A) \[\frac{R}{(\gamma -1)}\]
B) P V
C) \[\frac{P\,V}{(\gamma -1)}\]
D) \[\frac{\gamma \,PV}{(\gamma -1)}\]
Correct Answer: C
Solution :
Change in internal energy is \[\Delta U=\frac{1}{(\gamma -1)}({{P}_{2}}{{V}_{2}}-{{P}_{1}}{{V}_{1}})\] Here, \[{{V}_{1}}=V,\,{{V}_{2}}=2V\] \[\therefore \Delta U=\frac{1}{\gamma -1}[P\times 2V-P\times V]\] \[=\frac{1}{\gamma -1}\times PV\] \[=\frac{PV}{\gamma -1}\] Note: The internal energy of an ideal gas depends only on its absolute temperature (T) and is directly proportional to T.
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