A) 1.20
B) 1.41
C) 1.67
D) 1.83
Correct Answer: B
Solution :
In isothermal process, temperature of the gas remains constant, so the gas obeys Boyle's law. That is, \[p\propto \frac{1}{V}\] \[\Rightarrow \] \[\frac{{{p}_{2}}}{{{p}_{1}}}=\frac{{{V}_{1}}}{{{V}_{2}}}\] \[\Rightarrow \] \[\frac{2p}{p}=\frac{{{V}_{1}}}{{{V}_{2}}}\] \[\therefore \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=2\] Now, the gas is expanded adiabatically, so \[p{{V}^{\gamma }}=cons\operatorname{tant}\] \[\frac{{{p}_{1}}}{{{p}_{2}}}={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma }}\] \[\Rightarrow \] \[\frac{2p}{0.75p}={{\left( \frac{2}{1} \right)}^{\gamma }}\] (since volume is restored) \[\Rightarrow \] \[\log \left( \frac{8}{3} \right)=\gamma \log 2\] \[\Rightarrow \] \[\log 8-log3=\gamma log2\] \[\therefore \] \[\gamma =1.41\]
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