Q68 / 90 · JEE Main 2022 (29 Jun Shift 2) · Physics

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2, if the process is purely adiabatic and W3 if the process is purely isobaric. Then, choose the correct option

  • Correct answer
    A
    W2<W1<W3
  • B
    W1<W2<W3
  • C
    W2<W3<W1
  • D
    W3<W1<W2

Solution

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.

Figure

Also, W= area under P-V curve.

Hence,

∴ W2<W1<W3

Hence, the correct answer is option: 'A'.

Relevant Questions