question_answer

An ideal gas expands isothermally from a volume\[{{V}_{1}}\]and \[{{V}_{2}}\]and then compressed to original volume adiabatically. Initial pressure is\[{{P}_{1}}\]and final ressure is\[{{P}_{3}}\]The total work done is W. Then:

A)  \[{{P}_{3}}>{{P}_{1}},W>0\]

B)  \[{{P}_{3}}>{{P}_{1}},W<0\]

C)  \[{{P}_{3}}>{{P}_{1}},W<0\]

D)  \[{{P}_{3}}={{P}_{1}},W=0\]

Correct Answer: C

Solution :

                 Slope of adiabatic process at a given state (P,V,T) is more than the slope of isothermal process. The corresponding P-V graph for the two processes is as shown in figure.  In the graph, AB is isothermal and BC is adiabatic. \[{{W}_{AB}}=\]positive (as volume is increasing) and\[{{W}_{BC}}=\]negative (as volume is decreasing) \[|{{W}_{BC}}|\,>\,|{{W}_{AB}}|,\]as area under P-V graph gives the work done. Hence, \[{{W}_{AB}}+{{W}_{BC}}=W<0\] From the graph itself, it is clear that \[{{p}_{3}}>{{p}_{1}}\]