Kinetic Theory Of Gases Question 160
Certain perfect gas is found to obey $ P{{V}^{3/2}}= $ constant during adiabatic process. If such a gas at initial temperature T is adiabatically compressed to half the initial volume, its final temperature will be $ T^{1/2} $
Options:
A) $ \sqrt{2},\cdot T $
2T
C) $ 2\sqrt{2},\cdot T $
4T
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ P{{V}^{3/2}}= $ Constant, $ PV=RT $ or $ P=\frac{RT}{V}; $ $ ( \frac{RT}{V} )\times {{V}^{3/2}}= $ Constant or $ T{{V}^{1/2}}= $ Constant.
$ \
Therefore, $ V $ changes to $ V/2 $, the temp. becomes $ \sqrt{2T} $.
BETA
