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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /647/CH2/EX2.10/Example2_10.sce | |
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-rwxr-xr-x | 647/CH2/EX2.10/Example2_10.sce | 24 |
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diff --git a/647/CH2/EX2.10/Example2_10.sce b/647/CH2/EX2.10/Example2_10.sce new file mode 100755 index 000000000..c533cb897 --- /dev/null +++ b/647/CH2/EX2.10/Example2_10.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 2.10
+// Page: 53
+
+printf("Example: 2.10 - Page: 53\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 300;// [K]
+V1 = 30;// [L]
+V2 = 3;// [L]
+Cv = 5;// [cal/mol]
+R = 2;// [cal/K mol]
+//*************//
+
+Cp = Cv + R;// [cal/mol]
+gama = Cp/Cv;
+// The relation between temperature and volume of ideal gas undergoing adiabatic change is given by:
+// (T2/T1) = (V1/V2)^(gama - 1)
+T2 = T1 * (V1/V2)^(gama - 1);// [K]
+printf("The final temperature is %.1f K\n",T2);
\ No newline at end of file |