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

---
 2090/CH3/EX3.19/Chapter3_Example19.sce | 21 +++++++++++++++++++++
 1 file changed, 21 insertions(+)
 create mode 100755 2090/CH3/EX3.19/Chapter3_Example19.sce

(limited to '2090/CH3/EX3.19')

diff --git a/2090/CH3/EX3.19/Chapter3_Example19.sce b/2090/CH3/EX3.19/Chapter3_Example19.sce
new file mode 100755
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+++ b/2090/CH3/EX3.19/Chapter3_Example19.sce
@@ -0,0 +1,21 @@
+clc
+clear
+//Input data
+p=1;//Initial pressure in atm
+T=300;//Initial temperature in K
+Tc=2400;//To calculate the molefraction of the products at this temperature in K
+KP1=3.866;//Natural logarithm of equilibrium constant at 2400 K for the equation 
+
+//Calculations
+K1=exp(KP1);//The value of equilibrium constant at 2400 K
+nr=1+0.5;//The number of moles of reactants
+Pp=(p*Tc)/(nr*T);//Pressure exercted on the products side per mole in atm/mole
+a=0.098;//The dissociation of 1 mole of CO2
+np=(a+2)/2;//The number of moles of products
+xco=[2*(1-a)]/(2+a);//Mole fraction of CO2
+xc=[2*a]/(2+a);//Mole fraction of CO
+xo=a/(2+a);//Mole fraction of O2
+PP=5.333*np;//Pressure of the product in bar
+
+//output
+printf('Mole fraction of the carbondioxide is %3.4f \n Mole fraction of the carbonmonoxide is %3.4f \n Mole fraction of oxygen is %3.4f \n Pressure of the product is %3.3f bar',xco,xc,xo,PP)
-- 
cgit