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diff --git a/611/CH13/EX13.1/Chap13_Ex1.sce b/611/CH13/EX13.1/Chap13_Ex1.sce
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+// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.

+

+//Chapter-13,Example 1,Page 478

+//Title:Depression in freezing point

+//================================================================================================================

+clear 

+clc

+

+//INPUT

+weight=10; //weight of NaCl in grams

+volume=1; //volume of water in litres

+weight_water=1000; //  weight of water in grams (Weight=Volume*Density, density of water =1g/cc=1g/ml=1000g/l)

+molwt_NaCl=58.5; //molecular weight of NaCl in grams

+molwt_water=18; //molecular weight of water in grams

+hf=6.002; //enthalpy change of fusion in kJ/mol at 0 degree celsius

+P=101.325; //pressure in kPa

+T=273.15; // freezing point temperature of water at the given pressure in K

+R=8.314; //universal gas constant in J/molK;

+

+//CALCULATION

+x2=(weight/molwt_NaCl)/((weight/molwt_NaCl)+(weight_water/molwt_water));//  calculation of mole fraction of solute NaCl (no unit)

+delt=(R*T^2*x2)/(hf*10^3);//calculation of depression in freezing point of water using Eq.(13.14)

+

+//OUTPUT

+mprintf('\n The depression in freezing point of water when 10g of NaCl solute is added = %0.2f K',delt);

+

+//===============================================END OF PROGRAM===================================================

diff --git a/611/CH13/EX13.2/Chap13_Ex2_R1.sce b/611/CH13/EX13.2/Chap13_Ex2_R1.sce
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+// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.

+

+//Chapter-13,Example 2,Page 480

+//Title:Elevation in Boiling Point

+//================================================================================================================

+clear 

+clc

+

+//INPUT

+weight=10; //weight of NaCl in grams

+volume=1; //volume of water in litres

+weight_water=1000; //  weight of water in grams (Weight=Volume*Density, density of water =1g/cc=1g/ml=1000g/l)

+molwt_NaCl=58.5; //molecular weight of NaCl in grams

+molwt_water=18; //molecular weight of water in grams

+lat_ht=2256.94; //latent heat of vaporization in kJ/kg at 100 degree celsius (obtained from steam tables)

+P=101.325; //pressure in kPa

+T=373.15; //boiling point temperature of water at the given pressure in K

+R=8.314; //universal gas constant in J/molK

+

+//CALCULATION

+x2=0.0031;//mole fraction of solute NaCl (From Example 13.1)(no unit)

+hv=(lat_ht*molwt_water)/1000; //conversion of latent heat from kJ/kg to kJ/mol

+delt=(R*T^2*x2)/(hv*10^3); //calculation of elevation in boiling point of water using Eq.(13.24)

+

+//OUTPUT

+mprintf('\n The elevation in boiling point of water when 10g of NaCl solute is added = %0.2f K',delt);

+

+//===============================================END OF PROGRAM===================================================

diff --git a/611/CH13/EX13.3/Chap13_Ex3_R1.sce b/611/CH13/EX13.3/Chap13_Ex3_R1.sce
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+// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.

+

+//Chapter-13,Example 3,Page 481

+//Title:Osmotic pressure

+//================================================================================================================

+clear 

+clc

+

+//INPUT

+weight=10; //weight of NaCl in grams

+weight_water=1000; //  weight of water in grams

+molwt_NaCl=58.5; //molecular weight of NaCl in grams

+molwt_water=18; //molecular weight of water in grams

+T=300; //prevailing temperature of water in K

+R=8.314; //universal gas constant in (Pa m^3)/(mol K);

+v=18*10^-6;//molar volume in m^3/mol

+//CALCULATION

+x2=0.0031;//mole fraction of solute NaCl (From Example 13.1)(no unit)

+pi=((R*T*x2)/v)*10^-3; // calulation of osmotic pressure using Eq(13.30)(in kPa)

+

+//OUTPUT

+mprintf('\n The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K = %0.2f kPa',pi);

+

+//===============================================END OF PROGRAM===================================================

diff --git a/611/CH13/EX13.4/Chap13_Ex4.sce b/611/CH13/EX13.4/Chap13_Ex4.sce
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+// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.

+

+//Chapter-13,Example 4,Page 483

+//Title:Ideal solubility

+//================================================================================================================

+clear 

+clc

+

+//INPUT

+temp=20; // prevailing tempearture in degree celsius

+melt_temp=80.05; // melting point of naphthalene in degree celsius

+hf=18.574; // enthalpy of fusion in kJ/mol

+R=8.314; // universal gas constant in J/molK

+

+//CALCULATION

+t=temp+273.15; // convesion of prevailing temperature to K

+melt_t=melt_temp+273.15; //conversion of melting point of naphtalene to K

+x2=exp(((hf*10^3)/R)*((1/melt_t)-(1/t))); //calculation of ideal solubility using Eq.(13.40)(no unit)

+

+//OUTPUT

+mprintf('\n The ideal solubility of naphthalene at 20 degree celsius= %0.4f',x2);

+

+//===============================================END OF PROGRAM===================================================

diff --git a/611/CH13/EX13.5/Chap13_Ex5.sce b/611/CH13/EX13.5/Chap13_Ex5.sce
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+// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.

+

+//Chapter-13,Example 5,Page 483

+//Title:Solubility of gas

+//================================================================================================================

+clear 

+clc

+

+//INPUT

+t=295.43; //prevailing temperature in K

+sat_p=6.05; //Sasturation pressure of carbon dioxide at the prevailing temperature in MPa

+p=0.1; //pressure at which solubility has to be determined in MPa

+

+//CALCULATION

+x2=p/sat_p; //calculation of solubility using Eq.(13.44)(no unit)

+

+//OUTPUT

+mprintf('\n The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution at 0.1MPa= %0.4f',x2);

+

+//===============================================END OF PROGRAM===================================================

+