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

---
 551/CH9/EX9.4/4.sce | 67 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 67 insertions(+)
 create mode 100755 551/CH9/EX9.4/4.sce

(limited to '551/CH9/EX9.4')

diff --git a/551/CH9/EX9.4/4.sce b/551/CH9/EX9.4/4.sce
new file mode 100755
index 000000000..81928213a
--- /dev/null
+++ b/551/CH9/EX9.4/4.sce
@@ -0,0 +1,67 @@
+clc
+p=1*10^5; //Pa
+T=293; //K
+n_CO2=1; //moles of CO2
+n=4; //moles of air
+M_CO2=44;
+M_N2=28;
+M_O2=32;
+
+//Let A be the volumeetric analysis
+A_O2=0.21;
+A_N2=0.79;
+
+n_O2=A_O2*n;
+n_N2=A_N2*n;
+
+disp("(i) The masses of CO2, O2 and N2, and the total mass")
+
+m_CO2=n_CO2*M_CO2;
+disp("Mass of CO2=")
+disp(m_CO2)
+disp("kg")
+
+m_O2=n_O2*M_O2;
+disp("Mass of O2=")
+disp(m_O2)
+disp("kg")
+
+m_N2=n_N2*M_N2;
+disp("Mass of N2=")
+disp(m_N2)
+disp("kg")
+
+m=m_CO2 + m_O2 + m_N2;
+disp("Total mass =")
+disp(m)
+disp("kg")
+
+
+disp("(ii) The percentage carbon content by mass")
+//Since the molecular weight of carbon is 12, therefore, there are 12 kg of carbon present for every mole of CO2
+m_C=12; //kg
+
+%C=m_C/m*100;
+disp("Percentage carbon in mixture")
+disp(%C)
+disp("%")
+
+
+disp("(iii) The apparent molecular weight and the gas constant for the mixture")
+n=n_CO2 + n_O2 + n_N2;
+M=n_CO2/n*M_CO2 + n_O2/n*M_O2 + n_N2/n*M_N2;
+disp("Apparent Molecular weight")
+disp(M)
+
+R0=8.314;
+R=R0/M;
+disp("Gas constant for the mixture=")
+disp(R)
+disp("kJ/kg K")
+
+
+disp("(iv) The specific volume of the mixture")
+v=R*10^3*T/p;
+disp("specific volume=")
+disp(v)
+disp("m^3/kg")
-- 
cgit