A) 0.5
B) 0.75
C) 0.99
D) 0.25
Correct Answer: B
Solution :
[b] The efficiency of cycle is \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] For adiabatic process \[T{{V}^{\gamma -1}}=\]constant For diatomic gas \[\gamma =\frac{7}{5}\] \[{{T}_{1}}{{V}_{1}}^{\gamma -1}=\,\,\,\,{{T}_{2}}{{V}_{2}}^{\gamma -1}\text{ }{{T}_{1}}={{T}_{2}}{{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}\] \[{{T}_{1}}={{T}_{2}}{{(32)}^{\frac{7}{5}-1}}={{T}_{2}}{{({{2}^{5}})}^{2/5}}={{T}_{2}}\times 4\] \[{{T}_{1}}=4{{T}_{2}}\] \[\therefore \,\,\,\eta =\left( 1-\frac{1}{4} \right)=\frac{3}{4}=0.75\]
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