
$P \propto T$
i.e., pressure will be doubled if temperature is doubled
$P = 2{P_0}$
Now let F be the tension in the wire. Then equilibrium of any one piston gives
$F = (P - {P_0})A = (2{P_0} - {P_0})A = {P_0}A$



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, $R =8.3 J K ^{-1} mol ^{-1}$ ]