From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 611/CH13/EX13.1/Chap13_Ex1.sce    | 27 +++++++++++++++++++++++++++
 611/CH13/EX13.2/Chap13_Ex2_R1.sce | 28 ++++++++++++++++++++++++++++
 611/CH13/EX13.3/Chap13_Ex3_R1.sce | 24 ++++++++++++++++++++++++
 611/CH13/EX13.4/Chap13_Ex4.sce    | 23 +++++++++++++++++++++++
 611/CH13/EX13.5/Chap13_Ex5.sce    | 21 +++++++++++++++++++++
 5 files changed, 123 insertions(+)
 create mode 100755 611/CH13/EX13.1/Chap13_Ex1.sce
 create mode 100755 611/CH13/EX13.2/Chap13_Ex2_R1.sce
 create mode 100755 611/CH13/EX13.3/Chap13_Ex3_R1.sce
 create mode 100755 611/CH13/EX13.4/Chap13_Ex4.sce
 create mode 100755 611/CH13/EX13.5/Chap13_Ex5.sce

(limited to '611/CH13')

diff --git a/611/CH13/EX13.1/Chap13_Ex1.sce b/611/CH13/EX13.1/Chap13_Ex1.sce
new file mode 100755
index 000000000..66b2bece0
--- /dev/null
+++ b/611/CH13/EX13.1/Chap13_Ex1.sce
@@ -0,0 +1,27 @@
+// 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
new file mode 100755
index 000000000..9f7787158
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+++ b/611/CH13/EX13.2/Chap13_Ex2_R1.sce
@@ -0,0 +1,28 @@
+// 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
new file mode 100755
index 000000000..325b8783d
--- /dev/null
+++ b/611/CH13/EX13.3/Chap13_Ex3_R1.sce
@@ -0,0 +1,24 @@
+// 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
new file mode 100755
index 000000000..922270d48
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+++ b/611/CH13/EX13.4/Chap13_Ex4.sce
@@ -0,0 +1,23 @@
+// 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
new file mode 100755
index 000000000..c2f90edbd
--- /dev/null
+++ b/611/CH13/EX13.5/Chap13_Ex5.sce
@@ -0,0 +1,21 @@
+// 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===================================================
+ 
-- 
cgit