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. $P_2{V}_2^r=P_1{V}_1^y$

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, $P{V}^{\gamma }$ = constant, is applicable only for ideal gas in reversible adiabatic process
Hence , $P_2{V}_2^{\gamma }=P_1{V}_1^{\gamma }$ equation is not applicable