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

---
 647/CH9/EX9.6/Example9_6.sce | 31 +++++++++++++++++++++++++++++++
 1 file changed, 31 insertions(+)
 create mode 100755 647/CH9/EX9.6/Example9_6.sce

(limited to '647/CH9/EX9.6/Example9_6.sce')

diff --git a/647/CH9/EX9.6/Example9_6.sce b/647/CH9/EX9.6/Example9_6.sce
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+clear;
+clc;
+
+// Example: 9.6
+// Page: 341
+printf("Example - 9.6 and Page number - 341\n\n");
+
+//Given
+T = 25+273.15;// [K]
+P = 1;// [atm]
+// Component 1 = water
+// Component 2 = methanol
+a = -3.2;// [cubic cm/mol]  A constant
+V2 = 40.7;// [cubic cm/mol]  Molar volume of pure component 2 (methanol)
+// V1_bar = 18.1 + a*x_2^(2)
+
+// From Gibbs-Duhem equation at constant temperature and pressure we have
+// x_1*dV1_bar + x_2*dV2_bar = 0
+// dV2_bar = -(x_1/x_2)*dV1_bar = -(x_1/x_2)*a*2*x_2*dx_2 = -2*a*x_1*dx_2 = 2*a*x_1*dx_1
+
+// At x_1 = 0: x_2 = 1 and thus V2_bar = V2
+// Integrating the above equation from x_1 = 0 to x_1 in the RHS, and from V2_bar = V2 to V2 in the LHS, we get
+// V2_bar = V2 + a*x_1^(2) -  Molar volume of component 2(methanol) in the mixture 
+
+printf("The expression for the partial molar volume of methanol(2) is\nV2_bar = V2 + a*x_1^(2) [cubic cm/mol]\n\n");
+
+// At infinite dilution, x_2 approach 0 and thus x_1 approach 1, therefore
+x_1 = 1;// Mole fraction of component 1(water) at infinite dilution
+V2_bar_infinite = V2 + a*(x_1^(2));//[cubic cm/mol]
+
+printf("The partial molar volume of methanol at infinite dilution is %.1f cubic cm/mol",V2_bar_infinite);
\ No newline at end of file
-- 
cgit