A) 64
B) 32
C) 16
D) 8
Correct Answer: B
Solution :
Root mean square velocity of gas molecules is given by \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] or \[{{v}_{rms}}\propto \sqrt{T}\] or \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] or \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{v}_{1}}}{{{v}_{2}}} \right)}^{2}}\] Here, \[{{v}_{2}}=\frac{{{v}_{1}}}{2}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=4\] ?(i) For adiabatic expansion, \[T{{V}^{\gamma -1}}=\text{constant,}\] or \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{1/\gamma -1}}\] or \[{{V}_{2}}={{V}_{1}}{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{1/\gamma -1}}\] \[\therefore \] \[{{V}_{2}}={{V}_{1}}{{(4)}^{1/\left( \frac{7}{5}-1 \right)}}\] \[\left( \because \,\gamma =\frac{7}{5} \right)\] or \[{{V}_{2}}=V{{(4)}^{5/2}}\] or \[{{V}_{2}}=32\,V\] Hence, gas should be expanded 32 times adiabatically.
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