initial commit / add all books

10 files changed, 145 insertions, 0 deletions

diff --git a/3856/CH7/EX7.1/Ex7_1.sce b/3856/CH7/EX7.1/Ex7_1.sce
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index 000000000..b9aa6c556
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+//Calculate the Gibbs Energy and Entropy of mixing of Argon and Nitrogen

+

+//Example 7.1

+

+clc;

+

+clear;

+

+n1=1.6;  //Number of moles of Argon at 1 atm

+

+n2=2.6;  //Number of moles of Nitrogen at 1 atm

+

+XAr=n1/(n1+n2);  //The mole fraction of Argon and Nitrogen

+

+XN2=n2/(n1+n2); // The mole fraction of Nitrogen and Argon

+

+n=n1+n2;  //Total moles of gas in mol

+

+R=8.314;  //Gas constant in J K^-1 mol^-1

+

+T=298;  //Temperature of the gas

+

+delmixG=n*R*T*(XAr*log(XAr)+XN2*log(XN2))/1000;  //The Gibbs Energy of mixing in kJ

+

+printf("Gibbs Energy of mixing = %.1f kJ",delmixG);

+

+delmixS=-(delmixG*1000)/T;  //Entropy of mixing in J K^-1

+

+printf("\nEntropy of mixing = %.0f J K^-1",delmixS);

+

diff --git a/3856/CH7/EX7.1/Ex7_1.txt b/3856/CH7/EX7.1/Ex7_1.txt
new file mode 100644
index 000000000..f95d0e0ff
--- /dev/null
+++ b/3856/CH7/EX7.1/Ex7_1.txt
@@ -0,0 +1,2 @@
+ Gibbs Energy of mixing = -6.9 kJ

+Entropy of mixing = 23 J K^-1
\ No newline at end of file
diff --git a/3856/CH7/EX7.2/Ex7_2.sce b/3856/CH7/EX7.2/Ex7_2.sce
new file mode 100644
index 000000000..5914e05b9
--- /dev/null
+++ b/3856/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,29 @@
+//Calculate the Composition of the Vapor in Equilibrium of Liquid A and Liquid B

+

+//Example 7.2

+

+clc;

+

+clear;

+

+XA=36/100;  //Number of mole of Liquid A

+

+XB=1-(36/100);  //Number of mole of liquid B

+

+PdegA=66;  //Vapor pressure of pure A in torr

+

+PdegB=88;  //Vapor pressure of pure B in torr

+

+PA=XA*PdegA;  //Vapor pressure of A in solution in torr

+

+PB=XB*PdegB;  //Vapor pressure of B in solution in torr

+

+PT=PA+PB;  //Total Vapor pressure of solution in torr

+

+XAv=PA/PT;  //Composition of Vapor of A in solution 

+

+printf("Composition of Vapor of A = %.2f ",XAv);

+

+XBv=PB/PT;  //Composition of Vapor B in solution

+

+printf("\nComposition of Vapor of B = %.2f",XBv);

diff --git a/3856/CH7/EX7.2/Ex7_2.txt b/3856/CH7/EX7.2/Ex7_2.txt
new file mode 100644
index 000000000..faf06331c
--- /dev/null
+++ b/3856/CH7/EX7.2/Ex7_2.txt
@@ -0,0 +1,2 @@
+ Composition of Vapor of A = 0.30 

+Composition of Vapor of B = 0.70
\ No newline at end of file
diff --git a/3856/CH7/EX7.3/Ex7_3.sce b/3856/CH7/EX7.3/Ex7_3.sce
new file mode 100644
index 000000000..28736f2d9
--- /dev/null
+++ b/3856/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,25 @@
+//Caculate the Molal Solubility of Carbon Dioxide in water

+

+//Example 7.3

+

+clc;

+

+clear;

+

+Pco2=3.3*10^-4*760;  //Partial pressure of CO2 in air in torr

+

+K=1.24*10^6;  //Henry's Law Constant in torr

+

+Xco2=Pco2/K;  //Mole raction of solue (CO2)

+

+nH2O=1000/18.01;  //Mole fraction of solvent (H2O)in mol^-1

+

+nCO2=Xco2*nH2O;  //Molal solubility of CO2 in mol/kg(H2O)

+

+printf("Molal Solubility of Carbon di Oxide = %.2f*10^-5 mol/kg",nCO2*10^5);

+

+Kdes=29.3;  //Henry's Law Constant in atm mol^-1 kg^-1

+

+m=(Pco2/760)/Kdes;  //Molal solubilty of CO2 in mol/kg(H2O)(Alternatively we can find out )(The answer vary  due to round off error)

+

+printf("\nMolal solubility = %.2f*10^-5 atm mol^-1 kg^-1",m*10^5);

diff --git a/3856/CH7/EX7.3/Ex7_3.txt b/3856/CH7/EX7.3/Ex7_3.txt
new file mode 100644
index 000000000..76593893d
--- /dev/null
+++ b/3856/CH7/EX7.3/Ex7_3.txt
@@ -0,0 +1,2 @@
+ Molal Solubility of Carbon di Oxide = 1.12*10^-5 mol/kg

+Molal solubility = 1.13*10^-5 atm mol^-1 kg^-1
\ No newline at end of file
diff --git a/3856/CH7/EX7.4/Ex7_4.sce b/3856/CH7/EX7.4/Ex7_4.sce
new file mode 100644
index 000000000..6bb634750
--- /dev/null
+++ b/3856/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,27 @@
+//Calculate the Boiling point and freezing poin of solution of  Sucrose in water

+

+//Example 7.4

+

+clc;

+

+clear;

+

+m1=45.20;  //Mass og the Sucrose in g

+

+m2=316.0/1000;  //Mass 0f the water in kg 

+

+n=m1/342.3;  // Molar mass of the Sucrose in mol

+

+Kb=0.51;  //Molal boiling point Elevation constant in K kg mol^-1

+

+m3=n/m2;  //Molality of the solution in mol kg^-1

+

+delT1=(Kb*m3)+373.15;  //Boiling point for solution of Sucrose in water

+

+printf("(a)Boiling point of Sucrose = %.2f K",delT1); 

+

+Kf=1.86;  //Molal freezing point depression constant in K kg mol^-1 

+

+delT2=273.15-(Kf*m3);  //Boiling point for solution of Sucrose in water

+

+printf("\n(b)Freezing point of Sucrose = %.2f K",delT2); 

diff --git a/3856/CH7/EX7.4/Ex7_4.txt b/3856/CH7/EX7.4/Ex7_4.txt
new file mode 100644
index 000000000..67287e417
--- /dev/null
+++ b/3856/CH7/EX7.4/Ex7_4.txt
@@ -0,0 +1,2 @@
+ (a)Boiling point of Sucrose = 373.36 K

+(b)Freezing point of Sucrose = 272.37 K
\ No newline at end of file
diff --git a/3856/CH7/EX7.5/Ex7_5.sce b/3856/CH7/EX7.5/Ex7_5.sce
new file mode 100644
index 000000000..ca19812d5
--- /dev/null
+++ b/3856/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,25 @@
+//What is the Molar mass of Hemoglobin 

+

+//Example 7.5

+

+clc;

+

+clear;

+

+h=77.8/1000;  //Height 0f the liquid in right column in m

+

+g=9.81;  //Acceleration due to gravity in m s^-2

+

+rho=1*10^3;  //Density of the solution in kg m^-3

+

+P=h*g*rho;  //Osmotic pressure of the solution in pascals (N m^-2)

+

+R=8.314;  //Gas constant in J K^-1 mol^-1

+

+T=298;  //Temperature of the solution in K

+

+c2=20;  //Concentration of the solute in kg m^-3

+

+mew2=(c2*R*T)/P;  //Molar mass of the Hemoglobin in kg mol^-1

+

+printf("Molar mass of Hemoglobin = %.0f kg mol^-1",mew2);

diff --git a/3856/CH7/EX7.5/Ex7_5.txt b/3856/CH7/EX7.5/Ex7_5.txt
new file mode 100644
index 000000000..0b9e865d8
--- /dev/null
+++ b/3856/CH7/EX7.5/Ex7_5.txt
@@ -0,0 +1 @@
+ Molar mass of Hemoglobin = 65 kg mol^-1
\ No newline at end of file