question_answer

'n' moles of an ideal gas undergoes a process \[A\to B\] as shown in the figure. The maximum temperature of the gas during the process will be:

A) \[\frac{9{{P}_{0}}{{V}_{0}}}{2nR}\]              

B) \[\frac{9{{P}_{0}}{{V}_{0}}}{nR}\]

C) \[\frac{9{{P}_{0}}{{V}_{0}}}{4nR}\]  

D)        \[\frac{3{{P}_{0}}{{V}_{0}}}{2nR}\]

Correct Answer: C

Solution :

The equation for the line is \[P=\frac{-{{P}_{0}}}{{{V}_{0}}}V+3P\,[slope\,\,=\frac{-{{P}_{0}}}{{{V}_{0}}},\,c=3{{P}_{0}}]\] \[P{{V}_{0}}+{{P}_{0}}V=3{{P}_{0}}{{V}_{0}}\]                ?.(i) But   \[PV=nRT\] \[\therefore \,\,P=\frac{nRT}{V}\]                                                            ....(ii) From (i) and (ii) \[\frac{nRT}{V}{{V}_{0}}+{{P}_{0}}V=3{{P}_{0}}{{V}_{0}}\] \[\therefore \,\,nRT\,\,{{V}_{0}}+{{P}_{0}}{{V}^{2}}=3{{P}_{0}}{{V}_{0}}\]           ?.(iii) For temperature to be maximum \[\frac{dT}{dV}=0\] Differentiating e.q. (iii) by 'V? we get \[nR{{V}_{0}}\frac{dT}{dV}+{{P}_{0}}(2V)=3{{P}_{0}}{{V}_{0}}\] \[\therefore \,\,nR{{V}_{0}}\frac{dT}{dV}=3{{P}_{0}}{{V}_{0}}-2{{P}_{0}}V\] \[\frac{dT}{dV}=\frac{3{{P}_{0}}{{V}_{0}}-2{{P}_{0}}V}{nR{{V}_{0}}}=0\] \[V=\frac{3{{V}_{0}}}{2}\]      \[\therefore \,\,P=\frac{3{{P}_{0}}}{2}\] [From (i)] \[\therefore \,\,{{T}_{\max }}=\frac{9{{P}_{o}}{{V}_{o}}}{4nR}\]        [From (iii)]