A liquid which is confined inside an adiabatic piston is suddenly taken from state 1 to state 2 by a single stage process. If the piston comes to rest at point 2 as shown, then the enthalpy change for the process will be:
.
- A$\Delta H=\frac{2\gamma P_0V_0}{\gamma-1}$
- B$\Delta H=\frac{3\gamma P_0V_0}{\gamma-1}$
- C$\Delta H=-P_0V_0$
- DNone of these
✓ Correct answer: C
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