A) 8 times
B) 16 times
C) \[\frac{1}{8}\] times
D) 4 times
Correct Answer: B
Solution :
\[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] ie, \[{{v}_{rms}}\propto \sqrt{T}\] To reduce rms velocity 2 times, temperature must be reduced 4 times, ie, \[{{T}_{2}}=\frac{1}{4}{{T}_{1}}\] In an adiabatic process, \[{{T}_{2}}V_{2}^{\gamma -1}={{T}_{1}}V_{1}^{\gamma -1}\] \[{{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}=\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{4}{1}\] \[\frac{{{V}_{2}}}{{{V}_{1}}}={{(4)}^{1/\gamma -1}}={{(4)}^{\frac{1}{1.5-1}}}={{(4)}^{2}}=16\] \[{{V}_{2}}=16{{V}_{1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
