Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$. It expands isothermally to volume $4V_1$. After this, the gas expands adiabatically and its volume becomes $32V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{iso}$ and $W_{adia}$ respectively. If the ratio $\frac{W_{iso}}{W_{adia}} = f \ln 2$, then $f$ is ______.

Step 1: Identify the properties of Helium gas.

Helium is a monatomic gas, which determines its heat capacities and adiabatic index $\gamma$.

$$C_V = \frac{3}{2}R$$ $$C_P = \frac{5}{2}R$$ $$\gamma = \frac{C_P}{C_V} = \frac{5}{3}$$

Step 2: Calculate the work done during the isothermal expansion.

The gas expands isothermally at temperature $T_1$ from volume $V_1$ to $4V_1$.

$$W_{\text{iso}} = nRT_1 \ln\left(\frac{4V_1}{V_1}\right)$$ $$W_{\text{iso}} = P_1V_1 \ln 4$$ $$W_{\text{iso}} = 2P_1V_1 \ln 2$$

Step 3: Determine the final temperature after the adiabatic expansion.

The gas expands adiabatically from volume $4V_1$ to $32V_1$.

$$T_1 (4V_1)^{\gamma-1} = T_2 (32V_1)^{\gamma-1}$$ $$T_2 = T_1 \left(\frac{4V_1}{32V_1}\right)^{\gamma-1}$$ $$T_2 = T_1 \left(\frac{1}{8}\right)^{2/3}$$ $$T_2 = \frac{T_1}{4}$$

Step 4: Calculate the work done during the adiabatic expansion.

Use the formula for work done in an adiabatic process.

$$W_{\text{adia}} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$ $$W_{\text{adia}} = \frac{nR\left(T_1 - \frac{T_1}{4}\right)}{\frac{5}{3} - 1}$$ $$W_{\text{adia}} = \frac{\frac{3}{4}nRT_1}{\frac{2}{3}}$$ $$W_{\text{adia}} = \frac{9}{8}P_1V_1$$

Step 5: Find the ratio of the work done to determine $f$.

Divide the isothermal work by the adiabatic work.

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

Comparing this to the given relation $\frac{W_{\text{iso}}}{W_{\text{adia}}} = f \ln 2$ yields the value of $f$.

$$f = \frac{16}{9}$$

Answer: $\dfrac{16}{9}$