Question

An ideal gas in 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 the gas are $p_2,V_{2}and{T}_2$ respectively.
For this expansion

• A. q = 0
• B. $T_2=T_1$
• C. $p_2{V}_2=p_1{V}_1$
• D. $p_2{V}_2^{\gamma }=p_1{V}_1^{\gamma }$

Answer 1

Answer: (A), (B), (C)

$p_2{V}_2=p_1{V}_1$

This problem includes concept of isothermal adiabatic irreversible expansion.
Process is adiabatic because of the use of thermal insolution therefore, q = 0
$\because P_{ext}=0$
$w=p_{ext}.\Delta V=0\times \Delta V=0$
Internal energy can be written as
$\Delta U=q+W=0$
The change in internal energy of an ideal gas depends only on temperature and change in internal energy ($\Delta U$) = 0 therefore, $\Delta T=0$ hence, process is isothermal and
$T_2=T_1$
and $p_2{V}_2=p_1{V}_1$
(d) $p_2{V}_2^{\gamma }=p_1{V}_1^{\gamma }$ is incorrect, it is valid for adiabatic reversible process.
Hence, only (a), (b) and (c) are correct choices.