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

---
 593/CH6/EX6.1/ex6_1.sce | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)
 create mode 100755 593/CH6/EX6.1/ex6_1.sce

(limited to '593/CH6/EX6.1')

diff --git a/593/CH6/EX6.1/ex6_1.sce b/593/CH6/EX6.1/ex6_1.sce
new file mode 100755
index 000000000..140b7a475
--- /dev/null
+++ b/593/CH6/EX6.1/ex6_1.sce
@@ -0,0 +1,27 @@
+clear;
+//clc();
+
+// Example 6.1
+// Page: 108
+printf("Example-6.1  Page no.-108\n\n");
+
+//***Data***//
+T = 20;//[C]
+m_1 = 0;//[molal]
+m_2 = 1;//[molal]
+// The data given in the figure 6.2 , as reported in book, can be repersented with excellent accuracy by a simple data fitting equation
+//V = 1.0019+0.054668*m-0.000418*m^(2);
+// Where 'V' is( solution volume, liters per 1000g of water ) and 'm' is the molality of ethanol in water
+//The partial molal volume is obtained by differentiating the expression of the 'V' with respect to 'm'
+// v_ethanol = dV/dm = 0.054668-2*0.000418*m
+// So that at zero molality 
+m = 0;//[molal]
+// the partial molal volume is 
+v_1 = 0.054668-2*0.000418*m;//[L/mol]
+// and at
+m = 1;//[molal]
+v_2 = 0.054668-2*0.000418*m;//[L/mol]
+v_1 = v_1*1000;//[cm^(3)/mol]
+v_2 = v_2*1000;//[cm^(3)/mol]
+printf("Partial molal volume of ethanol in water at zero molality is  %f cm^(3)/mol\n",v_1);
+printf(" Partial molal volume of ethanol in water at unity molality is %f cm^(3)/mol",v_2);
-- 
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