Discussion Forum

 An ideal gas in a thermally insulated vessel at internal pressure =p₁. Volume =V₁. and absolute temperature =T₁ expands irreversibly  against  zero  external pressure, as shown in the diagram.

 The final internal pressure, volume  and absolute temperature of the gas are p₂,V₂ and T₂  respectively. For this expansion

Answer:

Explanation:

Plan:This problem includes concept of isothermal adiabatic irreversible  expansion.

 Process is adiabatic  because of the use of thermal insolution therefore ,q=0

$\because   p_{ext}=0$

$W= p_{ext}.\triangle V$

$=0\times \triangle V=0$

Internal energy can be written as

  $\triangle$ U=q+w=0

 The change in internal  energy of an ideal gas depends only on temperature and change in internal energy ($\triangle U$)=0 therefore , $\triangle T=0$  hence , process is isothermal and

                         T₂=T₁

 and     $p_{2}V_{2}=p_{1}V_{1}$

 (d)   $p_{2}V_{2}^{r}=p_{1}V_{1}^{r}$ is incorrect. It is valid for adiabatic reversible  process.

 Hence, only (a) ,(b) and (c) are correct choices.