Thermodynamics

JEE Main level, single correct, medium

180 of 190

Question 180 of 190

The work done on the gas in taking it from DD to AA is

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Answer: (A) −414R-414 \mathrm{R}

At constant temperature (isothermal process)
WDA=nRTln⁡(P1P2)=2.303×2R×300log⁡(1052×105)=2.303×600Rlog⁡(12)=0.693×600R=−414R.\begin{aligned} W_{D A}=n R T \ln \left(\frac{P_1}{P_2}\right)= & 2.303 \times 2 \mathrm{R} \times 300 \log \left(\frac{10^5}{2 \times 10^5}\right) \\ & =2.303 \times 600 \mathrm{R} \log \left(\frac{1}{2}\right) \\ & =0.693 \times 600 \mathrm{R}=-414 \mathrm{R} . \end{aligned}

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