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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /647/CH6/EX6.5/Example6_5.sce | |
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| -rwxr-xr-x | 647/CH6/EX6.5/Example6_5.sce | 23 |
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diff --git a/647/CH6/EX6.5/Example6_5.sce b/647/CH6/EX6.5/Example6_5.sce new file mode 100755 index 000000000..53136b998 --- /dev/null +++ b/647/CH6/EX6.5/Example6_5.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 6.5
+// Page: 206
+
+printf("Example: 6.5 - Page: 206\n\n");
+
+// Solution
+
+// *****Data******//
+Tb = -103.9 + 273;// [K]
+deff('[P] = f1(T)','P = 10^(-(834.13/T) + 1.75*log10(T) - 8.375*10^(-3)*T + 5.324)');
+R = 8.314;// [J/mol K]
+//***************//
+
+// From Eqn. 6.60, we get:
+// d(ln(P))/dT = deltaH/(R*T^2)
+deff('[P] = f2(T)','P = exp(2.303*log10(f1(T)))');
+// Differentiating it with respect to T
+// d(ln(P))/dT = (834.13*2.303/Tb^2 + 1.75/Tb - 2.303*8.375*10^(-3))
+deltaH = R*Tb^2*(834.13*2.303/Tb^2 + 1.75/Tb - 2.303*8.375*10^(-3))/1000;// [kJ/mol]
+printf("The enthalpy of vaporisation is %.2f kJ/mol\n",deltaH);
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