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

Option: 1

\mathrm{P}_{3}>\mathrm{P}_{1}, \mathrm{~W}>0


Option: 2

\mathrm{P}_{3}<\mathrm{P}_{1}, \mathrm{~W}<0


Option: 3

\mathrm{P}_{3}>\mathrm{P}_{1}, \mathrm{~W}<0


Option: 4

\mathrm{P_{3}=P_{1}, W=0}


Answers (1)

best_answer

Slope of adiabatic process at a given state (\mathrm{P}, \mathrm{V}, \mathrm{T}) is more than the slope of isothermal process. The corresponding \mathrm{P}-\mathrm{V}graph for the two process is as shown in figure.

 

In the graph, \mathrm{AB} is isothermal and \mathrm{BC}  is adiabatic.
\mathrm{\mathrm{W}_{\mathrm{AB}}=} positivie (as volume is increasing)
And \mathrm{\mathrm{W}_{\mathrm{BC}}=} negative (as volume is decreasing) plus,

\mathrm{\left|\mathrm{W}_{\mathrm{BC}}\right|>\left|\mathrm{W}_{\mathrm{AB}}\right|,} as area under \mathrm{\mathrm{P}-\mathrm{V}}graph gives the work done.
  \mathrm{Hence, \mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}=\mathrm{W}<0}

From the graph itself, it is clear that \mathrm{\mathrm{P}_{3}>\mathrm{P}_{1}}.
Hence, the correct option is (c).

 

Posted by

Ajit Kumar Dubey

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