initial commit / add all books

92 files changed, 1874 insertions, 0 deletions

diff --git a/944/CH2/EX2.1/example2_1_TACC.sce b/944/CH2/EX2.1/example2_1_TACC.sce
new file mode 100755
index 000000000..039c33727
--- /dev/null
+++ b/944/CH2/EX2.1/example2_1_TACC.sce
@@ -0,0 +1,16 @@
+//example 2.1

+

+clear;

+clc;

+

+//Given:

+m=25;//weight of water vapour [grams]

+w=18;//molecular weight of water vapour [grams/mol]

+T=9.69;//increase in temperature [K]

+Qp=0.45;//heat supplied at constant pressure[KJ]

+

+

+//To find the molar constant pressure specific heat

+n=m/w;//no. of moles of water vapour

+Cp=Qp/(n*T);//specific heat capacity at constant pressure[KJ]

+printf("The specific heat capacity at constant pressure = %f KJ/K/mol",Cp)
\ No newline at end of file
diff --git a/944/CH2/EX2.2/example2_2_TACC.sce b/944/CH2/EX2.2/example2_2_TACC.sce
new file mode 100755
index 000000000..6a1428d38
--- /dev/null
+++ b/944/CH2/EX2.2/example2_2_TACC.sce
@@ -0,0 +1,17 @@
+//example 2.2

+

+clear;

+clc;

+

+//Given:

+m=16;//weight of oxygen [grams]

+w=32;//molecular weight of oxygen [grams/mol]

+T=300;//Temperature during compression [K]

+P1=1;//initial pressure of process [atm]

+P2=100;//final pressure of process[atm]

+R=8.314;//Universal gas constant [J/K/mol]

+

+//To find the minimum work of compression 

+n=m/w;//no. of moles of oxygen

+W=-n*R*T*log(P1/P2);

+printf("Mininmum work done to compress oxygen = %f J " , W )
\ No newline at end of file
diff --git a/944/CH3/EX3.1/example3_1_TACC.sce b/944/CH3/EX3.1/example3_1_TACC.sce
new file mode 100755
index 000000000..174d4cbda
--- /dev/null
+++ b/944/CH3/EX3.1/example3_1_TACC.sce
@@ -0,0 +1,16 @@
+//example 3.1

+

+clear;

+clc;

+

+//given:

+V1=14;//initial volume of cylinder in m3

+V2=9;//final volume of cylinder in m3

+P=2000;//pressure during the operation in N/m2

+U=(-6000);//internal energy of the system in J

+

+//to find energy transfered in form of heat:

+W=-P*(V2-V1);//work done during the operation in J

+Q=U-W;//energy tranfered in form of heat in J

+printf("energy tranfered in form of heat is %f J",Q);

+ 
\ No newline at end of file
diff --git a/944/CH3/EX3.10/example3_10_TACC.sce b/944/CH3/EX3.10/example3_10_TACC.sce
new file mode 100755
index 000000000..4795d7f85
--- /dev/null
+++ b/944/CH3/EX3.10/example3_10_TACC.sce
@@ -0,0 +1,13 @@
+//example 3.10

+

+clear;

+clc;

+disp("N2(g)+3H2(g)-> 2NH3(g)");

+

+//Given:

+H=-92.22;//standard heat of reaction [KJ/mol]

+

+//to find the standard heat of formation of one mole of product

+H1=H/2;//standard heat of formation of 1 mole [KJ/mol]

+printf("H(heat of formation of 1 mole of product)= %f KJ/mol" , H1);

+

diff --git a/944/CH3/EX3.11/example3_11_TACC.sce b/944/CH3/EX3.11/example3_11_TACC.sce
new file mode 100755
index 000000000..1e4bf843c
--- /dev/null
+++ b/944/CH3/EX3.11/example3_11_TACC.sce
@@ -0,0 +1,16 @@
+//example 3.11

+

+clear;

+clc;

+

+disp("C2H5OH(l)+3O2(g)->2CO2(g)+3H2O(l)");

+//Given:

+T=298;//temperature during the reaction[K]

+Hw=-285.83;//standard heat of formation of liquid water [KJ/mol]

+He=-277.69;//standard heat of formation of liquid ethanol[KJ/mol]

+Hco2=-393.51;//standard heat of formation of carbon dioxide[KJ/mol]

+Ho2=0;//standard heat of formation of oxygen gas[KJ/mol]

+

+//to find the standard heat of reaction

+H=2*Hco2+3*Hw-He-3*Ho2;//standard heat of reaction

+printf("H(standard heat of reaction)=%f KJ/mol",H);

diff --git a/944/CH3/EX3.12/example3_12_TACC.sce b/944/CH3/EX3.12/example3_12_TACC.sce
new file mode 100755
index 000000000..f43d390f8
--- /dev/null
+++ b/944/CH3/EX3.12/example3_12_TACC.sce
@@ -0,0 +1,16 @@
+//example 3.12

+

+clear;

+clc;

+

+disp("CO(g)+NO(g)->0.5N2(g)+CO2(g)");

+//Given:

+Hrxn=-374;//standard heat of reaction[KJ/mol]

+Hno=90.25;//standard heat of formation of NO[KJ/mol]

+Hco2=-393.51;//standard heat of formation of CO2[KJ/mol]

+Hn2=0;//standard heat of formation of N2[KJ/mol]

+T=298;//temperature of reaction [K]

+

+//to find the value of standard heat of formation of CO

+Hco=0.5*Hn2+Hco2-Hno-Hrxn;//standard heat of formation of CO[KJ/mol]

+printf("Hco(standard heat of formation)=%f KJ/mol",Hco);
\ No newline at end of file
diff --git a/944/CH3/EX3.13/example3_13_TACC.sce b/944/CH3/EX3.13/example3_13_TACC.sce
new file mode 100755
index 000000000..dd1bec523
--- /dev/null
+++ b/944/CH3/EX3.13/example3_13_TACC.sce
@@ -0,0 +1,23 @@
+//example 3.13

+

+clear;

+clc;

+

+//Given:

+//("C3H6(g)+H2(g)->C3H8(g);H1=-29.6 Kcal");

+//("C3H8(g)+5O2(g)->3CO2(g)+4H2O(l)  ;     H2=-530.6 Kcal");

+//("C(s)+O2(g)->CO2(g);H3=-94.0 Kcal");

+//("H2(g)+0.5O2(g)->H2O(l);H4=-68.3 Kcal. ");

+H1=-29.6;//the standard heat of hydrogenation of gaseous propylene to propane[Kcal]

+H2=-530.6;//the heat of combustion of propane[Kcal] 

+H3=-94.0;//the heat of formation of carbon dioxide[Kcal]

+H4=-68.3;//the heat of formation of liquid water[Kcal]

+

+

+//to find the heat of combustion and formation of propylene

+disp("C3H6(g)+4.5O2(g)->3CO2(g)+3H2O(l)");

+H5=[3*H3+4*H4]-[H1+H2];//[Kcal]

+printf("\n H(standard heat of combustion)=%f Kcal.\n",H5);

+disp("3C(s)+3H2(g)->C3H6(g)");

+H6=-H5+3*H3+3*H4;//[Kcal]

+printf("\n H(standard heat of formation)=%f Kcal. \n",H6);

diff --git a/944/CH3/EX3.14/example3_14_TACC.sce b/944/CH3/EX3.14/example3_14_TACC.sce
new file mode 100755
index 000000000..b948111f1
--- /dev/null
+++ b/944/CH3/EX3.14/example3_14_TACC.sce
@@ -0,0 +1,15 @@
+//example 3.14

+

+clear;

+clc;

+

+//Given:

+H1=-114.1;//standard heat of reaction:2NO(g)+O2(g)->2NO2(g);[KJ/mol]

+H2=-110.2;//standard heat of reaction:4NO2(g)+O2(g)->2N2O5(g);[KJ/mol]

+H3=180.5;//standard heat of reaction:N2(g)+O2(g)->2NO(g);[KJ/mol]

+

+

+//to find the heat of formation of N2O5

+//reacton:N2(g)+2.5O2(g)->N2O5(g)

+H4=(2*H1+H2+2*H3)/2;//standard heat of formation of N2O5[KJ/mol]

+printf("H(standard heat of formation of N2O5)=%f KJ/mol",H4);
\ No newline at end of file
diff --git a/944/CH3/EX3.15/example3_15_TACC.sce b/944/CH3/EX3.15/example3_15_TACC.sce
new file mode 100755
index 000000000..515df245c
--- /dev/null
+++ b/944/CH3/EX3.15/example3_15_TACC.sce
@@ -0,0 +1,15 @@
+//example 3.15

+

+clear;

+clc;

+

+//Given:

+Hc=-5645;//standard enthalpy of combustion of reaction:C12H22O11(s)+12O2(g)->12CO2(g)+11H2O(l) [KJ/mol]

+Hf1=-393.51;//standard heat of formation of CO2: C(s)+O2(g)->CO2(g) [KJ/mol]

+Hf2=-285.83;//standard heat of formation of H2O: H2(g)+0.5O2(g)->H2O(l) [KJ/mol]

+

+

+//to find the standard heat of formaton of solid sucrose

+//reaction:12C(s)+11H2(g)+5.5O2(g)->C12H22O11(s)

+Hf=12*Hf1+11*Hf2-Hc;//[KJ/mol]

+printf("Hf(standard heat of formation of solid sucrose)=%f KJ/mol",Hf);
\ No newline at end of file
diff --git a/944/CH3/EX3.16/example3_16_TACC.sce b/944/CH3/EX3.16/example3_16_TACC.sce
new file mode 100755
index 000000000..9e61c9877
--- /dev/null
+++ b/944/CH3/EX3.16/example3_16_TACC.sce
@@ -0,0 +1,20 @@
+//example 3.16

+

+

+clear;

+clc;

+

+//Given:

+Hf1=-46.11;//standard heat of formation of NH3 at 298K //reaction:0.5N2(g)+1.5H2(g)->NH3(g) [KJ/mol]

+Cp1=29.125;//molar heat capacity at constant pressure for N2(g)[J/K/mol]

+Cp2=28.824;//molar heat capacity at constant pressure for H2(g)[J/K/mol]

+Cp3=35.06;//molar heat capacity at constant pressure for NH3(g)[J/K/mol]

+T1=298;//initial temperature[K]

+T2=400;//final temperature[K]

+

+

+//to find the standard heat of formation of NH3 at 400K for same reaction

+Cp=Cp3-0.5*Cp1-1.5*Cp2;//[J/K/mol]

+T=T2-T1;//[K]

+Hf2=Hf1+Cp*0.001*T;//standard heat of formation for NH3 at 400K[KJ/mol]

+printf("\n Hf2(standard heat of formation for NH3 at 400K = %f KJ/mol. \n",Hf2);
\ No newline at end of file
diff --git a/944/CH3/EX3.17/example3_17_TACC.sce b/944/CH3/EX3.17/example3_17_TACC.sce
new file mode 100755
index 000000000..6fcf8fa4d
--- /dev/null
+++ b/944/CH3/EX3.17/example3_17_TACC.sce
@@ -0,0 +1,21 @@
+//example 3.17

+

+clear;

+clc;

+

+//Given:

+Cp1=poly([29.07, -0.836*10^-3, 20.1*10^-7],'T','c');//heat capacity for H2(g)

+Cp2=poly([25.72, 12.98*10^-3, -38.6*10^-7],'T','c');//heat capacity for O2(g)

+Cp3=poly([30.36, 9.61*10^-3, 11.8*10^-7],'T','c');//heat capacity for H2O(g)

+Hf1=-241820;//standard heat of formation H2O(g) at 298K :                H2(g)+ 0.5O2(g)->H2O(g) [J/mol]

+T1=298;//initial temperature[K]

+T2=1273;//final temperature[K]

+

+//to find the standard heat of formation of H2O(g) at 1273K

+Cp=Cp3-0.5*Cp2-Cp1;//heat capacity for the formation[J/K/mol]

+i=horner(Cp,1273)-horner(Cp,298);//[J/mol]

+Hf2=(Hf1-i)*0.001;//[KJ/mol]

+printf("The heat of formation of H2O at 1273K = %f KJ/mol", Hf2);

+

+

+

diff --git a/944/CH3/EX3.18/example3_18_TACC.sce b/944/CH3/EX3.18/example3_18_TACC.sce
new file mode 100755
index 000000000..cc246e53d
--- /dev/null
+++ b/944/CH3/EX3.18/example3_18_TACC.sce
@@ -0,0 +1,15 @@
+//example 3.18

+

+clear;

+clc;

+

+//given:

+H1=435;//bond dissociation energy for: CH4->CH3+H [KJ/mol]

+H2=364;//bond dissociation energy for:CH3->CH2+H [KJ/mol]

+H3=385;//bond dissociation energy for:CH2->CH+H [KJ/mol] 

+H4=335;//bond dissociation energy for:CH->C+H [KJ/mol]

+

+

+//to find the C-H bond energy of CH4 from the above bond energies

+H=(H1+H2+H3+H4)/4;//the bond energy for C-H bond in CH4 [KJ/mol]

+printf("\n H(the C-H bond energy in CH4)=%f KJ/mol. \n",H);

diff --git a/944/CH3/EX3.19/example3_19_TACC.sce b/944/CH3/EX3.19/example3_19_TACC.sce
new file mode 100755
index 000000000..54ec43556
--- /dev/null
+++ b/944/CH3/EX3.19/example3_19_TACC.sce
@@ -0,0 +1,16 @@
+//example 3.19

+

+clear;

+clc;

+

+//Given:

+H1=-84.68;//heat of formation : 2C(s)+3H2(g)->C2H6(g) [KJ/mol]

+H2=2*716.68;//heat of formation : 2C(s)->2C(g) [KJ]

+H3=3*436;//heat of formation : 3H2(g)->6H(g) [KJ]

+H4=412;//taking it as bond energy for one C-H bond[KJ/mol]

+

+

+//to find the C-C bond energy in ethane bond

+H=H2+H3-H1;//heat of reaction : C2H6(g)->2C(g)+6H(g) [KJ/mol]

+H5=H-6*H4;//bond energy for one C-C bond in ethane bond [KJ/mol]

+printf("\n Hc-c(bond energy for one C-C bond in ethane bond)=%f KJ/mol.\n",H5);

diff --git a/944/CH3/EX3.2/example3_2_TACC.sce b/944/CH3/EX3.2/example3_2_TACC.sce
new file mode 100755
index 000000000..d1293e8ea
--- /dev/null
+++ b/944/CH3/EX3.2/example3_2_TACC.sce
@@ -0,0 +1,17 @@
+//example 3.2

+

+clear;

+clc;

+

+//given:

+R=8.314;//universal gas constant [J/K/mol]

+T=300;//temperture for the process [K]

+U=0;//change in internal energy [J]

+V1=2.28;//initial volume [m3]

+V2=4.56;//final volume[m3]

+

+

+//to find the heat lost or gained by the system:

+W=2.303*R*T*log10(V2/V1);//work done during the process[J]

+Q=W;//heat lost or gained by the system[J]

+printf("the heat gained by the system is %f J",Q);
\ No newline at end of file
diff --git a/944/CH3/EX3.20/example3_20_TACC.sce b/944/CH3/EX3.20/example3_20_TACC.sce
new file mode 100755
index 000000000..df0078807
--- /dev/null
+++ b/944/CH3/EX3.20/example3_20_TACC.sce
@@ -0,0 +1,17 @@
+//example 3.20

+

+clear;

+clc;

+

+//Given:

+//MgBr2(s)-->Mg(s)+Br2(l)-->Mg(g)+Br2(l)-->Mg(g)+Br2(g)-->Mg(g)+2Br(g)-->Mg+2(g) + 2e(g) + 2Br(g)-->Mg+2(g) + 2Br-(g)

+H1=-524;//enthalpy of formation of MgBr2(s) from its element [KJ/mol]

+H2=148;//enthalpy of sublimation of Mg(s) [KJ/mol]

+H3=31;//enthalpy of vaporization of Br2(l) [KJ/mol]

+H4=193;//enthalpy of dissociation Br2 to 2Br(g) [KJ/mol]

+H5=2187;//enthalpy of ionization of Mg(g) to Mg+2(g) [KJ/mol]

+H6=-650;//enthalpy of formation of Br-(g) [KJ/mol]

+

+//to find the lattice enthalpy of magnesium bromide

+H=-H1+H2+H3+H4+H5+H6;//lattice enthalpy [KJ/mol]

+printf("\n H(lattice enthalpy of magnesium bromide)=%f KJ/mol.\n",H);

diff --git a/944/CH3/EX3.21/example3_21_TACC.sce b/944/CH3/EX3.21/example3_21_TACC.sce
new file mode 100755
index 000000000..c62e80231
--- /dev/null
+++ b/944/CH3/EX3.21/example3_21_TACC.sce
@@ -0,0 +1,25 @@
+//example 3.21

+

+clear;

+clc;

+

+//Given:

+Cp1=poly([26.0, 43.5*10^-3, -148.3*10^-7],'T','c');//heat capacity for CO2(g)[J/K/mol]

+Cp2=poly([30.36, 9.61*10^-3, 11.8*10^-7],'T','c');//heat capacity for H2O(g)[J/K/mol]

+Cp3=poly([27.30, -5.23*10^-3, -0.04*10^-7],'T','c');//heat capacity for N2(g)[J/K/mol]

+H1=-881.25;//heat of combustion of methane at 298K[KJ/mol]:                CH4 + 2O2(g) --> CO2(g) + 2H2O(l)

+H2=43.6;//heat of vaporization of water at 298K[KJ/mol]:        H2O(l)-->H2O(g)  

+T1=298;//initial temperature [K]

+

+

+//to find the maximum flame temperature when one mole of methane is burnt completely  in calculated amount of air(N2 to O2 ratio 4),at constant pressure

+H=(H1+2*H2);//enthalpy of reaction: CH4 + 2O2(g) --> CO2(g) + 2H2O(g) [KJ]

+printf("H(enthalpy of reaction)=%f KJ/mol.\n",H);

+printf("H(enthalpy of heating)=%f KJ/mol. \n",-H);

+Cp=Cp1+2*Cp2+8*Cp3;//[J/K/mol]

+P=poly([0 , 305.12 , 52.28*10^-3 , -41.66*10^-7] , 'T' , 'c');

+i=horner(P,298);

+c1=-H*1000+i;

+P1=poly([-c1 , 305.12 , 52.28*10^-3 , -41.66*10^-7] , 'T' , 'c');

+T2=roots(P1)

+printf("T2 (maximum flame temperature)= %f K",T2(1));
\ No newline at end of file
diff --git a/944/CH3/EX3.3/example3_3_TACC.sce b/944/CH3/EX3.3/example3_3_TACC.sce
new file mode 100755
index 000000000..bed7e2a68
--- /dev/null
+++ b/944/CH3/EX3.3/example3_3_TACC.sce
@@ -0,0 +1,21 @@
+//example 3.3

+

+clear;

+clc;

+

+//given:

+H=29.2;//latent heat of vaporisation[KJ/mol]

+T=332;//temperature of the system[K]

+R=8.314;//universal gas constant [J/K/mol]

+

+

+//to find Q,W,U for 1 mole of bromine vaporizes

+//where Q is heat absorbed or evolved

+//W is the work done by system

+//U is the change in internak energy

+Qp=H;//at constant pressure [KJ]

+W=-R*0.001*T;//workdone [KJ]

+U=Qp+W;//change in internal energy[KJ]

+printf("heat absorbed by the bromine vapours is %f KJ",Qp);

+printf("\nworkdone during the process is %f KJ",W);

+printf("\nchange in internal energy of the system is %f KJ",U);

diff --git a/944/CH3/EX3.4/example3_4_TACC.sce b/944/CH3/EX3.4/example3_4_TACC.sce
new file mode 100755
index 000000000..88261fcc1
--- /dev/null
+++ b/944/CH3/EX3.4/example3_4_TACC.sce
@@ -0,0 +1,18 @@
+//example 3.4

+

+clear;

+clc;

+

+//given:

+disp("C7H16(l) + 11O2(g) -> 7CO2(g) + 8H2O(l)");

+n=-4;//change in no. of moles when reaction proceeds from reactants to products

+T=298;//temperature of the process [K]

+R=8.314;//universal gas constant [J/K/mol]

+Qv=-4800;//heat energy at constant volume [KJ]

+

+

+//to find change in enthalpy of the process:

+U=Qv;//change in internal energy of system [KJ]

+H=U+n*R*0.001*T;//change in enthalpy of the system[KJ]

+printf("the change in enthalpy of system is %f KJ",H);

+

diff --git a/944/CH3/EX3.5/example3_5_TACC.sce b/944/CH3/EX3.5/example3_5_TACC.sce
new file mode 100755
index 000000000..bdb5d09e2
--- /dev/null
+++ b/944/CH3/EX3.5/example3_5_TACC.sce
@@ -0,0 +1,24 @@
+//example 3.5

+

+clear;

+clc;

+

+//given:

+n=1;//number of moles of an given ideal gas

+T=298;//temperature for the process[K]

+V1=8.3;//initial volume of the ideal gas[m3]

+V2=16.8;//final volume of the ideal gas[m3]

+R=8.314;//universal gas constant[J/K/mol]

+

+

+//to find the Q,W,H

+//where Q is heat absorbed or evolved by the system

+//W is the net workdone 

+//H is the change in enthalpy of system

+W=-2.303*R*T*log10(V2/V1);//[J]

+Q=-W;//[J]

+disp("H=U+PV ,where U is change in internal energy which is zero due to isothermal process");

+disp("PV where V is change in volume of system ,PV=RT & RT==0 since T i.e change in temp is zero for system");

+disp("Therefore,the change in enthalpy is 0J");

+printf("the workdone by system is %f J",W);

+printf("\nthe heat evolved is %f J",Q);

diff --git a/944/CH3/EX3.6/example3_6_TACC.sce b/944/CH3/EX3.6/example3_6_TACC.sce
new file mode 100755
index 000000000..fc4bf00e4
--- /dev/null
+++ b/944/CH3/EX3.6/example3_6_TACC.sce
@@ -0,0 +1,16 @@
+//example 3.6

+

+clear;

+clc;

+

+//given:

+T1=323;//intial temperature of water[K]

+T2=373;//final temperature of water[K]

+Cp=75.29;//specific heat of water[J/K/mol]

+w=100;//weight of water[g]

+mol.wt=18;//molecular weight of water[g/mol]

+

+//to find the change in enthalpy(H) of water 

+n=w/mol.wt;//no. of moles of water[moles]

+H=(n*Cp*(T2-T1))*0.001;//change in enthalpy of water[J]

+printf("The change in enthalpy of water is %f J",H);

diff --git a/944/CH3/EX3.7/example3_7_TACC.sce b/944/CH3/EX3.7/example3_7_TACC.sce
new file mode 100755
index 000000000..069c9e78b
--- /dev/null
+++ b/944/CH3/EX3.7/example3_7_TACC.sce
@@ -0,0 +1,16 @@
+//example 3.7

+

+clear;

+clc;

+

+disp("SO2 + 0.5O2 -> SO3");

+//given:

+U=-97030;//heat of reaction[J]

+n=1-(1+0.5);//change in no. of moles 

+R=8.314;//universal gas constant[J/K/mol]

+T=298;//temperature during the reaction[K]

+

+

+//to find the change in enthalpy of reaction(H)

+H=U+n*R*T;//change in enthalpy of reaction[J]

+printf("The change in enthalpy of reaction is %f J ",H);

diff --git a/944/CH3/EX3.8/example3_8_TACC.sce b/944/CH3/EX3.8/example3_8_TACC.sce
new file mode 100755
index 000000000..16d13d024
--- /dev/null
+++ b/944/CH3/EX3.8/example3_8_TACC.sce
@@ -0,0 +1,27 @@
+//example 3.8

+

+clear;

+clc;

+

+disp("i.C(s) + O2(g) -> CO2(g)");

+//Given:

+H1=-393.5;//change in enthalpy [KJ/mol]

+T1=298;//temperature [K]

+n1=0;//change in no. of moles in reaction moving in forward direction

+R=0.008314;//universal gas constant [KJ/K/mol]

+

+//to find the change in internal energy (U) of given reaction

+U1=H1-n1*R*T1;//change in internal energy [KJ]

+printf("The change in internal energy is %f KJ/mol",U1);

+

+disp("ii.C(s) + 0.5O2 -> CO(g)");

+//Given:

+H2=-110.5;//change in enthalpy[KJ/mol]

+T2=298;//temperature[K]

+n2=1-0.5;//change in no. of moles in reaction moving in forward direction

+R=0.008314;//universal gas constant [KJ/K/mol]

+

+//to find the change in internal energy (U) of given reaction

+U2=H2-n2*R*T2;//change in internal energy [KJ]

+printf("The change in internal energy is %f KJ/mol",U2);

+

diff --git a/944/CH3/EX3.9/example3_9_TACC.sce b/944/CH3/EX3.9/example3_9_TACC.sce
new file mode 100755
index 000000000..558fdb051
--- /dev/null
+++ b/944/CH3/EX3.9/example3_9_TACC.sce
@@ -0,0 +1,15 @@
+//example 3.9

+

+clear;

+clc;

+

+disp("The standard heat of combustion of");

+disp("2C6H6(l)+ 15O2(g)-> 12 CO2(g)+ 6 H2O(l)");

+disp("H1(standard heat of combustion)=-6536 KJ/mol");

+//Given:

+H1=-6536;//standard heat of combustion [KJ/mol]

+

+//to find the standard heat of rxn for combustion of 1 mole of C6H6

+disp("C6H6(l)+ 7.5 O2(g)-> 6 CO2(g)+ 6 H2O(l)");

+H2=H1/2;//standard heat of combustion[KJ/mol]

+printf(" H2(standard heat of combustion for 1 mole of C6H6)= %f KJ/mol",H2);

diff --git a/944/CH4/EX4.1/example4_1_TACC.sce b/944/CH4/EX4.1/example4_1_TACC.sce
new file mode 100755
index 000000000..f5857d992
--- /dev/null
+++ b/944/CH4/EX4.1/example4_1_TACC.sce
@@ -0,0 +1,16 @@
+//example 4.1

+

+clear;

+clc;

+

+//Given:

+M=2.5;//mass of the substance[Kg]

+x=0.6;//mass fraction for vapour phase 

+P=7;//pressure [atm]

+T=438;//temperature[K]

+

+//to find the mass of substance present in liquid and vapour phase 

+Ml=(1-x)*M;//mass fraction of liquid phase[Kg]

+Mg=x*M;//mass fraction of vapour phase[Kg]

+

+printf("M(liquid phase)=%f Kg , M(vapour phase)=%f Kg",Ml,Mg)
\ No newline at end of file
diff --git a/944/CH4/EX4.2/example4_2_TACC.sce b/944/CH4/EX4.2/example4_2_TACC.sce
new file mode 100755
index 000000000..c2fcb8bc6
--- /dev/null
+++ b/944/CH4/EX4.2/example4_2_TACC.sce
@@ -0,0 +1,16 @@
+//example 4.2

+

+clear;

+clc;

+

+//Given:

+Vl=0.0177;//specific volume of saturated liquid[m3/Kg]

+Vg=4.43;//specific volume of saturated vapour[m3/Kg]

+P=7;//pressure[atm]

+T=438;//temperature[K]

+x=0.6;//fraction of vapour phase

+M=2.5;//mass of the substance[Kg]

+

+//to find the total volume occupied by  the mixture 

+V=[(1-x)*Vl+x*Vg]*M;//total volume occupied [m3]

+printf("Total volume occupied =%f m3", V)

diff --git a/944/CH4/EX4.3/example4_3_TACC.sce b/944/CH4/EX4.3/example4_3_TACC.sce
new file mode 100755
index 000000000..b3b1177f9
--- /dev/null
+++ b/944/CH4/EX4.3/example4_3_TACC.sce
@@ -0,0 +1,16 @@
+//example 4.3

+

+clear;

+clc;

+

+//Given:

+M=2.5;//mass of a substance[Kg]

+x=0.6;//fraction of vapour phase 

+Ug=1105;//specific internal energy of saturated vapour[J/Kg]

+Ul=298;//specific internal energy of saturated liquid[J/Kg] 

+

+

+//to find the total internal energy of the mixture

+U=M*[(1-x)*Ul+x*Ug];

+printf("The total internal energy of the mixture = %f J", U)

+ 
\ No newline at end of file
diff --git a/944/CH5/EX5.1/example5_1_TACC.sce b/944/CH5/EX5.1/example5_1_TACC.sce
new file mode 100755
index 000000000..e64de8835
--- /dev/null
+++ b/944/CH5/EX5.1/example5_1_TACC.sce
@@ -0,0 +1,18 @@
+//example 5.1

+

+clear;

+clc;

+

+//given:

+T1=373;//initial temperature [K]

+T2=573;//final temperature [K]

+Q2=750;//Heat absorbed by carnot engine[J]

+

+//to find the workdone , heat rejected , and efficiency of the engine

+e=(T2-T1)/T2;//efficiency of the engine

+W=e*Q2;//Workdone by the engine[J]

+Q1=T1*Q2/T2;//Heat rejected by the engine[J]

+

+printf("Efficiency of the engine = %f ", e);

+printf("\n Workdone by the engine = %f J", W);

+printf("\n Heat rejected by the engine = %f J", Q1);

diff --git a/944/CH5/EX5.10/example5_10_TACC.sce b/944/CH5/EX5.10/example5_10_TACC.sce
new file mode 100755
index 000000000..27c8c22e3
--- /dev/null
+++ b/944/CH5/EX5.10/example5_10_TACC.sce
@@ -0,0 +1,18 @@
+//example 5.10

+

+clear;

+clc;

+

+//Given:

+mass=32;//weight of methane gas[gm]

+P1=6*10^5;//Initial temperature[N/m2]

+P2=3*10^5;//Final pressure[N/m2]

+mol.wt=16;//molecular weight of methane gas[g/mol]

+T=298;//Temperature[K]

+//Isothermal process

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the change in entropy of the gas

+n=mass/mol.wt;//no. of moles

+S=n*R*log(P1/P2);//change in entropy of gas[J/K]

+printf("The change in entropy of the gas is %f J/K",S);
\ No newline at end of file
diff --git a/944/CH5/EX5.11/example5_11_TACC.sce b/944/CH5/EX5.11/example5_11_TACC.sce
new file mode 100755
index 000000000..f50b96dec
--- /dev/null
+++ b/944/CH5/EX5.11/example5_11_TACC.sce
@@ -0,0 +1,22 @@
+//example 5.11

+

+clear;

+clc;

+

+//Given:

+N=3;//Total no. of balls

+Nb=2;//Black balls

+Nw=1;//white ball

+

+//To find the total no. possible configuration

+w=prod(1:N)/prod(1:Nb)/prod(1:Nw);

+

+printf("The total no. of possible configurations are %f ",w);

+

+

+

+

+

+

+

+

diff --git a/944/CH5/EX5.12/example5_12_TACC.sce b/944/CH5/EX5.12/example5_12_TACC.sce
new file mode 100755
index 000000000..e5c02c411
--- /dev/null
+++ b/944/CH5/EX5.12/example5_12_TACC.sce
@@ -0,0 +1,19 @@
+//example 5.12

+

+clear;

+clc;

+

+//Given:

+n=1;//no. of moles

+T=273;//temperature [K]

+Hf=6000;//enthalpy of fusion at 273K [J/mol]

+k=1.38*(10^-23);//boltzmann constant[J/K]

+

+//To find the relative number of distinguishable quantum states in 1 mole of water and ice at 273K 

+

+p=Hf/(k*T)/2.303;

+w=10^(p);//w is the relative no. of distinguishable quantum states

+printf("The relative no. of distinguishable quantum states in 1 mole of water and ice at 273K is %f",w);

+

+

+

diff --git a/944/CH5/EX5.13/example5_13_TACC.sce b/944/CH5/EX5.13/example5_13_TACC.sce
new file mode 100755
index 000000000..3c7bc5453
--- /dev/null
+++ b/944/CH5/EX5.13/example5_13_TACC.sce
@@ -0,0 +1,15 @@
+//example 5.13

+

+clear;

+clc;

+

+//Given:

+T=300;//temperature[K]

+n=4;//no. of moles of an ideal gas

+P1=2.02*10^5;//initial pressure[N/m2]

+P2=4.04*10^5;//final pressure[N/m2]

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the value of Gibb's free energy

+G=n*R*T*log(P2/P1);//[J]

+printf(" The change in Gibbs free energy is %f J",G);

diff --git a/944/CH5/EX5.14/example5_14_TACC.sce b/944/CH5/EX5.14/example5_14_TACC.sce
new file mode 100755
index 000000000..222526ca6
--- /dev/null
+++ b/944/CH5/EX5.14/example5_14_TACC.sce
@@ -0,0 +1,16 @@
+//example 5.14

+

+clear;

+clc;

+

+//Given:

+n=1;//no. of moles

+T=300;//temperature[K]

+V1=2;//initial volume[m3]

+V2=20;//final volume[m3]

+R=8.314;//Universal gas constant[J/K/mol]

+

+

+//To find the work function or Helmholts free energy

+A=-n*R*T*log(V2/V1);//Change in work function[J/mol]

+printf("The change in Helmholts free energy is %f J/mol",A);
\ No newline at end of file
diff --git a/944/CH5/EX5.15/example5_15_TACC.sce b/944/CH5/EX5.15/example5_15_TACC.sce
new file mode 100755
index 000000000..a30fc396e
--- /dev/null
+++ b/944/CH5/EX5.15/example5_15_TACC.sce
@@ -0,0 +1,23 @@
+//example 5.15

+

+clear;

+clc;

+

+disp("C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l)");

+//Given:

+T=298;//Temperature[k]

+R=8.314;//Universal gas constant[J/K/mol]

+S=182.45;//standard entropy change at 298K [J/K]

+U=-2808;//change in internal energy at 298K[KJ/mol]

+//reaction is taking place in bomb calorimeter so no volume change 

+//therefore U=Q at constant volume

+

+//To find the energy change that can be extracted as heat and work 

+A=U-T*S*0.001;//Energy extracted as heat[KJ/mol]

+Wmax=A;//work done [KJ/mol]

+dn=6-6;//change in no. of moles

+H=U+dn*R*T;//Change in enthalpy of the bomb calorimeter[KJ]

+printf("The energy change that can be extracted as heat is %f KJ/mol", A);

+printf("\nThe energy change that can be extracted as work is %f KJ/mol", -A);

+printf("\nThe change in enthalpy of bomb calorimeter is %f KJ/mol",H);

+

diff --git a/944/CH5/EX5.16/example5_16_TACC.sce b/944/CH5/EX5.16/example5_16_TACC.sce
new file mode 100755
index 000000000..729b242ed
--- /dev/null
+++ b/944/CH5/EX5.16/example5_16_TACC.sce
@@ -0,0 +1,20 @@
+//example 5.16

+

+clear;

+clc;

+

+disp("C8H18(g)+12.5O2(g)-->8CO2(g)+9H2O(l)");

+//Given:

+T=298;//temperature[K]

+S=421.5;//change in entropy[J/K]

+H=-5109000;//Heat of reaction[J]

+R=8.314;//Universal gas constant[J/K/mol]

+dn=8-(1+12.5);//change in no. of moles

+

+

+//To find the helmholts free energy and Gibbs free energy

+U=H;//[J]

+A=U-T*S;//Change in helmholts free energy[J]

+G=A+dn*R*T;//Change in Gibbs free energy[J]

+printf("The change in Helmholts free energy is %f J",A);

+printf("\nThe change in Gibbs free energy is %f J",G);
\ No newline at end of file
diff --git a/944/CH5/EX5.17/example5_17_TACC.sce b/944/CH5/EX5.17/example5_17_TACC.sce
new file mode 100755
index 000000000..0944668df
--- /dev/null
+++ b/944/CH5/EX5.17/example5_17_TACC.sce
@@ -0,0 +1,28 @@
+//example 5.17

+

+clear;

+clc;

+

+disp("C3H6(g)+4.5O2(g)-->3CO2(g)+3H2O(l)");

+//Given:

+S=-339.23;//standard change in entropy [J/K]

+T=298;//temperature[K]

+Hf1=20.42;//enthalpy of formation of C3H6(g)[J]

+Hf2=-393.51;//enthalpy of formation of CO2(g)[J]

+Hf3=-285.83;//enthalpy of formation of H2O(l)[J]

+dn=3-4.5-1;//change in no. of moles

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the Helmholts free energy and Gibbs free energy

+H=3*Hf2+3*Hf3-Hf1;//Enthalpy of the reaction[J]

+U=H-dn*R*0.001*T;//Change in internal energy of the reaction[J]

+A=U-T*S*0.001;//Helmholts free energy change[J]

+G=A+dn*R*0.001*T;//Gibbs free energy change[J]

+printf("The change in Helmholts free energy is %f J",A);

+printf("\nThe change in Gibbs free energy is %f J ",G);

+

+

+

+

+

+

diff --git a/944/CH5/EX5.19/example5_19_TACC.sce b/944/CH5/EX5.19/example5_19_TACC.sce
new file mode 100755
index 000000000..8a3a387e5
--- /dev/null
+++ b/944/CH5/EX5.19/example5_19_TACC.sce
@@ -0,0 +1,19 @@
+//example 5.19

+

+clear;

+clc;

+

+disp("CH4(g)+2O2(g)-->CO2(g)+2H2O(l)");

+

+//Given:

+S1=-242.98;//standard entropy change for the combustion reaction[J/K]

+Hf1=-74.81;//Enthalpy of formation of CH4(g)[KJ/mol]

+Hf2=-393.51;//Enthalpy of formation of CO2(g)[KJ/mol]

+Hf3=-285.83;//Enthalpy of formation of H2O(l)[KJ/mol]

+T=298;//temperature[K]

+

+//To find the total entropy change 

+H=Hf2+2*Hf3-Hf1;//Change in enthalpy of reaction[KJ]

+S2=-H*1000/T;//Change in entropy of the surrounding[J/K]

+Stotal=(S1+S2)*0.001;//Total entropy change 

+printf("The total change in entropy is %f KJ/K",Stotal);

diff --git a/944/CH5/EX5.2/example5_2_TACC.sce b/944/CH5/EX5.2/example5_2_TACC.sce
new file mode 100755
index 000000000..b22f97a64
--- /dev/null
+++ b/944/CH5/EX5.2/example5_2_TACC.sce
@@ -0,0 +1,13 @@
+//example 5.2

+

+clear;

+clc;

+

+//given:

+T1=250;//temperature of heat rejection[K]

+T2=1000;//temperature of heat absorption[K]

+

+//to analyse the efficiency of the engine 

+e=1-(T1/T2);

+printf("Efficiency of the corresponding carnot engine = %f\n",e);

+disp(" Therefore , the inventors claim of 80% efficiency is absurd.The patent application should be rejected");

diff --git a/944/CH5/EX5.20/example5_20_TACC.sce b/944/CH5/EX5.20/example5_20_TACC.sce
new file mode 100755
index 000000000..79a2b8b6a
--- /dev/null
+++ b/944/CH5/EX5.20/example5_20_TACC.sce
@@ -0,0 +1,17 @@
+//example 5.20

+

+clear;

+clc;

+

+disp("2H2(g)+O2(g)-->2H2O(l)");

+//Given:

+Hf1=-285.83;//standard enthalpy of formation of H2O(l)[KJ/mol]

+S=-327;//Standard entropy change for the same reaction[J/K]

+T=298;//temperature[K]

+

+

+//To find the spontanity of the reaction

+H=2*Hf1-0-0;//Enthalpy of the reaction[KJ/mol]

+G=H-T*S*0.001;//Change in Gibbs free energy[KJ]

+printf("The change in Gibbs free energy is %f KJ\n ",G);

+disp("As change in Gibbs free energy is negative.Therefore,the reaction is spontaneous")

diff --git a/944/CH5/EX5.21/example5_21_TACC.sce b/944/CH5/EX5.21/example5_21_TACC.sce
new file mode 100755
index 000000000..0f6e02820
--- /dev/null
+++ b/944/CH5/EX5.21/example5_21_TACC.sce
@@ -0,0 +1,24 @@
+//example 5.21

+

+clear;

+clc;

+

+disp("CH4(g)+2O2(g)-->CO2(g)+2H2O(l)");

+//Given:

+S=-242.98;//standard entropy change for reaction [J/K]

+T=298;//temperature[K]

+Gf1=-50.72;//standard Gibbs free energy of formation for CH4(g)[KJ/mol]

+Gf2=-394.36;//standard Gibbs free energy of formation for CO2(g)[KJ/mol]

+Gf3=-237.13;//standard Gibbs free energy of formation for H2O(l)[KJ/mol]

+

+

+//To find the standard enthalpy of reaction 

+G=Gf2+2*Gf3-Gf1;//Standard Gibbs free energy for reaction[KJ/mol]

+H=G+T*S*0.001;//Standard enthalpy of reaction [KJ]

+printf("The standard enthalpy of reaction is %f KJ",H);

+

+

+

+

+

+

diff --git a/944/CH5/EX5.22/example5_22_TACC.sce b/944/CH5/EX5.22/example5_22_TACC.sce
new file mode 100755
index 000000000..b214f92b4
--- /dev/null
+++ b/944/CH5/EX5.22/example5_22_TACC.sce
@@ -0,0 +1,19 @@
+//example 5.22

+

+clear;

+clc;

+

+disp("C6H12O6(s)+6O2(g)-->6CO2(g)+6H2O(l)");

+//Given:

+mass=25;//mass of glucose for combustion under standard condition[gm]

+T=298;//temperature[K]

+Gf1=-910;//Standard Gibbs free energy of formation for C6H12O6[KJ/mol]

+Gf2=-394.4;//Standard Gibbs free energy of formation for CO2(g)[KJ/mol]

+Gf3=-237.13;//Standard Gibbs free energy of formation for H2O(l)[KJ/mol]

+mol.wt=180;//molecular weight of glucose[gm/mol]

+

+//To find the maximum energy that can be extracted as non-expansion work is equal to the change in free energy of the system

+G=6*Gf2+6*Gf3-Gf1

+n=mass/mol.wt;//no. of moles

+Gactual=G*n;//Gibbs free energy for the combustion of 0.139mol of glucose 

+printf("The energy that can be extracted as non-expansion work is %f KJ",-Gactual);

diff --git a/944/CH5/EX5.23/example5_23_TACC.sce b/944/CH5/EX5.23/example5_23_TACC.sce
new file mode 100755
index 000000000..1d7f9a995
--- /dev/null
+++ b/944/CH5/EX5.23/example5_23_TACC.sce
@@ -0,0 +1,13 @@
+//example 5.23

+

+clear;

+clc;

+

+//Given:

+a=1.39*10^-2;//constant for a vanderwaal's gas[lit2.atm/mol2]

+b=3.92*10^-2;//constant for a vanderwaal's gas[lit2.atm/mol2]

+R=0.082;//Universal gas constant[lit.atm/deg/mol]

+

+//To find the value of the inversion temperature for the gas

+Ti=(2*a)/(R*b);//inversion temperature [K]

+printf("The inversion temperature for the gas is %f K",Ti);

diff --git a/944/CH5/EX5.26/example5_26_TACC.sce b/944/CH5/EX5.26/example5_26_TACC.sce
new file mode 100755
index 000000000..13204e17c
--- /dev/null
+++ b/944/CH5/EX5.26/example5_26_TACC.sce
@@ -0,0 +1,22 @@
+//example 5.26

+

+

+clear;

+clc;

+

+//Given:

+T=169.25;//Boiling point[K]

+R=8.314;//Universal gas constant[J/K/mol]

+disp("dlnP/dT=He/R*T^2");

+disp("dlnP/dT=(2.303*834.13/T^2)+(1.75/T)-(2.30*8.375*10^-3)");

+disp("Therefore using these two equations we calculate the He(enthalpy) of ethylene");

+

+//To find the Enthalpy of vaporization of ethylene

+x=(2.303*834.13/T^2)+(1.75/T)-(2.30*8.375*10^-3);//it is dlnP/dT

+He=R*0.001*T^2*x;//Enthalpy of vaporization[J/mol]

+printf("\n\nThe Enthalpy of vaporization of ethylene at its boiling point is %f KJ/mol",He);

+

+

+

+

+

diff --git a/944/CH5/EX5.27/example5_27_TACC.sce b/944/CH5/EX5.27/example5_27_TACC.sce
new file mode 100755
index 000000000..eaaa9e10f
--- /dev/null
+++ b/944/CH5/EX5.27/example5_27_TACC.sce
@@ -0,0 +1,16 @@
+//example 5.27

+

+clear;

+clc;

+

+//Given:

+P1=101.3;//Initial Pressure[KPa]

+P2=60;//Final Pressure[KPa]

+He=31.8;//Enthalpy of vaporization[KJ/mol]

+R=8.314;//Universal gas constant[J/K/mol]

+T1=353.2;//boiling point of benzene at 101.3KPa[K]

+

+//To find the boiling point of b/enzene at 60KPa

+x=(T1^-1)-(R*0.001*log(P2/P1)/He);

+T2=x^-1;//Boiling point of benzene at 60KPa

+printf("The boiling point of benzene at 60KPa is %f K",T2);
\ No newline at end of file
diff --git a/944/CH5/EX5.28/example5_28_TACC.sce b/944/CH5/EX5.28/example5_28_TACC.sce
new file mode 100755
index 000000000..e8f35cf17
--- /dev/null
+++ b/944/CH5/EX5.28/example5_28_TACC.sce
@@ -0,0 +1,21 @@
+//example 5.28

+

+clear;

+clc;

+

+//Given:

+P1=0.016;//Vapour pressure of pure ethanol at 273K[bar]

+P2=0.470;//Vapour pressure of pure ethanol at 333K[bar]

+T1=273;//initial temperature [K]

+T2=333;//final temperature[K]

+R=8.314;//Universal gas constant[J/K/mol]

+P=1.01;//vapour pressure at normal boiling point[bar]

+

+

+//To find the molar enthalpy of vapourization

+x=(T2^-1)-(T1^-1);

+He=-R*0.001*log(P2/P1)/x;//molar enthalpy of vaporization[J/mol]

+t=(T2^-1)-(R*0.001*log(P/P2)/He);

+T=(t^-1)-273;//normal boiling point [C]

+printf("The molar enthalpy of vapourization is %f J/mol",He);

+printf("\n\nThe normal boiling point for pure ethanol is %f C",T);

diff --git a/944/CH5/EX5.29/example5_29_TACC.sce b/944/CH5/EX5.29/example5_29_TACC.sce
new file mode 100755
index 000000000..d29393999
--- /dev/null
+++ b/944/CH5/EX5.29/example5_29_TACC.sce
@@ -0,0 +1,20 @@
+//example 5.29

+

+clear;

+clc;

+

+//Given:

+T2=353.2;//normal boiling point of benzene at 1.01325bar[K]

+T1=298;//temperature [K]

+R=8.314;//Universal gas constant[J/K/mol]

+P2=1.01325;//Vapour pressure of benzene[bar]

+//benzene obey's Trouton's rule

+disp(" from Troutons rule , ");

+disp(" He/Tb=85J/K/mol");

+

+//To find the vapour pressure of benzene at 298K

+He=85*T2;//molar enthalpy of vapourization[J/K/mol]

+x=(T2^-1)-(T1^-1);

+t=-He*x/R;

+P1=P2/exp(t);

+printf("\nThe vapour pressure of benzene at 298K is %f bar",P1);

diff --git a/944/CH5/EX5.3/example5_3_TACC.sce b/944/CH5/EX5.3/example5_3_TACC.sce
new file mode 100755
index 000000000..7bc5161de
--- /dev/null
+++ b/944/CH5/EX5.3/example5_3_TACC.sce
@@ -0,0 +1,15 @@
+//example 5.3

+

+clear;

+clc;

+

+//Given:

+T1=323;//temperature [K]

+T2=423;//temperature [K]

+W=1.3;//work [KJ]

+

+

+//To find the minimum heat required from heat source to yield the above work

+e=(T2-T1)/T2;//efficiency

+Q2=W/e;//minimum heat withdrawal from heat source[KJ]

+printf("Minimum heat withdrawal from heat source=%f KJ",Q2);
\ No newline at end of file
diff --git a/944/CH5/EX5.30/example5_30_TACC.sce b/944/CH5/EX5.30/example5_30_TACC.sce
new file mode 100755
index 000000000..a103f03d5
--- /dev/null
+++ b/944/CH5/EX5.30/example5_30_TACC.sce
@@ -0,0 +1,13 @@
+//example 5.30

+

+clear;

+clc;

+

+//Given:

+c=1;//no. of components(only CO2)

+p=2;//no. of phases(liquid + gas)

+

+//To find the degrees of freedom 

+F=c-p+2;//degree of freedom

+printf("Degrees of freedom is %f \n\n",F);

+disp("Degrees of freedom 1 means that either pressure or temperature can be varied independently,i.e.when temperature is fixed,pressure is automatically fixed");
\ No newline at end of file
diff --git a/944/CH5/EX5.31/example5_31_TACC.sce b/944/CH5/EX5.31/example5_31_TACC.sce
new file mode 100755
index 000000000..3e01eb3fe
--- /dev/null
+++ b/944/CH5/EX5.31/example5_31_TACC.sce
@@ -0,0 +1,16 @@
+//example 5.31

+

+clear;

+clc;

+

+//Given:

+c=1;//no. of components

+p=1;//no. of phases

+

+//To find the values of degrees of freedom

+F=c-p+2;//Degrees of freedom

+printf("Degrees of freedom is %f\n\n",F);

+disp("Degrees of freedom 2 means both the pressure and temperature can be varied independently");

+

+

+

diff --git a/944/CH5/EX5.32/example5_32_TACC.sce b/944/CH5/EX5.32/example5_32_TACC.sce
new file mode 100755
index 000000000..26ebf23d7
--- /dev/null
+++ b/944/CH5/EX5.32/example5_32_TACC.sce
@@ -0,0 +1,14 @@
+//example 5.32

+

+clear;

+clc;

+

+//Given:

+P=1.75*10^-5;//Vapour pressure of pure water at 293K[torr]

+dP=1.1*10^-7;//Lowering in vapour pressure of water

+//To find the mole fraction of sucrose,so that the vapour pressure of water will be lowered by dP

+x=dP/P;//mole fraction of sucrose

+disp(x,'The mole fraction of sucrose is ');

+

+

+

diff --git a/944/CH5/EX5.33/example5_33_TACC.sce b/944/CH5/EX5.33/example5_33_TACC.sce
new file mode 100755
index 000000000..123871334
--- /dev/null
+++ b/944/CH5/EX5.33/example5_33_TACC.sce
@@ -0,0 +1,14 @@
+//example 5.33

+

+clear;

+clc;

+

+//Given:

+P=94.6;//The vapour pressure of pure benzene at 298K[torr]

+n1=20;//no. of moles of pure benzene

+n2=5;//no. of moles of pure naphthalene

+

+//To find the partial vapour pressure of benzene over the solution

+x=n1/(n1+n2);//(mole fraction of benzene)

+p=x*P;//the partial vapour pressure of benzene[torr]

+printf("The partial vapour pressure of benzene is %f torr",p);
\ No newline at end of file
diff --git a/944/CH5/EX5.34/example5_34_TACC.sce b/944/CH5/EX5.34/example5_34_TACC.sce
new file mode 100755
index 000000000..44bb57ffe
--- /dev/null
+++ b/944/CH5/EX5.34/example5_34_TACC.sce
@@ -0,0 +1,18 @@
+//example 5.34

+

+clear;

+clc;

+

+//Given:

+x=0.28;//mole fraction of solute

+R=8.314;//Universal gas constant[J/K/mol]

+T=298;//temperature[K]

+

+//To find the reduction in chemical potential

+du=R*T*log(1-x);//reduction in chemical potential[J/mol]

+printf("The reduction in chemical potential is %f J/mol",-du);

+

+

+

+

+

diff --git a/944/CH5/EX5.35/example5_35_TACC.sce b/944/CH5/EX5.35/example5_35_TACC.sce
new file mode 100755
index 000000000..246e58f62
--- /dev/null
+++ b/944/CH5/EX5.35/example5_35_TACC.sce
@@ -0,0 +1,17 @@
+//example 5.35

+

+clear;

+clc;

+

+//Given:

+Kb=0.51;//ebullioscopic constant of water [K*Kg/mol]

+n=155/180;//no. of moles of glucose

+m=n/1;//[mol/Kg]

+Ti=373;//Boiling point temperature of water[K]

+

+//To find the boiling point of the solution which is made by dissolving 155g of glucose in 1000g of water

+Tf=(Ti+Kb*m)-273;//boiling point temperature of the solution[C]

+printf("The boiling point of the solution is %f C",Tf);

+

+

+

diff --git a/944/CH5/EX5.36/example5_36_TACC.sce b/944/CH5/EX5.36/example5_36_TACC.sce
new file mode 100755
index 000000000..004bebe67
--- /dev/null
+++ b/944/CH5/EX5.36/example5_36_TACC.sce
@@ -0,0 +1,19 @@
+//example 5.36

+

+clear;

+clc;

+

+//Given:

+Ti=5.44;//freezing point of pure benzene[K]

+Tf=4.63;//freezing point of solution[K]

+m1=2.12;//mass of the solute[gm]

+m2=125;//mass of the benzene[gm]

+Kf=5.12;//cryoscopic constant of pure benzene[K*Kg/mol]

+

+//To find the molar mass of solute(M2)

+dTf=Ti-Tf;//depression in freezing point[K]

+M2=(m1*1000*Kf)/(m2*dTf);//molar mass of solute

+printf("The molar mass of solute is %f",M2);

+

+

+

diff --git a/944/CH5/EX5.38/example5_38_TACC.sce b/944/CH5/EX5.38/example5_38_TACC.sce
new file mode 100755
index 000000000..e3d26e843
--- /dev/null
+++ b/944/CH5/EX5.38/example5_38_TACC.sce
@@ -0,0 +1,16 @@
+//example 5.38

+

+clear;

+clc;

+

+disp("N2(g)+3H2(g)<=>2NH3(g)");

+//Given:

+T=298;//Temperature[K]

+Gf1=-16450;//Gibb's free energy of formation for NH3(g)[J/mol]

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the Kp value of the above reaction

+Gf=2*Gf1//Gibb's free energy for the reaction[KJ]

+x=Gf/R/T

+Kp=exp(-x);

+disp(Kp,'The Kp for above reaction is ');
\ No newline at end of file
diff --git a/944/CH5/EX5.39/example5_39_TACC.sce b/944/CH5/EX5.39/example5_39_TACC.sce
new file mode 100755
index 000000000..3b234bdfb
--- /dev/null
+++ b/944/CH5/EX5.39/example5_39_TACC.sce
@@ -0,0 +1,20 @@
+//example 5.39

+

+clear;

+clc;

+

+disp("0.5N2(g)+1.5H2(g)<=>NH3(g)");

+//Given:

+T=298;//Temperature[K]

+Kp=900;//Equilibrium constant for above reaction

+P1=0.32;//partial pressure of N2(g)[bar]

+P2=0.73;//partial pressure of H2(g)[bar]

+P3=0.98;//partial pressure of NH3(g)[bar]

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the reaction Gibb's energy

+G=-R*T*log(Kp);

+x=(P1^0.5)*(P2^1.5);

+p=P3/x;

+Gr=(G+R*T*log(p))*0.001;

+printf("The reaction Gibbs free energy is %f KJ/mol ",Gr);

diff --git a/944/CH5/EX5.40/example5_40_TACC.sce b/944/CH5/EX5.40/example5_40_TACC.sce
new file mode 100755
index 000000000..4ec8929ca
--- /dev/null
+++ b/944/CH5/EX5.40/example5_40_TACC.sce
@@ -0,0 +1,28 @@
+//example 5.40

+

+clear;

+clc;

+

+disp("N2(g)+3H2(g)<=>2NH3(g)");

+

+//Given:

+Kp1=5.85*10^5;//equilibrium constant at 298K

+H1=-46.11;//standard enthalpy of formation of NH3(g)[KJ/mol]

+T1=298;//Initial temperature[K]

+T2=423;//Final temperature[K]

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the Kp at 423K temperature

+H=2*H1;//enthalpy for reaction;[KJ]

+t=(T1^-1)-(T2^-1);

+x=-H*t/(R*0.001);

+Kp2=Kp1*exp(x);

+disp(Kp2,'The Equilibrium constant for reaction  at 423K is ');

+

+

+

+

+

+

+

+

diff --git a/944/CH5/EX5.41/example5_41_TACC.sce b/944/CH5/EX5.41/example5_41_TACC.sce
new file mode 100755
index 000000000..e1582f216
--- /dev/null
+++ b/944/CH5/EX5.41/example5_41_TACC.sce
@@ -0,0 +1,34 @@
+//example 5.41

+

+clear;

+clc;

+

+

+disp("Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)");

+//For Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)

+//Given:

+T=298;//Temperature[K]

+R=8.314;//Universal gas constant[J/K/mol]

+E1=-0.7618;//Standard electrode potential for Zn2+/Zn [volts]

+E2=-0.403;//Standard electrode potential for Cd2+/Cd [volts]

+F=96500;//Faraday's constant[coulomb/mol]

+n=2;//no. of electrons balancing

+//To find the Standard Free energy and equilibrium constant 

+Ei=E2-E1;//Standard potential for the reaction[volts]

+Gi=-n*F*Ei;//Standard Gibb's Free Energy [KJ/mol] 

+Ki=exp(-Gi/R/T);//Equilibrium constant

+printf("The Free energy for the rection is %f KJ/mol",Gi*0.001);

+disp(Ki,'The value of equilibrium constant is ');

+

+

+disp("Cd(s)|CdSO4(aq),Hg2SO4(s)|Hg(l)");

+//For Cd(s)|CdSO4(aq),Hg2SO4(s)|Hg(l)

+//Given:

+E3=0.6141;//Standard electrode potential for Hg2SO4(s),SO4^2-/Hg(l) [volts]

+//To find the standard free energy and equilibrium constant

+Eii=E3-E2;//Standard potantial for the reaction[volts]

+Gii=-n*F*Eii;//Standard Gibb's free energy[KJ/mol]

+Kii=exp(-Gii/R/T);//Equilibrium constant

+printf("The Free energy for the rection is %f KJ/mol",Gii*0.001);

+disp(Kii,'The value of equilibrium constant is');

+

diff --git a/944/CH5/EX5.42/example5_42_TACC.sce b/944/CH5/EX5.42/example5_42_TACC.sce
new file mode 100755
index 000000000..5e5cea63a
--- /dev/null
+++ b/944/CH5/EX5.42/example5_42_TACC.sce
@@ -0,0 +1,31 @@
+//example 5.42

+

+clear;

+clc;

+

+//Given:

+disp("Zn(s)|ZnCl2(soln)||AgCl(s)|Ag-Ag|AgCl(s)|ZnCl2(soln)|Zn(s)");

+m1=0.02;//concentration[M]

+Y1=0.65;//mean ionic activity coefficient

+m2=1.5;//concentration[M]

+Y2=0.29;//mean ionic activity coefficient 

+R=8.314;//Universal gas constant[J/K/mol]

+T=298;//Temperature [K]

+F=96500;//Faraday's constant[coulomb/mol]

+

+//To find the overall e.m.f of the above cell

+E=R*T*(log(m2*Y2/m1/Y1))*3/2/F;//[volts]

+printf("The overall e.m.f of the cell is %f volts",E);

+

+

+

+

+

+

+

+

+

+

+

+

+

diff --git a/944/CH5/EX5.43/example5_43_TACC.sce b/944/CH5/EX5.43/example5_43_TACC.sce
new file mode 100755
index 000000000..de6ac8496
--- /dev/null
+++ b/944/CH5/EX5.43/example5_43_TACC.sce
@@ -0,0 +1,27 @@
+//example 5.43

+

+clear;

+clc;

+

+//Given:

+disp("H2(g,1atm)|HCl(aq)|HCl(aq)|H2(g,1atm)");

+m1=0.02;//concentration[M]

+Y1=0.88;//mean ionic activity coefficient

+m2=1;//concentration[M]

+Y2=0.81;//mean ionic activity coefficient

+R=8.314;//universal gas constant[J/K/mol]

+T=298;//Temperature[K]

+F=96487;//Faraday's constant[coulombs/mol]

+t=0.178;//Tranference number of Cl-1

+

+//To find the e.m.f of the cell

+E=-2*t*R*T*(log(m1*Y1/m2/Y2))/F;//e.m.f of the cell[volts]

+printf("The e.m.f of the cell is %f volts",E);

+

+

+

+

+

+

+

+

diff --git a/944/CH5/EX5.44/example5_44_TACC.sce b/944/CH5/EX5.44/example5_44_TACC.sce
new file mode 100755
index 000000000..9b9a7f948
--- /dev/null
+++ b/944/CH5/EX5.44/example5_44_TACC.sce
@@ -0,0 +1,29 @@
+//example 5.44

+

+clear;

+clc;

+

+disp("The values for reaction that goes on within the cadmium cell");

+//Given:

+n=2;//no. of moles

+E=1.01463;//standard cadmium cell potential[volts]

+d=-5*10^-5;//i.e d=dE/dT[V/K]

+F=96500;//[coulomb/mol]

+T=298;//Temperature [K]

+

+//To find the values of dG,dS and dH

+dG=-n*E*F;//Change in Gibb's free energy[J]

+dS=n*F*d;//Change in entropy [J/K]

+dH=dG+T*dS;//change in enthalpy[J]

+printf(" dG=%f J",dG);

+printf("\n dS=%f J/K",dS);

+printf("\n dH=%f J",dH);

+

+

+

+

+

+

+

+

+

diff --git a/944/CH5/EX5.5/example5_5_TACC.sce b/944/CH5/EX5.5/example5_5_TACC.sce
new file mode 100755
index 000000000..f37567929
--- /dev/null
+++ b/944/CH5/EX5.5/example5_5_TACC.sce
@@ -0,0 +1,14 @@
+//example 5.5

+

+clear;

+clc;

+//Given:

+T=298;//Temperature [K]

+n=1;//no. of moles

+V1=500;//initial volume [cm3]

+V2=1000;//final volume [cm3]

+R=8.314;//Universal gas constant [J/mol/K]

+

+//to find the molar entropy change 

+S=R*log(V2/V1)//molar entropy change at constant temperature[J/K]

+printf("Molar entropy change of argon = %f J/K", S);

diff --git a/944/CH5/EX5.6/example5_6_TACC.sce b/944/CH5/EX5.6/example5_6_TACC.sce
new file mode 100755
index 000000000..bf4b73ef5
--- /dev/null
+++ b/944/CH5/EX5.6/example5_6_TACC.sce
@@ -0,0 +1,13 @@
+//example 5.6 

+

+clear;

+clc;

+

+//Given:

+W=1728;//Isothermal and reversible work done[J/mol]

+T=298;//Isothermal temperature[K]

+

+//to find the change in molar entropy when the gas expands isothermally and reversibly

+

+S=W/T;//change in molar entropy for isothermal and reversible process

+printf("The change in molar entropy = %f J/mol/K", S );

diff --git a/944/CH5/EX5.7/example5_7_TACC.sce b/944/CH5/EX5.7/example5_7_TACC.sce
new file mode 100755
index 000000000..31e338324
--- /dev/null
+++ b/944/CH5/EX5.7/example5_7_TACC.sce
@@ -0,0 +1,15 @@
+//example 5.7

+

+clear;

+clc;

+

+//Given:

+H=-92.22;//Standard reaction enthalpy[KJ]

+T=298;//Temperature [K]

+

+

+//To find the change in entropy of the surroundings at 298K

+//standard reaction enthalpy is H.Therefore, heat gained by the surroundings at 298K is -H

+S=-H*1000/T;//Change in entropy[J/K]

+printf("Change in entropy of the surroundings at 298k = %f J/K",S);

+

diff --git a/944/CH5/EX5.8/example5_8_TACC.sce b/944/CH5/EX5.8/example5_8_TACC.sce
new file mode 100755
index 000000000..b663828b8
--- /dev/null
+++ b/944/CH5/EX5.8/example5_8_TACC.sce
@@ -0,0 +1,17 @@
+//example 5.8

+

+clear;

+clc;

+

+//Given:

+T1=298;//Initial Temperature[K]

+T2=573;//Final Temperature[K]

+Cv=29.1;//Specific Heat capacity of argon gas [J/K/mol]

+n=1;//no. of moles

+

+

+//To find the change in entropy for argon gas

+S=n*Cv*log(T2/T1);//Change in entropy [J/K]

+printf("The change in entropy of the argon gas is %f J/K",S);

+

+

diff --git a/944/CH5/EX5.9/example5_9_TACC.sce b/944/CH5/EX5.9/example5_9_TACC.sce
new file mode 100755
index 000000000..a6d421b10
--- /dev/null
+++ b/944/CH5/EX5.9/example5_9_TACC.sce
@@ -0,0 +1,28 @@
+//example 5.9

+

+clear;

+clc;

+

+//Given:

+T1=276;//Initial temperature[K]

+Tf=278.7;//Freezing point temperature[K]

+Tb=353.3;//Boiling point temperature[K]

+T2=373;//Final temperature[K]

+Hf=9870;//Standard enthalpy of fusion[J/mol]

+Hv=30800;//Standard enthalpy of vaporization[J/mol]

+Cp=136.1;//Specific heat capacity of benzene[J/K/mol]

+mol.wt=78;//molecular weight of benzene[g/mol]

+mass=200;//weight of solid benzene[g]

+disp("     Cp doesnot change within this temp limit");

+

+//To find the total entropy change of solid

+n=mass/mol.wt;//no. of moles

+S1=n*Cp*log(Tf/T1);//entropy change in heating [J/K]

+S2=n*Hf/Tf;//entropy change in melting[J/K] 

+S3=n*Cp*log(Tb/Tf);//entropy change in heating[J/K]

+S4=n*Hv/Tb;//entropy change in vaporization[J/K]

+S5=n*Cp*log(T2/Tb);//entropy change in heating[J/K]

+S=S1+S2+S3+S4+S5;//total entropy change in heating from 276 to 373K

+

+printf("Total entropy change in heating 200g benzene from 3 to 100`C is %f J/K",S);

+ 
\ No newline at end of file
diff --git a/944/CH6/EX6.10/example6_10_TACC.sce b/944/CH6/EX6.10/example6_10_TACC.sce
new file mode 100755
index 000000000..d295ee754
--- /dev/null
+++ b/944/CH6/EX6.10/example6_10_TACC.sce
@@ -0,0 +1,27 @@
+//example 6.10

+

+clear;

+clc;

+

+//Given:

+m1=0.03;//mass of CO2(g)[gm]

+w1=44.01;//molecular weight of CO2(g)[gm/mol]

+m2=250;//mass of water[gm]

+w2=18.02;//molecular weight of water[gm/mol]

+k=1.25*10^6;//Henry's law constant[Torr]

+T=298;//Temperature[K]

+

+//To find the partial pressure of CO2 gas 

+n1=m1/w1;//no. of moles of CO2

+n2=m2/w2;//no. of moles of water

+x1=n1/(n1+n2);//mole fraction of CO2

+Pco2=k*x1;//Partial pressure of CO2[Torr]

+printf("The partial pressure of CO2 gas is %f Torr",Pco2);

+

+

+

+

+

+

+

+

diff --git a/944/CH6/EX6.11/example6_11_TACC.sce b/944/CH6/EX6.11/example6_11_TACC.sce
new file mode 100755
index 000000000..9996f489f
--- /dev/null
+++ b/944/CH6/EX6.11/example6_11_TACC.sce
@@ -0,0 +1,26 @@
+//example 6.11

+

+clear;

+clc;

+

+//Given:

+W=1000;//Total mass of a solution[gm]

+x1=0.5;//mole fraction of Chloroform

+x2=0.5;//mole fraction of Acetone

+V1m=80.235;//Partial molar volume of chloroform[cm3/mol]

+V2m=74.166;//Partial molar volume of Acetone[cm3/mol]

+M1=119.59;//molecular weight of chloroform[gm/mol]

+M2=58;//molecular weight of Acetone[gm/mol]

+

+//To find the Volume of the solution

+nT=W/(x1*M1+x2*M2);//Total no. of moles

+V=nT*(x1*V1m+x2*V2m);//Total volume[cm3]

+printf("The volume of the solution is %f cm3",V); 

+

+

+

+

+

+

+

+

diff --git a/944/CH6/EX6.12/example6_12_TACC.sce b/944/CH6/EX6.12/example6_12_TACC.sce
new file mode 100755
index 000000000..ac272095f
--- /dev/null
+++ b/944/CH6/EX6.12/example6_12_TACC.sce
@@ -0,0 +1,12 @@
+//example 6.12

+

+clear;

+clc;

+

+//Given:

+x1=0.5;//mole fraction of chloroform

+x2=0.5;//mole fraction of p-xylene

+T=298;//Temperature[K]

+//to find the excess volume 

+Ve=x1*x2*[0.585+0.085*(x1-x2)-0.165*(x1-x2)^2];//Excess volume measured by using a dilatometer

+printf("Ve/(cm3.mol^-1) = %f ",Ve);
\ No newline at end of file
diff --git a/944/CH6/EX6.14/example6_14_TACC.sce b/944/CH6/EX6.14/example6_14_TACC.sce
new file mode 100755
index 000000000..008ea9ba4
--- /dev/null
+++ b/944/CH6/EX6.14/example6_14_TACC.sce
@@ -0,0 +1,34 @@
+//example 6.14

+

+clear;

+clc;

+

+//Given:

+m1=0.01;//molality[m]

+v11=1;

+v12=2;

+Y1=0.71;

+m2=0.005;//molality[m]

+v21=1;

+v22=1;

+Y2=0.53;

+

+

+//To find the activity , molality of the electrolytes

+v1=(v11)+(v12);

+v2=(v21)+(v22);

+a1=(m1^v1)*(v11^v11)*(v12^v12)*(Y1^v1);

+a2=(m2^v2)*(v21^v21)*(v22^v22)*(Y2^v2);

+x=1/v1;

+a1m=a1^x;

+m1m=m1*(v11^v11*v12^v12)^x;//molality[m]

+y=1/v2;

+m2m=m2*(v21*v21*v22^v22)^y;//molality[m]

+a2m=a2^y;

+disp(a1,'The activity of the electrolyte ZnCl2 is');

+disp(a1m,'The mean activity of ZnCl2 is ');

+disp(m1m,'The mean molality of ZnCl2 in [m] ');

+disp(a2,'The activity of the electrolyte CuSO4 is ');

+disp(a2m,'The mean activity of CuSO4 is ');

+disp(m2m,'The mean molality of CuSO4 in [m] ');

+

diff --git a/944/CH6/EX6.15/example6_15_TACC.sce b/944/CH6/EX6.15/example6_15_TACC.sce
new file mode 100755
index 000000000..a2cbc0056
--- /dev/null
+++ b/944/CH6/EX6.15/example6_15_TACC.sce
@@ -0,0 +1,16 @@
+//example 6.15

+

+clear;

+clc;

+

+//Given:

+m2=3;//mass of the sucrose[gm]

+m1=0.1;//mass of water [Kg]

+Kf=1.86;//cryoscopic constant of water[K*Kg/mol]

+dTf=0.16;//Lowering in freezing point[K]

+

+//To find the molecular weight of sucrose

+a=m1*dTf;

+b=Kf*m2;

+M2=b/a;//molecular weight

+printf("M2=molecular weight ,    then M2=%f",M2); 
\ No newline at end of file
diff --git a/944/CH6/EX6.16/example6_16_TACC.sce b/944/CH6/EX6.16/example6_16_TACC.sce
new file mode 100755
index 000000000..fe7b2348d
--- /dev/null
+++ b/944/CH6/EX6.16/example6_16_TACC.sce
@@ -0,0 +1,18 @@
+//example 5.16

+

+clear;

+clc;

+

+//Given:

+dTf=0.088;//Lowering in freezing point[K]

+m2=0.45;//mass of sulphur[gm]

+m1=0.09955;//mass of benzene[gm]

+Kf=5.07;//cryoscopic constant for benzene[K*Kg/mol]

+

+//To find the molecular formula of sulphur

+a=m1*dTf;

+b=Kf*m2;

+M2=b/a;//molecular weight of sulphur

+printf("The molecular weight of sulphur is %f",M2);

+x=M2/32;//no. of sulphur atoms

+printf("\n The molecular formula of sulphur is S%f",x);
\ No newline at end of file
diff --git a/944/CH6/EX6.17/example6_17_TACC.sce b/944/CH6/EX6.17/example6_17_TACC.sce
new file mode 100755
index 000000000..4a2df57c2
--- /dev/null
+++ b/944/CH6/EX6.17/example6_17_TACC.sce
@@ -0,0 +1,26 @@
+//example 6.17

+

+clear;

+clc;

+

+//Given:

+m2=1.35;//mass of a macromolecule[gm]

+V=100;//volume of solution[cm^3]

+R=82;//Universal gas constant[atm.cm^3.K^-1]

+T=300;//Temperature[K]

+II=9.9;//osmotic pressure of the solution[cm]

+d=1;//density

+p=1013250;//Atmospheric pressure

+g=980.67;//gravitational field

+

+

+//To find the molar mass of macromolecule

+a=m2*R*T*p;

+b=V*9.9*d*g;

+M2=a/b;//molar mass of macromolecule

+printf(" M2 = molar mass of macromolecule , therefore M2 = %f g.mol^-1",M2);

+

+

+

+

+

diff --git a/944/CH6/EX6.18/example6_18_TACC.sce b/944/CH6/EX6.18/example6_18_TACC.sce
new file mode 100755
index 000000000..8a5cc5830
--- /dev/null
+++ b/944/CH6/EX6.18/example6_18_TACC.sce
@@ -0,0 +1,17 @@
+//example 6.18

+

+clear;

+clc;

+

+//Given:

+R=82;//Universal gas constant[atm.ml.K^-1.mol^-1]

+T=298;//Temperature[K]

+V=250;//volume of water[ml]

+m2=2.6;//mass of the protein

+M2=85000;//molar mass of protein[g.mol^-1]

+

+

+//To find the osmotic pressure of a solution

+n2=m2/M2;//no. of moles of protein

+II=(n2*R*T)/V;//Osmotic pressure of a solution[atm]

+printf("The osmotic pressure is %f atm ",II);

diff --git a/944/CH6/EX6.19/example6_19_TACC.sce b/944/CH6/EX6.19/example6_19_TACC.sce
new file mode 100755
index 000000000..712d4eeff
--- /dev/null
+++ b/944/CH6/EX6.19/example6_19_TACC.sce
@@ -0,0 +1,21 @@
+//example 6.19

+

+clear;

+clc;

+

+//Given:

+R=8.314;//Universal gas constant[J.K^-1.mol^-1]

+Tb=373.15;//Boiling point temperature[K]

+M1=0.018;// mass of water[kg]

+Hvap=40.7;//Enthalpy of vaporization[KJ.mol^-1]

+

+//To find the Ebullioscopic constant of water

+a=R*0.001*Tb^2*M1;

+b=Hvap;

+Kb=a/b;//Ebullioscopic constant of water[K.Kg.mol^-1]

+printf("The Ebullioscopic constant of water is %f K.Kg.mol-1",Kb);

+

+

+

+

+

diff --git a/944/CH6/EX6.2/example6_2_TACC.sce b/944/CH6/EX6.2/example6_2_TACC.sce
new file mode 100755
index 000000000..d549c654c
--- /dev/null
+++ b/944/CH6/EX6.2/example6_2_TACC.sce
@@ -0,0 +1,24 @@
+//example 6.2

+

+clear;

+clc;

+

+//Given:

+T=500;//Temperature[K]

+P=100;//Pressure[atm]

+a=3.61;//van der waals constant for CO2[atm.L^2.mol^-2]

+b=0.0429;//van der waals constant for CO2[L.mol^-1]

+R=0.082;//Universal gas constant[atm.K-1.mol^-1]

+

+//To find the molar volume of CO2

+x=b+(R*T/P);

+y=a/P;

+z=a*b/P;

+p2 = poly([-z y -x 1], 'Vm', 'c');

+t=roots(p2);

+printf("The value of molar volume of CO2 is %f L.mol^-1",t(3));

+

+

+

+

+

diff --git a/944/CH6/EX6.20/example6_20_TACC.sce b/944/CH6/EX6.20/example6_20_TACC.sce
new file mode 100755
index 000000000..ee330ae50
--- /dev/null
+++ b/944/CH6/EX6.20/example6_20_TACC.sce
@@ -0,0 +1,17 @@
+//example 6.20

+

+clear;

+clc;

+

+disp("CaF2(s)<=>CaF2(aq)<=>Ca+2(aq) + 2F-(aq)");

+

+//Given:

+Ksp=4*10^-11;//Solubility product of sparingly soluble salt CaF2

+

+//To find the value of activity coefficient

+x=Ksp/4;

+Cs=x^0.33;//Solubility 

+y=Cs^2;

+Y=(x/y)^0.33;//activity coefficient

+printf("The activity coefficient is %f ",Y);

+  
\ No newline at end of file
diff --git a/944/CH6/EX6.21/example6_21_TACC.sce b/944/CH6/EX6.21/example6_21_TACC.sce
new file mode 100755
index 000000000..d8ef30476
--- /dev/null
+++ b/944/CH6/EX6.21/example6_21_TACC.sce
@@ -0,0 +1,19 @@
+//example 6.21

+

+clear;

+clc;

+

+//Given:

+R=8.314;//Universal gas constant[J/K/mol]

+T=298;//Temperature[K]

+F=96500;//Faraday's constant

+Eo=0.98;//Standard e.m.f of the cell[Volts]

+E=1.16;//e.m.f of the cell[Volts]

+m=0.01;

+

+//To find the mean activity coefficient of ZnCl2 solution

+a=R*T;

+b=2*F;

+x=a/b;

+Y=exp((Eo-E-(x*log(4*m*m*m)))/(3*x));//mean activity coefficient

+printf("The mean activity coefficient is %f",Y);
\ No newline at end of file
diff --git a/944/CH6/EX6.22/example6_22_TACC.sce b/944/CH6/EX6.22/example6_22_TACC.sce
new file mode 100755
index 000000000..bccff608e
--- /dev/null
+++ b/944/CH6/EX6.22/example6_22_TACC.sce
@@ -0,0 +1,30 @@
+//example 6.22

+

+clear;

+clc;

+

+//Given:

+M1=0.01;//no. of moles of KCl

+M2=0.005;//no. of moles of MgCl2

+M3=0.002;//no. of moles of MgSO4

+M=0.1;//mass of water[Kg]

+z11=1;

+z12=1;

+z21=2;

+z22=1;

+z31=2;

+z32=2;

+

+//To find the ionic strength in a solution

+m1=M1/M;//molality of KCL[m]

+m2=M2/M;//molality of MgCl2[m]

+m3=M3/M;//molality of MgSO4[m]

+

+I=0.5*[(m1*z11^2+m1*z12^2+m2*z21^2+2*m2*z22^2+m3*z31^2+m3*z32^2)];//[mol/Kg]

+

+printf("The Ionic strength of a solution is %f mol/Kg",I);

+

+

+

+

+

diff --git a/944/CH6/EX6.23/example6_23_TACC.sce b/944/CH6/EX6.23/example6_23_TACC.sce
new file mode 100755
index 000000000..6d9180bcf
--- /dev/null
+++ b/944/CH6/EX6.23/example6_23_TACC.sce
@@ -0,0 +1,28 @@
+//example 6.23

+

+clear;

+clc;

+

+//Given:

+T=298;//Temperature[K]

+P=1;//pressure [atm]

+m=0.02;//Ionic strength of HCl solution in CH3OH[mol/Kg]

+E=32.6;//Di-electric constant

+d=0.787;//Density[gm/cm3]

+

+//To find the mean activity coefficient

+I=0.5*(0.02*1*1+0.02*1*1);//Ionic strength of HCl solution[mol/Kg]

+a=I*d;

+b=(E^3)*(298^3);

+x=(a/b)^0.5;

+Y=10^(-1.825*1000000*1*1*x);//mean activity coefficient

+printf("The mean activity coefficient is %f ",Y);

+

+

+

+

+

+

+

+

+

diff --git a/944/CH6/EX6.5/example6_5_TACC.sce b/944/CH6/EX6.5/example6_5_TACC.sce
new file mode 100755
index 000000000..f5842cfcb
--- /dev/null
+++ b/944/CH6/EX6.5/example6_5_TACC.sce
@@ -0,0 +1,26 @@
+//example 6.5

+

+clear;

+clc;

+

+//Given:

+b=0.0391;//Van der waals constant[dm3/mol]

+R=0.082;//Universal gas constant[dm3*atm/mol]

+P2=1000;//pressure [atm]

+P1=0;//pressure [atm]

+T=1273;//Temperature [K]

+

+//To find the fugacity and fugacity coefficient

+x=b*(P2-P1);

+y=R*T;

+fc=exp(x/y);//fugacity coefficient

+

+f=P2*fc;//fugacity[atm]

+printf("The fugacity is %f atm",f);

+printf("\n The fugacity coefficient is %f ",fc);

+

+

+

+

+

+

diff --git a/944/CH7/EX7.1/example7_1_TACC.sce b/944/CH7/EX7.1/example7_1_TACC.sce
new file mode 100755
index 000000000..191ef6e97
--- /dev/null
+++ b/944/CH7/EX7.1/example7_1_TACC.sce
@@ -0,0 +1,29 @@
+//example 7.1

+

+clear;

+clc;

+

+//Given:

+N=20;//no, of particles

+N1=4;//no. of particles in E1 energy level

+N2=4;//no. of particles in E2 energy level

+N3=6;//no. of particles in E3 energy level

+N4=3;//no. of particles in E4 energy level

+N5=3;//no. of particles in E5 energy level

+//To find the number of ways of distributing N particles

+N!=prod(1:N);

+N1!=prod(1:N1);

+N2!=prod(1:N2);

+N3!=prod(1:N3);

+N4!=prod(1:N4);

+N5!=prod(1:N5);

+n=N1!*N2!*N3!*N4!*N5!;

+W=N!/n;//no. of ways of distributing

+disp(W,'The no. of ways of distributing the particles is ');

+

+

+

+

+

+

+

diff --git a/944/CH7/EX7.12/example7_12_TACC.sce b/944/CH7/EX7.12/example7_12_TACC.sce
new file mode 100755
index 000000000..135c7b33a
--- /dev/null
+++ b/944/CH7/EX7.12/example7_12_TACC.sce
@@ -0,0 +1,21 @@
+//example 7.12

+

+clear;

+clc;

+

+//Given:

+R=1.99;//Universal gas constant [cal/K]

+e=2.718;

+V=22414;//volume[cm3]

+L=6.023*10^23;

+h=6.626*10^-27;//Planck's constant [erg.sec]

+m=6.63*10^-23;//mass[gm]

+k=1.381*10^-16;//Boltzmann constant[erg/K]

+T=273.2;//Temperature[K]

+

+//To find the Entropy of argon at 273K and 1 atmospheric pressure

+x=V*(e^2.5);

+y=L*(h^3);

+z=(2*3.14*m*k*T)^1.5;

+S=R*log(x*z/y);//Entropy [cal/degree/mol]

+printf("The Entropy of argon at 273K and 1 atm is %f cal/degree/mol",S);
\ No newline at end of file
diff --git a/944/CH7/EX7.14/example7_14_TACC.sce b/944/CH7/EX7.14/example7_14_TACC.sce
new file mode 100755
index 000000000..20f6a446f
--- /dev/null
+++ b/944/CH7/EX7.14/example7_14_TACC.sce
@@ -0,0 +1,19 @@
+//example 7.14

+

+clear;

+clc

+

+//Given:

+T=298;//Temperature[K]

+I=1.9373*10^-46;//moment of inertia of O2 gas [Kg/m2]

+h=6.626*10^-34;//Planck's constant[J.s]

+k=1.381*10^-23;//Boltzmann constant[J/K]

+R=8.314;//Universal gas constant[J/K/mol]

+u=2;//Homonuclear diatomic molecule

+

+

+//To find the rotational entropy and free energy for O2 gas

+Sr=R+R*log(8*3.14*3.14*I*k*T/(u*h*h));//[J/K/mol]

+Gr=-R*0.001*T*log(8*3.14*3.14*I*k*T/(u*h*h));//[KJ/mol]

+printf("The rotational entropy for O2 gas is %f J/K/mol",Sr);

+printf("\nThe rotational free energy for O2 gas is %f KJ/mol",Gr);

diff --git a/944/CH7/EX7.15/example7_15_TACC.sce b/944/CH7/EX7.15/example7_15_TACC.sce
new file mode 100755
index 000000000..08433e6d2
--- /dev/null
+++ b/944/CH7/EX7.15/example7_15_TACC.sce
@@ -0,0 +1,30 @@
+//example 7.15

+

+clear;

+clc;

+

+//Given:

+T=298;//Temperature[K]

+v=892.1*3*10^10;//frequency[sec-1]

+h=6.626*10^-27;//Planck's constant [J.s]

+k=1.381*10^-16;//Boltzmann constant[erg/K]

+e=2.718;

+R=1.998;//Universal gas constant[cal/K]

+//To find the vibrational contribution to the entropy of F2 at 298K

+x=h*v/(k*T);

+a=R*x*e^-x/(1-e^-x);//a=E-Eo/T

+b=R*log(1-e^-x);//b=G-Eo/T

+S=a-b;//[cal/deg]

+printf("The vibrational contribution to the entropy of F2 is %f cal/deg",S);

+

+

+

+

+

+

+

+

+

+

+

+

diff --git a/944/CH7/EX7.16/example7_16_TACC.sce b/944/CH7/EX7.16/example7_16_TACC.sce
new file mode 100755
index 000000000..d2bcdb095
--- /dev/null
+++ b/944/CH7/EX7.16/example7_16_TACC.sce
@@ -0,0 +1,25 @@
+//example 7.16

+

+clear;

+clc;

+

+//Given:

+T=1273;//Temperature[K]

+h=6.26*10^-27;//Planck's constant[J.s]

+k=1.381*10^-16;//Boltzmann constant[erg/K]

+T=1000;//Temperature[degrees]

+m=3.82*10^-23;//mass of Na [gm]

+I=(1.91*10^-23)*(3.078*10^-8)^2;//moment of inertia[gm.cm2]

+dE=0.73*1.602*10^-12;//[erg]

+v=159.23*(3*10^10);//frequency [s-1]

+R=82;//universal gas constant[cm3.atm/deg]

+u=2;//symmetry number

+L=6.023*10^23;//avogadro's number

+

+//To find the equilibrium constant

+p=((3.14*m*k*T)^1.5)/h/h/h;

+s=R*u*h*h/L/8/3.14/3.14/I/k;

+q=1-(exp(-h*v/k/T));

+r=exp(-dE/k/T);

+Kp=p*s*q*r;//Equilibrium constant 

+printf("The equilibrium constant is %f",Kp);
\ No newline at end of file
diff --git a/944/CH7/EX7.17/example7_17_TACC.sce b/944/CH7/EX7.17/example7_17_TACC.sce
new file mode 100755
index 000000000..8f38ebf0b
--- /dev/null
+++ b/944/CH7/EX7.17/example7_17_TACC.sce
@@ -0,0 +1,32 @@
+//example 7.17

+

+clear;

+clc;

+

+//Given:

+T=298;//Temperature[K]

+m1=32;

+m2=36;

+m3=34;

+u1=8;

+u2=9;

+u3=16*18/34;

+z1=0.99924;

+z2=0.99951;

+z3=0.99940;

+h=6.26*10^-27;//Planck's constant[J.s]

+c=3*10^10;//Speed of light[m/s]

+k=1.38*10^-16;//Boltzman's constant[erg/K]

+vo1=1535.8;//vibration frequency of 16O18O [cm-1]

+vo2=1580.4;//vibration frequency of 16O2 [cm-1]

+vo3=1490.0;//vibration frequency of 18O2 [cm-1]

+dE=0.5*h*c*(2*vo1-vo2-vo3);//[erg]

+r=dE/k/T;

+

+

+//To find the equilibrium constant for isotopic exchange reaction

+a=m3^3/m2^1.5/m1^1.5;

+b=(u3^2)*4/u2/u1;

+c=z3^2/z2/z1;

+Kp=a*b*c*exp(-r);

+printf("The value of equilibrium constant for isotopic exchange reaction is %f",Kp);
\ No newline at end of file
diff --git a/944/CH7/EX7.2/example7_2_TACC.sce b/944/CH7/EX7.2/example7_2_TACC.sce
new file mode 100755
index 000000000..3172f8822
--- /dev/null
+++ b/944/CH7/EX7.2/example7_2_TACC.sce
@@ -0,0 +1,35 @@
+//example 7.2

+

+clear;

+clc;

+

+//Given:

+T=298;//Temperature [K]

+v=6.5*10^13;//Frequency in [sec-1]

+//Consider zero point energy = 0.

+h=6.627*10^-34;//planck's constant[J.s]

+k=1.381*10^-23;//Boltzmann constant 

+N=1;//Since N=summation(gj*exp(-Ej/kT))

+

+//To find the fraction of molecules present in vibrational level

+E1=h*v;//for energy level 1[J]

+E2=2*h*v;//for energy level 2[J]

+x=k*T;

+g1=1;

+g2=1;

+N1=[g1*exp(-E1/x)];//molecules present in energy level 1

+N2=[g2*exp(-E2/x)];//molecules present in energy level 2

+n1=N1/N;//fraction of molecules present in energy level 1

+n2=N2/N;//fraction of molecules present in energy level 2

+printf("The fraction of molecule  s present in energy level 1 is %f",n1);

+

+

+disp(n2,'The fraction of molecules present in energy level 2 is '); 

+

+

+

+

+

+

+

+

diff --git a/944/CH7/EX7.3/example7_3_TACC.sce b/944/CH7/EX7.3/example7_3_TACC.sce
new file mode 100755
index 000000000..16ef2d2da
--- /dev/null
+++ b/944/CH7/EX7.3/example7_3_TACC.sce
@@ -0,0 +1,27 @@
+//example 7.3

+

+clear;

+clc;

+

+//Given:

+dE=4.3*10^-20;//difference in energy levels[J]

+T1=0.000001;//Initial Temperature[K](approximately zero , needed for execution)

+T2=300;//Final Temperature[K]

+k=1.381*10^-23;//Boltzmann constant [J/K]

+

+//To find the ratio of no. of particle per state for two states separated by an energy dE

+x1=k*T1;

+r1=exp(-dE/x1);

+x2=k*T2;

+r2=exp(-dE/x2);

+disp(r1,'The ratio of no. of particles per state at 0K is ');

+disp(r2,'The ratio of no. of particles per state at 300K is ');

+

+

+

+

+

+

+

+

+

diff --git a/944/CH7/EX7.4/example7_4_TACC.sce b/944/CH7/EX7.4/example7_4_TACC.sce
new file mode 100755
index 000000000..daae7caee
--- /dev/null
+++ b/944/CH7/EX7.4/example7_4_TACC.sce
@@ -0,0 +1,28 @@
+//example 7.4

+

+

+clear;

+clc;

+

+//Given:

+T1=273;//[K]

+T2=14273;//[K]

+E1=-13.6;//Energy of ground state [eV]

+k=8.617*10^-5;//Boltzmann constant[eV/K]

+g2=8;//total no. of states with energy E2

+g1=2;//total no. of states with energy E1

+

+

+//To find the no. of atoms in first-excited singlet state

+E2=E1/(2^2);//Energy for n=2 (i.e.E2=E1/n2)

+x1=k*T1;

+r1=(g2/g1)*exp(-(E2-E1)/x1);

+x2=k*T2;

+r2=(g2/g1)*exp(-(E2-E1)/x2);

+disp(r1,'The fraction of atoms present in level n=2 at 273K is ');

+disp("Therefore total 3*10^25 atoms we say   that all are present at ground     state");

+printf("\n\nThe fraction of atoms present in level n=2 at 14273 is %f\n",r2);

+x=r2*3*10^25;

+disp(x,'Therefore no. of atoms in level n=2 is ');

+

+

diff --git a/944/CH7/EX7.5/example7_5_TACC.sce b/944/CH7/EX7.5/example7_5_TACC.sce
new file mode 100755
index 000000000..12c06c8b1
--- /dev/null
+++ b/944/CH7/EX7.5/example7_5_TACC.sce
@@ -0,0 +1,15 @@
+//example 7.5

+

+clear;

+clc;

+

+//Given:

+r1=0.001;//the population of the states at a higher energy to that at a lower energy 

+dE=8*10^-20;//The difference in energy[J]

+k=1.381*10^-23;//Boltzmann constant [J/K]

+

+

+//To find the Temperature at this condition 

+x=k*log(r1);

+T=-dE/x;//[K]

+printf("The Temperature at this condition is %f K",T);
\ No newline at end of file
diff --git a/944/CH7/EX7.6/example7_6_TACC.sce b/944/CH7/EX7.6/example7_6_TACC.sce
new file mode 100755
index 000000000..70a4601a4
--- /dev/null
+++ b/944/CH7/EX7.6/example7_6_TACC.sce
@@ -0,0 +1,28 @@
+//example 7.6

+

+clear;

+clc;

+

+

+//section(1)

+//Given:

+//The energy levels are not degenerate 

+w=1;//no. of ways of distributing the molecules

+k=1.381*10^-23;//Boltzmann constant[J/K]

+//To find the entropy of the system

+S1=k*log(w);//Entropy of system at 0K

+printf("The Entropy of System at 0K and non-degenerate eng level is %f J/K/mol",S1);

+

+//section(2)

+//Here the energy levels are degenerate

+n=2;

+R=8.314;//Universal gas constant[J/K/mol]

+

+//To find the entropy of the system

+//S=klog(n^N)=>S=R*log(n)

+S2=R*log(n);//Entropy of the system[J/K/mol]

+printf("\nThe Entropy of system at 0K and degenerete eng level is %f J/K/mol",S2);

+

+

+

+

diff --git a/944/CH7/EX7.9/example7_9_TACC.sce b/944/CH7/EX7.9/example7_9_TACC.sce
new file mode 100755
index 000000000..60b1fa3c8
--- /dev/null
+++ b/944/CH7/EX7.9/example7_9_TACC.sce
@@ -0,0 +1,20 @@
+//example 7.9

+

+clear;

+clc;

+

+//Given:

+V=0.001;//Volume of vessel[m3]

+T=300;//Temperature [K]

+k=1.381*10^-23;//Boltzmann constant[J/K]

+mol.wt=32;//molecular mass of oxygen molecule

+h=6.626*10^-34;//planck's constant[J.s}

+

+

+

+//To find the Transitional partition function of an oxygen molecule confined in a 1-litre vessel at 300K

+m=32*1.66*(10^-27);//mass of oxygen molecule[Kg]

+x=[(2*3.14*m*k*T)^(3/2)]*V;

+y=h^3;

+zt=x/y;//Transitional partition function of an oxygen molecule

+disp(zt,'The Transitional partition function of an oxygen molecule confined in a 1-litre vessel at 300K is')
\ No newline at end of file