A liquid which is confined inside an adiabatic piston is suddently taken from state `-1` to state `-2` by a single stage irreversible process. If the piston comes to rest at point 2 as shown, then the enthalpy change for the process will be `:`

Options

A. `DeltaH=(2gammaP_(0)V_(0))/(gamma-1)`
B. `DeltaH=(3gammaP_(0)V_(0))/(gamma-1)`
C. `DeltaH=-P_(0)V_(0)`
D. None of these

Answer

3

Explanation

Since liquid is expanding against external pressure `P_(0)` hence work done
`w=-P_(0)(4V_(0)-V_(0))=-3P_(0)V_(0)`
`DeltaU=w=-3P_(0)V_(0)`
`rArr" "DeltaH=DeltaU+P_(2)V_(2)-P_(1)V_(1)=-3P_(0)V_(0)+4P_(0)V_(0)-2P_(0)V_(0)`