JEE Challenger
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PhyMedAdv 2023 • Paper 2

Ratio of Work Done in Cyclic Processes I and II

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 ________.

Question Diagram 1
Official Numerical Answer2

Topics & Concepts

ThermodynamicsFirst Law of Thermodynamics

Step-by-Step Solution

To find the ratio of the total work done in cycle I (WIW_I) to that in cycle II (WIIW_{II}), we calculate the work done in each individual thermodynamic process for both cycles.

1. Work Done in Cycle I (WIW_I)

Cycle I consists of four processes:

  • Process aa (Isobaric expansion): The gas expands from volume V0V_0 to 2V02V_0 at a constant pressure P=4P0P = 4P_0. Wa=PΔV=4P0(2V0−V0)=4P0V0W_a = P \Delta V = 4P_0 (2V_0 - V_0) = 4P_0 V_0

  • Process bb (Isothermal expansion): The state changes from (2V0,4P0)(2V_0, 4P_0) to (4V0,2P0)(4V_0, 2P_0). The product PV=(4P0)(2V0)=8P0V0=nRT1PV = (4P_0)(2V_0) = 8P_0 V_0 = nRT_1 is constant throughout this process. Wb=nRT1ln⁡(VfVi)=8P0V0ln⁡(4V02V0)=8P0V0ln⁡2W_b = nRT_1 \ln\left(\frac{V_f}{V_i}\right) = 8P_0 V_0 \ln\left(\frac{4V_0}{2V_0}\right) = 8P_0 V_0 \ln 2

  • Process cc (Isobaric compression): The gas is compressed from 4V04V_0 to V0V_0 at a constant pressure P=2P0P = 2P_0. Wc=PΔV=2P0(V0−4V0)=−6P0V0W_c = P \Delta V = 2P_0 (V_0 - 4V_0) = -6P_0 V_0

  • Process dd (Isochoric process): The volume remains constant at V=V0V = V_0 while the pressure increases from 2P02P_0 to 4P04P_0. Wd=0W_d = 0

Summing the work done in all four processes for cycle I: WI=Wa+Wb+Wc+WdW_I = W_a + W_b + W_c + W_d WI=4P0V0+8P0V0ln⁡2−6P0V0+0W_I = 4P_0 V_0 + 8P_0 V_0 \ln 2 - 6P_0 V_0 + 0 WI=8P0V0ln⁡2−2P0V0=2P0V0(4ln⁡2−1)W_I = 8P_0 V_0 \ln 2 - 2P_0 V_0 = 2P_0 V_0 (4\ln 2 - 1)

2. Work Done in Cycle II (WIIW_{II})

Cycle II consists of four processes:

  • Process a′a' (Isothermal expansion): The state changes from (V0,4P0)(V_0, 4P_0) to (2V0,2P0)(2V_0, 2P_0). The product PV=(4P0)(V0)=4P0V0=nRT2PV = (4P_0)(V_0) = 4P_0 V_0 = nRT_2 is constant throughout this process. Wa′=nRT2ln⁡(VfVi)=4P0V0ln⁡(2V0V0)=4P0V0ln⁡2W_{a'} = nRT_2 \ln\left(\frac{V_f}{V_i}\right) = 4P_0 V_0 \ln\left(\frac{2V_0}{V_0}\right) = 4P_0 V_0 \ln 2

  • Process b′b' (Isochoric process): The volume remains constant at V=2V0V = 2V_0 while pressure drops from 2P02P_0 to P0P_0. Wb′=0W_{b'} = 0

  • Process c′c' (Isobaric compression): The gas is compressed from 2V02V_0 to V0V_0 at a constant pressure P=P0P = P_0. Wc′=PΔV=P0(V0−2V0)=−P0V0W_{c'} = P \Delta V = P_0 (V_0 - 2V_0) = -P_0 V_0

  • Process d′d' (Isochoric process): The volume remains constant at V=V0V = V_0 while pressure rises from P0P_0 to 4P04P_0. Wd′=0W_{d'} = 0

Summing the work done in all four processes for cycle II: WII=Wa′+Wb′+Wc′+Wd′W_{II} = W_{a'} + W_{b'} + W_{c'} + W_{d'} WII=4P0V0ln⁡2+0−P0V0+0W_{II} = 4P_0 V_0 \ln 2 + 0 - P_0 V_0 + 0 WII=P0V0(4ln⁡2−1)W_{II} = P_0 V_0 (4\ln 2 - 1)

3. Ratio of WIW_I to WIIW_{II}

WIWII=2P0V0(4ln⁡2−1)P0V0(4ln⁡2−1)=2\frac{W_I}{W_{II}} = \frac{2P_0 V_0 (4\ln 2 - 1)}{P_0 V_0 (4\ln 2 - 1)} = 2

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