Question

A liquid confined inside as adiabatic container is suddenly taken from state 1 to state 2 by single stage process as shown, then choose the correct options(s)

• A. $\Delta H=\frac{2\gamma P_0{V}_0}{\gamma -1}$
• B. $\Delta H=-P_0{V}_0$
• C. $w=-3P_0{V}_0$
• D. $\Delta U=\frac{2P_0{V}_0}{\gamma -1}$

Answer 1

Answer: (B), (C)

$w=-3P_0{V}_0$

Since the process is adiabatic, Q = 0
According to first law of thermodynamic
$\Delta U=Q+w=w$
$\Delta U=w=-P\Delta V$
$w=-P_0(4V_0-V_0)=-3P_0{V}_0$
$\Delta H=\Delta U+\Delta \left(PV\right)$
$\Delta H=\Delta U+(P_2{V}_2-P_1{V}_1)$
$\Delta H=-3P_0{V}_0+(4P_0{V}_0-2P_0{V}_0)$
$\Delta H=-3P_0{V}_0+2P_0{V}_0$
$\Delta H=-P_0{V}_0$
Thus option (b) and (c) are correct.