Physics - Heat & Thermo Question with Solution | TestHub

At constant temperature (isothermal process) $$\begin{array}{rl}{W}_{DA}=nRT\ln \left(\frac{P_1}{P_2}\right)=& 2.303\times 2\mathrm{R}\times 300\log \left(\frac{10^5}{2\times 10^5}\right)\\ & =2.303\times 600\mathrm{R}\log \left(\frac{1}{2}\right)\\ & =0.693\times 600\mathrm{R}=-414\mathrm{R}.\end{array}$$