Kinetic Theory Of Gases Question 3
Question: Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity $ v $ . The molecular mass of gas is $ M $ . The rise in temperature of the gas when the vessel is suddenly stopped is $ (\gamma =C _{P}/C _{V}) $
Options:
A) $ \frac{Mv^{2}(\gamma -1)}{2R(\gamma +1)} $
B) $ \frac{Mv^{2}(\gamma -1)}{2R} $
C) $ \frac{Mv^{2}}{2R(\gamma +1)} $
D) $ \frac{Mv^{2}}{2R(\gamma -1)} $
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Answer:
Correct Answer: B
Solution:
[b] If m is the total mass of the gas, then its kinetic energy $ =1/2mv^{2} $ when the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because process will be adiabatic), i.e., $ \frac{1}{2}mv^{2}=\mu C _{v}\Delta T=\frac{m}{M}C _{v}\Delta T $
$ \Rightarrow \frac{m}{M}\frac{R}{\gamma -1}\Delta T=\frac{1}{2}mv^{2} $
$ ( As,C _{V}=\frac{R}{^{\gamma -1}} ) $
$ \Rightarrow \Delta T=\frac{Mv^{2}(\gamma -1)}{2R} $
BETA
