag+ + Cl-\nCell reaction 1/2H2 + AgCl = Ag+ + H+ + Cl-');
+dG=-n*F*E/c;//free energy change in the cell in cal per mol//
+printf('\nFree energy change in the cell=dG=%fcal per mol',dG);
+dH=dG+(n*F*T*dE/c);//Enthalpy change in the cell//
+printf('\nEnthalpy change in the cell=dH=%fcal per mol',dH);
+dS=(dH-dG)/T;//Entropy change in the cell in cal per deg//
+printf('\nEntropy change in the cell=dS=%fcal per deg',dS);
diff --git a/821/CH7/EX7.4/7_4.sce b/821/CH7/EX7.4/7_4.sce new file mode 100755 index 000000000..dc89c52ef --- /dev/null +++ b/821/CH7/EX7.4/7_4.sce @@ -0,0 +1,10 @@ +C=2.768*10^-3;//conductivity of the cell in ohm^-1 cm^-1//
+R=82.4;//resistance with KCl solution filled in ohms//
+K=C*R;//cell constant in cm^-1//
+printf('Cell constant=K=%fcm^-1',K);
+R1=326;//resistance with K2SO4 solution filled in ohms//
+c=K/R1;//Equivalent conductance of the KCl solution in ohm^-1 cm^-1//
+printf('\nEquivalent conductance of the KCl solution=c=%f=7*10^-4ohm^-1 cm^-1',c);
+printf('\n0.0025M K2SO4 solution=0.005N of K2SO4.');
+EC=1000*c/0.005;//equivalent conductance of K2SO4 solution in ohm^-1 cm^2//
+printf('\nEquivalent conductance of K2SO4 solution=EC=%fohm^-1cm^2',EC);
diff --git a/821/CH7/EX7.5/7_5.sce b/821/CH7/EX7.5/7_5.sce new file mode 100755 index 000000000..ecd2da750 --- /dev/null +++ b/821/CH7/EX7.5/7_5.sce @@ -0,0 +1,18 @@ +wAg=1.424;//weight of Ag deposited in the coulometer in grams//
+MW=108;//molecular weight of AgNO3 in grams//
+w1=90.25;//weight of silver nitrate in grams//
+w2=5.039;//weight of AgNO3 in grams//
+w=w1-w2;
+n=wAg/MW;
+printf('No. of equivalents of Ag deposited in the silver coulometer=n=%f.',n);
+printf('\nThis amount of Ag+ and NO3- ions would have discharged at the cathode and at the anode respectively.');
+printf('\n Anolyte solution\nBefore electrolysis 85.21(90.25-5.039)grams of water contained 0.02965 equivalents of AgNO3 or Ag+.');
+BEAg=0.007202;//no. of equivalents of Ag+ before electrolysis//
+printf('\nAfter electrolysis 20.893-0.193 i.e 20.7grams of water contains 0.001136equivalents of AgNO3 or Ag+.');
+AEAg=0.01136;//no. of equivalents of Ag+ after electrolysis//
+printf('\n20.7grams of water,before electrolysis would have contained 0.007202 equivalents of Ag+.');
+DC=BEAg-AEAg;//decrease in the conc. of anolyte//
+printf('\nDecrease in the conc. of anolyte=DC=0.006066equivalents.');
+tAg=n/DC;//transport number//
+printf('\ntAg=ratio of Decrease in anolyte conc. and No. of gram equivalents deposited at either electrode=tAg=0.460100');
+
diff --git a/821/CH7/EX7.6/7_6.sce b/821/CH7/EX7.6/7_6.sce new file mode 100755 index 000000000..4edf96d21 --- /dev/null +++ b/821/CH7/EX7.6/7_6.sce @@ -0,0 +1,25 @@ +wCu=0.0230;//weight of Cu deposited in the coulometer in grams//
+MW=63.54;//molecular weight of Cu in grams//
+n=wCu*2/MW;//no. of equivalents of Cu deposited//
+printf('In the coulometer,wt of Cu deposited=0.0230grams');
+printf('\nNumber of equivalents of Cu deposited=n=%fequivalents or Farads.',n);
+printf('\nThis would have resulted in deposition of 7.24*10^-4equivalents of Ag+ at the cathode \nand dissolution of the same amount at the anode.');
+wAgNO3=7.39;
+w1AgNO3=0.2360;//after electrolysis weight of AgNO3//
+MWAgNO3=170;//molecular weight of AgNO3//
+BEAgNO3=wAgNO3/MWAgNO3;
+printf('\n Anolyte solution \nBefore electrolysis 1000grams of water contains %fequivalents of AgNO3',BEAgNO3);
+AEAgNO3=w1AgNO3/MWAgNO3;
+printf('\nAfter electrolysis 23.14grams of water contains %fequivalents of AgNO3.',AEAgNO3);
+w=23.14;
+BE=w*BEAgNO3/1000;
+printf('\n23.14grams of water before electrolysis would have contained %fequivalents of AgNO3',BE);
+IC=AEAgNO3-BE;
+printf('\nIncrease in the concentration of anolyte=IC=%fequivalents.',IC);
+printf('\n0.000382gram equivalents of NO3- ions must have migrated into the anode compartment.\nAs a result of passin 7.24*10^-4Farads into the solution.\n0.000724equivalents of Ag should have dissolved to give the same amount of Ag+ ion.\nOut of this 0.000382gram equivalents are present in the anolyte.');
+ME=n-IC;//no of equivalents of migrated anodes//
+printf('\n%fgram equivalents of Ag+ ions must have migrated from the anode.',ME);
+tAg=ME/n;//transport number//
+printf('\nTransport number of Ag=tAg=%f',tAg);
+tSO3=1-tAg;
+printf('\nTransport number of SO3=tSO3=%f',tSO3);
diff --git a/821/CH7/EX7.7/7_7.sce b/821/CH7/EX7.7/7_7.sce new file mode 100755 index 000000000..5486e7076 --- /dev/null +++ b/821/CH7/EX7.7/7_7.sce @@ -0,0 +1,12 @@ +EC=426;//equivalent conductance of HCl in ohm^-1cm^2//
+tH=0.82;//transport number of H+//
+tCl=0.18;//transport number of Cl-//
+ICH=EC*tH;//ionic conductance of H+ in ohm^-1cm^2//
+printf('Ionic conductance of H+=ICH=%fohm^-1cm^2',ICH);
+ICCl=EC*tCl;//ionic conductance of Cl- in ohm^-1cm^2//
+printf('\nIonic conductance of Cl-=ICCl=%fohm^-1cm^2',ICCl);
+F=96500;
+IMH=ICH/F;//ionic mobility of H+ in cm^2v^-1s^-1
+printf('\nIonic mobility of H+=IMH=%f=36.20*10^-4cm^2v^-1s^-1',IMH);
+IMCl=ICCl/F;//ionic mobility of H+ in cm^2v^-1s^-1
+printf('\nIonic mobility of H+=IMCl=%f=7.95*10^-4cm^2v^-1s^-1',IMCl);
diff --git a/821/CH7/EX7.9/7_9.sce b/821/CH7/EX7.9/7_9.sce new file mode 100755 index 000000000..4dcfd75f0 --- /dev/null +++ b/821/CH7/EX7.9/7_9.sce @@ -0,0 +1,13 @@ +printf('A solution of NH3 is alkaline due to the following hydrolysis\nNH3+H2O = NH4+ + OH-');
+printf('\nKb=(NH4+)*(OH-)/(NH3)=(c*a^2)/(1-a). ');
+EC=3.7;//equivalent conductance of NH3 in water in ohm^-1cm^2//
+EC0NH4Cl=149.9;//equivalent conductance of NH4Cl in ohm^-1cm^2//
+EC0BaCl2=139.9;//equivalent conductance of 1/2BaCl2 in ohm^-1cm^2//
+EC0BaOH2=262.2;//equivalent conductance of 1/2Ba(OH)2 in ohm^-1cm^2//
+EC0=EC0NH4Cl-EC0BaCl2+EC0BaOH2;//effective Equivalent conductance in ohm^-1cm^2//
+printf('\nEC0=%fohm^-1cm^2',EC0);
+a=EC/EC0;//dissociation constant of the solution//
+printf('\nDissociation constant of the solution=a=%f',a);
+C=0.1;//normality of the solution//
+Kb=(C*a^2)/(1-a);
+printf('\nIonization constant=Kb=%f',Kb);//here the values of a and Kb are slightly different from textbook but that is ok//
diff --git a/821/CH8/EX8.10/8_10.sce b/821/CH8/EX8.10/8_10.sce new file mode 100755 index 000000000..8316372ac --- /dev/null +++ b/821/CH8/EX8.10/8_10.sce @@ -0,0 +1,7 @@ +k=8.676*10^-3;//average value of k in per min//
+printf('Average value of k=8.676*10^-3per min');
+r0=22.4;
+rt=0;
+rinfinite=-11.1;
+t=(2.303/k)*log10((r0-rinfinite)/(rt-rinfinite));
+printf('\nThe time at which the mixture is optically inactive=t=%fmin',t);//here in textbook the answer is given wrong,but by solving we get the same result as executed//
diff --git a/821/CH8/EX8.13/8_13.sce b/821/CH8/EX8.13/8_13.sce new file mode 100755 index 000000000..ca83e0255 --- /dev/null +++ b/821/CH8/EX8.13/8_13.sce @@ -0,0 +1,6 @@ +k=1;
+a=10;
+thalf=10^-1.88;//half time of the reaction//
+n=1-(log10(thalf/k)/log10(a));//order of the reaction//
+printf('order of the reaction after solving is n=%f',n);
+printf('\nHence the order of the reaction=n=3');
diff --git a/821/CH8/EX8.14/8_14.sce b/821/CH8/EX8.14/8_14.sce new file mode 100755 index 000000000..698863a19 --- /dev/null +++ b/821/CH8/EX8.14/8_14.sce @@ -0,0 +1,5 @@ +thalf1=49;//half life for 0.02M compound//
+thalf2=100;//half life for 0.04M compound//
+M1=0.02;//initial concentration of the compound//
+M2=0.01;//Final concentration of the compound//
+printf('Since thalf is directly proportional to concentration(doubling the concentration increases thalf by a factor 2),\nthe reaction is Zero order.\nThe half time corresponding to a concentration of 0.01M will be 24.5mins');
diff --git a/821/CH8/EX8.18/8_18.sce b/821/CH8/EX8.18/8_18.sce new file mode 100755 index 000000000..ca9c1dcf6 --- /dev/null +++ b/821/CH8/EX8.18/8_18.sce @@ -0,0 +1,7 @@ +R=1.987;//universal gas constant//
+T1=293;//initial temperature in kelvin//
+T2=303;//Final temperature in kelvin//
+K1=6.68*10^-3;//rate constant corresponding to T1 in per min//
+K2=1.31*10^-2;//rate constant corresponding to T2 in per min//
+E=(2.303*R*T1*T2*log10(K2/K1))/(T2-T1);//energy of activation in Kcal per mol//
+printf('Energy of activation=E=%fcal per mol=11.88Kcal per mol',E);
diff --git a/821/CH8/EX8.19/8_19.sce b/821/CH8/EX8.19/8_19.sce new file mode 100755 index 000000000..1f5ef0ba8 --- /dev/null +++ b/821/CH8/EX8.19/8_19.sce @@ -0,0 +1,6 @@ +R=1.987;//universal gas constant//
+T1=350;//initial temperature in kelvin//
+T2=360;//Final temperature in kelvin//
+E=40000;//energy of activation in cal per mol//
+K=10^(E*((T2-T1)/(T1*T2))/(2.303*R));//ratio of v2/v1//
+printf('The ratio of V2 and V1 is K=V2/V1=%f',K);
diff --git a/821/CH8/EX8.20/8_20.sce b/821/CH8/EX8.20/8_20.sce new file mode 100755 index 000000000..593ee6297 --- /dev/null +++ b/821/CH8/EX8.20/8_20.sce @@ -0,0 +1,8 @@ +R=1.987;//universal gas constant//
+T1=313;//initial temperature in kelvin//
+T2=333;//Final temperature in kelvin//
+t1=15;//time for 20% reaction at 313K in mins//
+t2=3;//time for 20%reaction at 333K in mins//
+K=t1/t2;//ratio of K2/K1//
+E=(2.303*R*T1*T2*log10(K))/(1000*(T2-T1));//energy of activation in Kcal per mol//
+printf('Energy of activation=E=%fKcal per mol',E);
diff --git a/821/CH8/EX8.21/8_21.sce b/821/CH8/EX8.21/8_21.sce new file mode 100755 index 000000000..170f4762e --- /dev/null +++ b/821/CH8/EX8.21/8_21.sce @@ -0,0 +1,7 @@ +R=1.987;//universal gas constant//
+printf('From the graph slope=(-0.92/0.30)=(-E/(2.303*R))\nGraphical evaluation of A requires the determination of the intercept on the y axis corresponding to 1/T=0\nOne can also calculate A from k=A*exp(-E/(R*T))');
+E=(0.92*R*2.303)/(0.30*10^3);
+printf('\nEnergy of activation=E=%f=14.04Kcal per mol',E);
+k=2.31*10^-2;
+T=273;//temperature in kelvin//
+printf('\nwe can find the value of A using log10(k)=log10(A)-(E/(2.303*R*T))\nUpon solving we get A=4.015*10^9litre per mol per second');
diff --git a/821/CH8/EX8.22/8_22.sce b/821/CH8/EX8.22/8_22.sce new file mode 100755 index 000000000..56fd876d2 --- /dev/null +++ b/821/CH8/EX8.22/8_22.sce @@ -0,0 +1,6 @@ +R=1.987;//universal gas constant//
+T=556;//temperature in kelvin//
+E=44;//Energy in Kcal//
+dS=-10;//entropy change in cal per deg//
+k=(exp(2)*exp(dS/R)*exp(-E/(R*T))*10^13.07);//rate constant for the reaction//
+printf('Rate constant for the reaction=k=3.47*10^-7litre per mol per sec');
diff --git a/821/CH8/EX8.23/8_23.sce b/821/CH8/EX8.23/8_23.sce new file mode 100755 index 000000000..6445d599f --- /dev/null +++ b/821/CH8/EX8.23/8_23.sce @@ -0,0 +1,10 @@ +R=1.987;//universal gas constant//
+T=473;//temperature in kelvin//
+A=2.75*10^15;//frequency factor in per sec//
+K=1.38*10^-16;//boltzmans constant//
+h=6.625*10^-27;//planks constant//
+dn=0;
+dS=4.57*(log10(A)-log10(exp(1))-log10(9.85*10^12));//entropy change in cal per deg//
+printf('The entropy of activation=dS=%f=9.19eu',dS);
+printf('\nSince A is independent of concentration units dS does not sepend on the concentration units used\nand hence the standard state.\nHowever if the time were expressed in different units A will assume a different value\nand consequently the value of dS will be different\nIf time were expressed in minutes A=2.75*10^15*60 per min\ndS=9.19+4.57*log10(60)=17.32eu\nfor bimolecular reaction e^2=7.4*10^10 \nso dS will result in dS=-10.1eu or mol per litre.');
+printf('\nIf the concentration were expressed in mol per millilitre A would be 7.4*10^13 \nso dS will result in dS=-10.1+13.6=3.5eu or mol per millilitre\nIf the concentration were expressed in molecules per millilitre the value of A will be multiplied by 6.023*10^23 \nso dS would result in as dS=-10.1-94.9=-105eu or -105molecules per millilitre');
diff --git a/821/CH8/EX8.3/8_3.sce b/821/CH8/EX8.3/8_3.sce new file mode 100755 index 000000000..5032cb34e --- /dev/null +++ b/821/CH8/EX8.3/8_3.sce @@ -0,0 +1,6 @@ +k=1.37*10^-4;//rate constant in per sec//
+thalf=5057;//half time of the reaction in sec//
+printf('If out of each mole of N2O5,x mole of it decomposes at any instant\nThe total pressure in the system is equal to that due to (1-x)moles of undecomposed N2O5,\nx moles of N2O4 and x/2moles of O2 i.e due to 1+(x/2)moles.');
+printf('\nThe increase in pressure is thus due to x/2moles\nSo the amount of N2O5 that has decomposed at any instant i.e x is proportional to twice the observed increase in pressure');
+printf('\nThese can be fitted into the kinetic equation for a first order reaction\nk=(2.303/t)*log10(a/(a-x))\nand the value of k can be obtained.The average value of k is 1.37*10^-4per sec');
+printf('\nk=0.693/t0.5 that will result in t0.5=5057seconds\nThis can also be obtained as the value corresponding to (a-x)=154.1mm,from a graph of t vs (a-x).');
diff --git a/821/CH8/EX8.4/8_4.sce b/821/CH8/EX8.4/8_4.sce new file mode 100755 index 000000000..5206bf7a4 --- /dev/null +++ b/821/CH8/EX8.4/8_4.sce @@ -0,0 +1,6 @@ +Vinfinite=58.3;//volume of nitrogen evolved at infinite time//
+V0=19.3;//volume of nitrogen evolved at initial time//
+printf('Let V0,Vt,Vinfinite be the volumes of N2 evolved at the beginning,at time t and at infinite time(no more collection of N2 is observed)respectively,');
+printf('\n(Vinfinite-V0)is a measure of the total amount of material that can decompose,\ni.e the initial concentration a.\n(Vinfinite-Vt) is a measure of the amount of material that remains unreacted at time t,i.e (a-x),\nbecause this volume corresponds to the amount of material that can still decompose between time t and infinite time.');
+printf('\nk=(2.303/t)*(log10((Vinfinte-V0)/(Vinfinite-Vt))).');
+printf('\nThe average value of k can be found to be 6.54*10^-2per min or 1.09*10^-3per sec');
diff --git a/821/CH8/EX8.5/8_5.sce b/821/CH8/EX8.5/8_5.sce new file mode 100755 index 000000000..2dd2a24d2 --- /dev/null +++ b/821/CH8/EX8.5/8_5.sce @@ -0,0 +1,12 @@ +t=12.8;//half life of the particle in hours//
+k=0.693/(t*60);//value of k for the experiment in per min//
+printf('Value of k for the experiment=k=%fper min',k);
+printf('\n-dN/dt=rate=100*k*N=(0.693/12.8*60)*N\nN,the number of copper atoms required to produce 100Beta particles per minute\n we get N=(100*12.8*60)/0.693=1.108*10^5');
+w=63.5;//atomic weight of Cu in grams//
+AN=6.023*10^23;
+N=1.108*10^5;
+W=(N*w)/AN;//weight of Cu in grams//
+printf('\nWeight of Cu=W=%f=1.17*10^-17grams',W);
+printf('\nSince the maximum activity is 100Beta particles per minute,N=(a-x) at the end of six hours,i.e t=6 and N=1.108*19^5atoms');
+printf('\nAt zero time N0=a\n a-x=a*exp(-k*t)\nUpon solving the above equation we get N0=a=1.533*10^5atoms\nWeight of Cu to start with=1.66*10^-17grams.');
+printf('\nInitial activity=138.30000 disintegrations per minute');
diff --git a/821/CH8/EX8.7/8_7.sce b/821/CH8/EX8.7/8_7.sce new file mode 100755 index 000000000..e054409b6 --- /dev/null +++ b/821/CH8/EX8.7/8_7.sce @@ -0,0 +1,5 @@ +printf('The kinetic equation for a second order reaction involving unequal concentrations of the reactants is\nk2=(2.303/(t*(a-b)))*log10((b*(a-x))/(a*(b-x)))');
+k=2.312*10^-4;
+p=4.94*10^-3;//value of (a-b)//
+k2=(k*2.303)/p;//second order reaction rate//
+printf('\nSecond order reaction rate=k2=%flitre per mol per second',k2);
diff --git a/821/CH8/EX8.8/8_8.sce b/821/CH8/EX8.8/8_8.sce new file mode 100755 index 000000000..0686c18f5 --- /dev/null +++ b/821/CH8/EX8.8/8_8.sce @@ -0,0 +1,6 @@ +t1half=37.00;//half time for the first order reaction//
+t2half=19.2;//half time for the second order reaction//
+t3half=9.45;//half time for the third order reaction//
+printf('to know the order of the equation we can use 2^(n-1)=t1half/t2half');
+printf('\nby solving for first and second order n=1.95\nby solving for second and third order n=2.02');
+printf('\nby solving for first and third order n=1.98\nSo the order of the reaction=n=2');
diff --git a/821/CH9/EX9.1/9_1.sce b/821/CH9/EX9.1/9_1.sce new file mode 100755 index 000000000..a151a6422 --- /dev/null +++ b/821/CH9/EX9.1/9_1.sce @@ -0,0 +1,26 @@ +printf('Let M be the Molecular weight of the dye.Original concentration is 30.1/M mol litre^-1');
+I0=100;
+I=50;
+b=1;
+A=log10(I0/I);
+printf('\nFrom Beers law=A=%f',A);
+x=A/30.1;
+printf('\na/M=%f',x);
+c=15.05;
+I=70.7;
+printf('\nPercentage of light transmitted=I=%f',I);
+AI=100-I;
+printf('\nPercentage of light absorbed=AI=%f',AI);
+c=60.2;
+I=25;
+printf('\nPercentage of light transmitted=I=%f',I);
+AI=100-I;
+printf('\nPercentage of light absorbed=AI=%f',AI);
+printf('\nIt must be noted that it is absorbance A that is linearly related to concentration and not percentage light transmitted or absorbed');
+b=2;
+c=30.1;
+I=25;
+printf('\nPercentage of light transmitted=I=%f',I);
+b=4.32;
+printf('\nb=%fcm',b);
+
diff --git a/821/CH9/EX9.2/9_2.sce b/821/CH9/EX9.2/9_2.sce new file mode 100755 index 000000000..3ead4f7f8 --- /dev/null +++ b/821/CH9/EX9.2/9_2.sce @@ -0,0 +1,14 @@ +c=3*10^10;//velocity of light in cm//
+h=6.625*10^-27;//plank's constant//
+L=6.02*10^23;//Avagadro number//
+l=3020*10^-8;//wavelength of light radiation in cm//
+E=(L*h*c)/l;//value of one einstein in ergs//
+printf('Value of one einstein=E=%f=3.96*10^12ergs',E);
+LA=15000;//light absorbed in ergs per second//
+NE=LA/E;//number of einsteins absorbed per second//
+printf('\nNumber of Einsteins absorbed per second=NE=3.788*10^-9');
+QY=0.54;//Quantum yield for CO formation//
+N=QY*NE;//number of moles of CO formed per sec//
+printf('\nNumber of moles of Co formed per sec=N=2.046*10^-9');
+R=2.046*10^-9;//Rate of formation of CO in moles per sec//
+printf('\nRate of formation of CO=R=2.046*10^-9 moles per sec');
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