An ideal gas in a thermally insulated vessel at internal pressure $=P_{1}$, volume $=V_{1}$ and abso

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,\

Professor PC Thomas Classes · Answer

( $\mathbf{a}, \mathrm{b}, \mathrm{c}$ ) : Since, vessel is thermally insulated, i.e. the process is adiabatic hence, $q=0$\ Also, $P_{	ext {ext }}=0$, hence $w=0$\ From $1^{	ext {st }}$ law of thermodynamics,\ $\Delta E=q+w$\ $	herefore \quad \Delta E=0$ (for ideal gas)\ $	herefore \quad \Delta T=0 \quad$ or $\quad T_{2}=T_{1}$\ $[\because$ Internal energy of an ideal gas is a function of temperature.]\ Applying ideal gas equation, $P V=n R T$ 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.\