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 and d are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes a', b', c' and 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 $$W_I/W_{II}$$ is ____.

Total work done in a cyclic process is the sum of the work done in each individual stage, where $$W = P\Delta V$$ for isobaric, $$W = nRT\ln\left(\frac{V_f}{V_i}\right)$$ for isothermal, and $$W = 0$$ for isochoric processes.

For Cycle I:

$$W_a = 4P_0(2V_0 - V_0) = 4P_0V_0$$

$$W_b = nRT_b\ln\left(\frac{4V_0}{2V_0}\right) = (4P_0)(2V_0)\ln(2) = 8P_0V_0\ln 2$$

$$W_c = 2P_0(V_0 - 4V_0) = -6P_0V_0$$

$$W_d = 0$$

$$W_I = 4P_0V_0 + 8P_0V_0\ln 2 - 6P_0V_0 = 8P_0V_0\ln 2 - 2P_0V_0$$

For Cycle II:

$$W_{a'} = nRT_{a'}\ln\left(\frac{2V_0}{V_0}\right) = (4P_0)(V_0)\ln(2) = 4P_0V_0\ln 2$$

$$W_{b'} = 0$$

$$W_{c'} = P_0(V_0 - 2V_0) = -P_0V_0$$

$$W_{d'} = 0$$

$$W_{II} = 4P_0V_0\ln 2 - P_0V_0$$

Finding the ratio:

$$\frac{W_I}{W_{II}} = \frac{2(4P_0V_0\ln 2 - P_0V_0)}{4P_0V_0\ln 2 - P_0V_0} = 2$$