Question

A liquid which is confined inside an adiabatic piston is suddenly taken from state $1$ to state $2$ by expanding it against constant external pressure of $P _{0}$. If the piston comes to rest at point $2$ as shown. Then the enthalpy change for the process will be :

• A. $\Delta H =\frac{2 \gamma P _{0} V _{0}}{\gamma-1}$
• B. $\Delta H=\frac{3 \gamma P_{0} V_{0}}{\gamma-1}$
• C. $\Delta H =- P _{0} V _{0}$
• D. None of these

Answer 1

Answer: (C)

Since liquid is expanding against external pressure $P _{0}$ hence work done
$w=-P_{0}\left(4 V_{0}-V_{0}\right)=-3 P_{0} V_{0}$
$\Delta U=w=-3 P_{0} V_{0}$
$\Rightarrow \Delta H=\Delta U+P_{2} V_{2}-P_{1} V_{1}$
$=-3 P_{0} V_{0}+4 P_{0} V_{0}-2 P_{0} V_{0}$