One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the P − V diagrams below. In cycle I, processes a, b, c, d are isobaric, isothermal, isobaric, and isochoric, respectively. In cycle II, processes a′, b′, c′, d′ are isothermal, isochoric, isobaric, and isochoric, respectively. The total work done during cycle I is W_I and that during cycle II is W_II. The ratio $\frac{W_{\mathrm{I}}}{W_{\mathrm{II}}}$ is .

The correct answer is 2.

Let us analyze the P - V diagrams for both cyclic processes step-by-step to calculate the total work done in each cycle.

1. Calculation of Work Done in Cycle I (W_I):

Cycle I consists of four processes: a, b, c, and d.

• Process a (Isobaric expansion):
Pressure is constant at P = 4P₀ as volume goes from V₀ to 2V₀.

$W_a=P·\Delta V=4P_0(2V_0-V_0)=4P_0{V}_0$

• Process b (Isothermal expansion):
The gas expands isothermally at initial state (4P₀, 2V₀) to final pressure 2P₀.
Since P₁V₁ = P₂V₂, we have:
(4P₀)(2V₀) = (2P₀)(V_final) ⇒ V_final = 4V₀.
Work done in isothermal expansion:

$W_b=nRT\ln \left(\frac{V_{\text{final}}}{V_{\text{initial}}}\right)=P_1{V}_1\ln \left(\frac{4V_0}{2V_0}\right)=8P_0{V}_0\ln \left(2\right)$

• Process c (Isobaric compression):
Pressure is constant at P = 2P₀ while volume decreases from 4V₀ back to V₀.

$W_c=2P_0(V_0-4V_0)=-6P_0{V}_0$

• Process d (Isochoric process):
Volume remains constant at V₀, so no work is done:

$W_d=0$

The total work done in Cycle I is:

$W_{\text{I}}=W_a+W_b+W_c+W_d=4P_0{V}_0+8P_0{V}_0\ln \left(2\right)-6P_0{V}_0+0$

$W_{\text{I}}=8P_0{V}_0\ln \left(2\right)-2P_0{V}_0$

2. Calculation of Work Done in Cycle II (W_II):

Cycle II consists of four processes: a′, b′, c′, and d′.

• Process a′ (Isothermal expansion):
Starts at state (4P₀, V₀) and expands isothermally to volume 2V₀.
Since P₁V₁ = P₂V₂, final pressure is 2P₀.
Work done:

$W_{a\prime }=P_1{V}_1\ln \left(\frac{2V_0}{V_0}\right)=4P_0{V}_0\ln \left(2\right)$

• Process b′ (Isochoric process):
Volume is constant at 2V₀ as pressure drops from 2P₀ to P₀.

$W_{b\prime }=0$

• Process c′ (Isobaric compression):
Pressure is constant at P = P₀ as volume goes from 2V₀ to V₀.

$W_{c\prime }=P_0(V_0-2V_0)=-P_0{V}_0$

• Process d′ (Isochoric process):
Volume remains constant at V₀.

$W_{d\prime }=0$

The total work done in Cycle II is:

$W_{\text{II}}=W_{a\prime }+W_{b\prime }+W_{c\prime }+W_{d\prime }=4P_0{V}_0\ln \left(2\right)-P_0{V}_0$

3. Calculating the Ratio:

Taking the ratio of W_I to W_II:

$\frac{W_{\text{I}}}{W_{\text{II}}}=\frac{8P_0{V}_0\ln \left(2\right)-2P_0{V}_0}{4P_0{V}_0\ln \left(2\right)-P_0{V}_0}=\frac{2\left[4P_0{V}_0\ln \left(2\right)-P_0{V}_0\right]}{4P_0{V}_0\ln \left(2\right)-P_0{V}_0}=2$

Hence, the ratio $\frac{W_{\mathrm{I}}}{W_{\mathrm{II}}}$ is equal to 2.