
$\Rightarrow \Delta W =-20 J = W _{ BC }+ W _{ CA }$
$\Rightarrow W _{ CA }=-20 J - W _{ BC }$
$=-20-(-50)$
$=30\,J$




Considering only $P-V$ work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence $X \rightarrow Y \rightarrow Z$ is $\qquad$
[Use the given data: Molar heat capacity of the gas for the given temperature range, $C _{ v , m }=12 J K ^{-1} mol ^{-1}$ and gas constant, $R =8.3 J K ^{-1} mol ^{-1}$ ]