Discussion Forum



1)

 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

2232021905_thermal.JPG


A) q=0

B) $T_{2}$=$T_{1}$

C) $ p_{2}V_{2}=p_{1}V_{1}$

D) $p_{2}V_{2}^{r}=p_{1}V_{1}^{r}$

Answer:

Option A,B,C

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.