JEE Advanced 2019 Paper 2 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

One mole of a monatomic ideal gas undergoes a cyclic process A→ B→ C→ A in the P-V plane. A is the state (V0,P0) with temperature T0=P0V0/R. The…

One or more options correct

One mole of a monatomic ideal gas undergoes a cyclic process

A\to B\to C\to A

in the

P

V

plane.

A

is the state

(V_0,P_0)

with temperature

T_0=P_0V_0/R

. The process

AB

is an isobaric expansion to volume

2V_0

BC

is an isochoric decrease of pressure to

P_0/2

, and

CA

is a straight line in the

P

V

plane that returns the gas to

A

. Which of the following statement(s) is(are) correct?

A. The net work done by the gas in one cycle is $\tfrac14P_0V_0$ .
B. The highest temperature attained by the gas during the process $CA$ is $\tfrac98T_0$ .
C. The efficiency of the cycle is exactly $10\%$ .
D. Although the net heat exchanged in the process $CA$ is negative, the gas absorbs heat during a part of $CA$ .

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Answer: A, B, D

The cycle is a right triangle with base

V_0

and height

P_0/2

, traversed clockwise, so

W=\tfrac12 V_0\cdot\tfrac{P_0}{2}=\tfrac14P_0V_0

and (A) is correct. On

CA

P=\tfrac32P_0-\dfrac{P_0V}{2V_0}

, so

PV

has its maximum at

V=\tfrac32V_0

with value

\tfrac98P_0V_0

, giving

T_{\max}=\tfrac98T_0

on this segment and (B) is correct. Heat in:

Q_{AB}=\tfrac52P_0V_0

. On

CA

dQ=dU+P\,dV\propto(3.75-2V/V_0)\,dV

changes sign at

V=1.875V_0

, so the gas absorbs a small positive heat

\tfrac{1}{64}P_0V_0

near

C

although the net

Q_{CA}=-\tfrac34P_0V_0

; (D) is correct. Hence

\eta=\dfrac{0.25}{2.5+0.0156}\approx9.94\%

, not exactly

10\%

; (C) is wrong. Answer: A, B, D.

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