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Diffstat (limited to '1019/CH5/EX5.4')
-rw-r--r-- | 1019/CH5/EX5.4/Example_5_4.sce | 27 |
1 files changed, 27 insertions, 0 deletions
diff --git a/1019/CH5/EX5.4/Example_5_4.sce b/1019/CH5/EX5.4/Example_5_4.sce new file mode 100644 index 000000000..fe910cd37 --- /dev/null +++ b/1019/CH5/EX5.4/Example_5_4.sce @@ -0,0 +1,27 @@ +//Example 5.4
+clear;
+clc;
+
+//Given
+n=1;//moles of gas
+V1=2;//initial volume in dm^3
+V2=20;//final volume in dm^3
+R=8.314;//gas constant in J K^-1 mol^-1
+T=300//temperature in K
+
+//To determine q,w,delE,delA,delG and delS
+w=-R*T*log(V2/V1);//work done in J
+delE=0;//isothermal expansion of ideal gas
+q=delE-w;//by 1st Law of thermodynamics
+delH=0;//delH=delE+del(n*R*T) and both are 0
+delA=-n*R*T*log(V2/V1);//helmoltz free energy in J
+delG=n*R*T*log(V1/V2);//Gibbs free energy in J
+delS=q/T;//entropy change in J K^-1
+mprintf('(i) w = %f J mol^-1',w);
+mprintf('\n (ii) delE = %f J since it is isothermal expansion of an ideal gas',delE);
+mprintf('\n (iii) q = %f J mol^-1',q);
+mprintf('\n (iv) delH = %f J mol^-1',delH);
+mprintf('\n (v) delA = %f J mol^-1',delA);
+mprintf('\n (vi) delG = %f J mol^-1',delG);
+mprintf('\n (vii) delS = %f J K^-1 mol^-1',delS);
+//end
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