An ideal gas initially at a state (P 1 , V 1 ) is allowed to expand isothermally to a state (P 2 , V 2 ). Then the gas is compressed adiabatically to its initial volume V 1 . Let the final pressure be P 3 and the work done by the gas during the whole process be W, then
Text Solution
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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 process
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) plus,
|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 .
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