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← JEE Advanced 2023 · Paper 2
PhysicsInteger+4Question 30 of 51

JEE Advanced 2023 · Paper 2, Physics question 13

One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the PP-VV diagrams below. In cycle I, processes a,b,ca, b, c and dd are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes a′,b′,c′a', b', c' and d′d' are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is WIW_I and that during cycle II is WIIW_{II}. The ratio WI/WIIW_I/W_{II} is _____.

Two P (Pa) against V (m^3) diagrams, both clockwise cycles. (I): a is isobaric at 4P0 from V0 to 2V0; b is an isothermal curve from (2V0, 4P0) down to (4V0, 2P0); c is isobaric at 2P0 back from 4V0 to V0; d is isochoric at V0 from 2P0 up to 4P0. (II): a' is an isothermal curve from (V0, 4P0) to (2V0, 2P0); b' is isochoric at 2V0 down from 2P0 to P0; c' is isobaric at P0 back from 2V0 to V0; d' is isochoric at V0 from P0 up to 4P0.
Answer and worked solution

Official answer: 2

Write nRTnRT as PVPV at any point of an isotherm.

Cycle I (clockwise):

Step Process Work
a 4P04P_0, V0→2V0V_0\to2V_0 4P0V04P_0V_0
b isotherm, PV=8P0V0PV=8P_0V_0, 2V0→4V02V_0\to4V_0 8P0V0ln⁡28P_0V_0\ln2
c 2P02P_0, 4V0→V04V_0\to V_0 −6P0V0-6P_0V_0
d isochoric 0

WI=P0V0(8ln⁡2−2)W_I=P_0V_0(8\ln2-2).

Cycle II (clockwise):

Step Process Work
a′ isotherm, PV=4P0V0PV=4P_0V_0, V0→2V0V_0\to2V_0 4P0V0ln⁡24P_0V_0\ln2
b′ isochoric 0
c′ P0P_0, 2V0→V02V_0\to V_0 −P0V0-P_0V_0
d′ isochoric 0

WII=P0V0(4ln⁡2−1)W_{II}=P_0V_0(4\ln2-1).

WIWII=2.\frac{W_I}{W_{II}}=2.