A) 2 times
B) 4 times
C) 8 times
D) 16 times
Correct Answer: D
Solution :
The root mean square velocity, \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\Rightarrow {{v}_{rms}}=\sqrt{T}\] So, to reduce \[{{v}_{rms}}\] velocity 2 times, then temperature of the gas is to be reduced four times. i.e., \[\frac{T}{T}=\frac{1}{4}\] During adiabatic process \[T{{V}^{\gamma -1}}=T{{V}^{\gamma -1}}\] \[\Rightarrow \] \[\frac{V}{V}={{\left( \frac{T}{T} \right)}^{\frac{1}{\gamma -1}}}={{(4)}^{\frac{1}{1.5-1}}}\] \[={{(4)}^{2}}=16\,\,\Rightarrow \,\,V=16\,V\]
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