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    				| 1. | 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 `:` 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» Correct Answer - 3 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)` | |