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Sirish Chanda
Asked: 2022-10-29T01:03:43+05:30 In:

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 `:`
image
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

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 `:`
image
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
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1 Answer

  1. 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)`

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