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 :

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) Δ H =(2 &gamma; P 0 V 0/&gamma;-1) (B) Δ H=(3 &gamma; P0 V0/&gamma;-1) (C) Δ H =- P 0 V 0 (D) None of these

Tardigrade Admin · Accepted Answer

Since liquid is expanding against external pressure P 0 hence work done w=-P0(4 V0-V0)=-3 P0 V0 Δ U=w=-3 P0 V0 ⇒ Δ H=Δ U+P2 V2-P1 V1 =-3 P0 V0+4 P0 V0-2 P0 V0

Q. 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 :

$Δ H = \frac{2 γ P _{0} V _{0}}{γ - 1}$

$Δ H = \frac{3 γ P _{0} V _{0}}{γ - 1}$

$Δ H = - P_{0} V_{0}$

None of these

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

Q. 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 P0​. If the piston comes to rest at point 2 as shown. Then the enthalpy change for the process will be :

ΔH=γ−12γP0​V0​​

ΔH=γ−13γP0​V0​​

ΔH=−P0​V0​