1. Given: one mole of helium gas.

For helium, being a monoatomic ideal gas, $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$

We need the ratio $\frac{W_{\text{iso}}}{W_{\text{adia}}} = f\ln 2$

2. Isothermal expansion from $V_1$ to $4V_1$.

For one mole of ideal gas, work done in isothermal expansion is $W_{\text{iso}} = nRT_1 \ln\left(\frac{V_f}{V_i}\right)$ Here $n=1$, and initially $P_1V_1=RT_1$, so $W_{\text{iso}} = P_1V_1\ln\left(\frac{4V_1}{V_1}\right)$ $W_{\text{iso}} = P_1V_1\ln 4 = 2P_1V_1\ln 2$

3. State after isothermal expansion.

Since the process is isothermal, $P_2V_2 = P_1V_1$ with $V_2 = 4V_1$ So, $P_2 = \frac{P_1V_1}{4V_1} = \frac{P_1}{4}$ Also, temperature remains $T_1$.

4. Adiabatic expansion from $4V_1$ to $32V_1$.

For adiabatic process, $TV^{\gamma-1} = \text{constant}$ Thus, $T_2(4V_1)^{\gamma-1} = T_3(32V_1)^{\gamma-1}$ So, $T_3 = T_2\left(\frac{4}{32}\right)^{\gamma-1} = T_1\left(\frac{1}{8}\right)^{2/3}$ Now, $\left(\frac{1}{8}\right)^{2/3} = \frac{1}{4}$ Hence, $T_3 = \frac{T_1}{4}$

5. Work done in adiabatic expansion.

For adiabatic expansion of ideal gas, $W_{\text{adia}} = nC_V(T_2-T_3)$ For one mole of monoatomic gas, $C_V = \frac{3R}{2}$ Therefore, $W_{\text{adia}} = \frac{3R}{2}\left(T_1-\frac{T_1}{4}\right)$ $W_{\text{adia}} = \frac{3R}{2}\cdot \frac{3T_1}{4} = \frac{9}{8}RT_1$ Using $RT_1=P_1V_1$, $W_{\text{adia}} = \frac{9}{8}P_1V_1$

6. Take the ratio.

$\frac{W_{\text{iso}}}{W_{\text{adia}}} = \frac{2P_1V_1\ln 2}{(9/8)P_1V_1}$ $\frac{W_{\text{iso}}}{W_{\text{adia}}} = \frac{16}{9}\ln 2$

Comparing with $\frac{W_{\text{iso}}}{W_{\text{adia}}} = f\ln 2$ we get $f = \frac{16}{9} \approx 1.78$

7. Comparison with stored answer.

Stored correct answer = $1.77$.

Our derived value is $f = \frac{16}{9} = 1.777\ldots$ which rounds to $1.78$, and to two decimal places may also be reported as $1.77$ depending on truncation.

So the stored answer is essentially consistent.