JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Heat & Thermodynamics

If one mole of an ideal gas at (

P_{1} V_{1}

) is allowed to expand reversibly and isothermally ( A to B ) its pressure is reduced to one- half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value

(B \to C)

. Then it is restored to its initial state by a reversible adiabatic compression ( C to A ). The net work done by the gas is equal to:

(A)0
(B) $RT ln 2$
(C) $RT (ln 2 - \frac{1}{2 ( γ - 1 )})$ ✓ Correct
(D) $- \frac{RT}{2 ( γ - 1 )}$

Answer: (C)

Solution

Along the isothermal

A \to B

the pressure halves, so the volume doubles to

2 V_{1}

and

W_{A B} = R T ln \frac{2 V _{1}}{V _{1}} = R T ln 2

The cooling

B \to C

is at constant volume, so

W_{B C} = 0

. At

C

the state is

(\frac{P _{1}}{4}, 2 V_{1})

, which gives

T_{C} = \frac{( P _{1} /4 ) ( 2 V _{1} )}{R} = \frac{P _{1} V _{1}}{2 R} = \frac{T}{2}

For the adiabatic compression

C \to A

the work done by the gas is

W_{C A} = \frac{R ( T _{C} - T _{A} )}{γ - 1} = \frac{R ( \frac{T}{2} - T )}{γ - 1} = - \frac{R T}{2 ( γ - 1 )}

Adding the three contributions, the net work done by the gas over the cycle is

R T (ln 2 - \frac{1}{2 ( γ - 1 )})

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Heat & Thermodynamics →

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Heat & Thermodynamics

Solution

More JEE Main questions on Heat & Thermodynamics