Starting with the same initial conditions, an ideal gas expands from volume V _ 1 to V _ 2 in three different ways. The work done by the gas is W _ 1 if the process is purely isothermal, W _ 2 , if the process is purely

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Correct answer: A. W _ 2 < W _ 1 < W _ 3

As we know, in isobaric process pressure remains constant and magnitude of slope of adiabatic curve will be more than the isothermal curve at the common point, therefore diagram for process will be as given below.

Starting with the same initial conditions, an ideal gas expands from volume V _ 1 to V _ 2 in three different ways. The

Also, $W =$ area under $P - V$ curve.

Hence,

$∴ {W}_{2} < {W}_{1} < {W}_{3}$

Starting with the same initial conditions, an ideal gas expands...

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