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  • Figure shows a cyclic process ABCA for an ideal diatomic gas. The ratio of the heat energy absorbed in the process A → Bto the work done on the gas in the in the process B → C is <

Figure shows a cyclic process ABCA for an ideal diatomic gas. The ratio of the heat energy absorbed in the process A → Bto the work done on the gas in the in the process B → C is

Option: 1

\frac{7}{4 \ln (2)}


Option: 2

\frac{7}{2 \ln (2)}


Option: 3

\frac{5 \ln (2)}{4}


Option: 4

\frac{5 \ln (2)}{2}


Answers (1)

best_answer

In the process A → B, V is proportional to T. Hence pressure P remains constant. Therefore, heat energy absorbed in this process is \mathrm{\left(\because C_p=7 R / 2 \text { for a diatomic gas }\right)}

\mathrm{\begin{aligned} & (Q)_{A \rightarrow B}=n C_p \Delta T=n \times \frac{7 R}{2} \times\left(2 T_0-T_0\right) \\ & =\frac{7}{2} n R T_0 \end{aligned}}

Process B → C is isothermal in which the gas is compressed. Hence, work done on the
gas in this process is

\mathrm{\begin{aligned} & =-2 n R T_0 \ln \left(\frac{1}{2}\right) \\ & =2 n R T_0 \ln (2) \\ & \therefore \frac{(Q)_{A \rightarrow B}}{(W)_{B \rightarrow C}}=\frac{7}{4 \ln (2)}, \text { Which is choice (a) } \end{aligned}}

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Gautam harsolia

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