Question

An ideal gas in a thermally insulated vessel at internal pressure $= P_{1}$, volume $= V_{1}$ and absolute temperature $= T_{1}$ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of gas are $P_{2}$, $V_{2}$ and $T_{2}$, respectively

For this expansion, which of the following option is not correct

• A. $q = 0$
• B. $T_{2} = T_{1}$
• C. $P_{2}V_{2} = P_{1}V_{1}$
• D. $P_{2}V_{2}^{r} =P_{1}V^{y}_{1}$

Answer 1

Answer: (D)

Since vessel is thermally insulated, i.e. the process is adiabatic hence, $q = 0$
Also, $P_{\text{ext}}=0$, hence $w = 0 $
From $1^{st}$ law of thermodynamics, $\Delta\,E =q+w$
$\therefore \Delta\,E=0$ (for ideal gas)
$\therefore \Delta T=0$
or $T_{2}=T_{1}$
$[\because$ Internal energy of an ideal gas is a function of temperature]
Applying ideal gas equation, $PV = nRT$
where $n, R$ and $T$ are constant,
then $P_{1}V_{1} = P_{2}V_{2}$
Equation, $PV^{\gamma}$ = constant, is applicable only for ideal gas in reversible adiabatic process
Hence , $P_{2}V_{2}^{\gamma} =P_{1}V^{\gamma}_{1}$ equation is not applicable