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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 /1919/CH5/EX5.13 | |
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initial commit / add all books
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| -rwxr-xr-x | 1919/CH5/EX5.13/Ex5_13.sce | 24 |
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diff --git a/1919/CH5/EX5.13/Ex5_13.sce b/1919/CH5/EX5.13/Ex5_13.sce new file mode 100755 index 000000000..f0d98174e --- /dev/null +++ b/1919/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,24 @@ +
+// Theory and Problems of Thermodynamics
+// Chapter 5
+//Second Law of Thermodynamcis
+// Example 13
+
+clear ;clc;
+
+//Given data
+N = 1 // number of moles of nitrogen
+r = 1.4 // ratio of specific heats of nitrogen at 0.1 MPa and 300 K
+P1 = 0.1 // initial state pressure in MPa
+T1 = 300 // initial state temperature in K
+P2 = 0.5 // final state pressure in MPa
+T2 = 600 // final state temperature in K
+R = 8.314 // universal gas constant J/mol/K
+
+// Calculation for determining the change in the entrophy of nitrogen
+// del_S = N{Cp ln(T2/T1) - R ln(P2/P1)}
+del_S = N*{(R*r/(r -1))*log(T2/T1) - R*log(P2/P1)}
+
+// Output Results
+mprintf('The change in entrophy of nitrogen = %5.3f kJ/mol K' ,del_S)
+
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