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Q.
The work done on the gas in taking it from $D$ to $A$ is
AIEEEAIEEE 2009Thermodynamics
Solution:
At constant temperature (isothermal process)
$W_{DA}=nRT\,In\left(\frac{P_{1}}{P_{2}}\right)=2.303\times2R\times300\,log\left(\frac{10^{5}}{2\times10^{5}}\right)$
$=2.303\times600R\,log\left(\frac{1}{2}\right)$
$=0.693\times600\,R=-414\,R.$
