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(B) For a liquid,the change in enthalpy is given by $\Delta H = \Delta U + \Delta(PV) = \Delta U + P_2V_2 - P_1V_1$.Since the container is adiabatic,t

Vedclass · Accepted Answer

$\Delta H = -2.5\,P_0V_0$

Vedclass · Answer

(B) For a liquid,the change in enthalpy is given by $\Delta H = \Delta U + \Delta(PV) = \Delta U + P_2V_2 - P_1V_1$.Since the container is adiabatic,the heat exchange $q = 0$.According to the first law of thermodynamics,$\Delta U = q + w = w$.The work done $w$ in a process is the negative of the area under the $P-V$ curve.Area under the curve (trapezium) $= \frac{1}{2} 	imes (P_1 + P_2) 	imes (V_2 - V_1) = \frac{1}{2} 	imes (2P_0 + P_0) 	imes (4V_0 - V_0) = \frac{1}{2} 	imes 3P_0 	imes 3V_0 = 4.5\,P_0V_0$.Since the process is an expansion,work is done by the system,so $w = -4.5\,P_0V_0$.Thus,$\Delta U = -4.5\,P_0V_0$.Now,$\Delta H = \Delta U + (P_2V_2 - P_1V_1) = -4.5\,P_0V_0 + (P_0 	imes 4V_0 - 2P_0 	imes V_0) = -4.5\,P_0V_0 + (4P_0V_0 - 2P_0V_0) = -4.5\,P_0V_0 + 2P_0V_0 = -2.5\,P_0V_0$.

$A$ liquid confined inside an adiabatic container is taken from state $1$ to state $2$ by a single-stage process as shown in the $P-V$ diagram. Then,$\Delta H$ is:

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