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 In2, then 𝑓 is ________.

One mole of helium behaves as an ideal, monatomic gas, therefore
$$\gamma=\frac{C_p}{C_v}=\frac{5}{3}, \qquad C_v=\frac{3}{2}R$$

Isothermal expansion (state-1 : $$V_1,\,T$$ → state-2 : $$4V_1,\,T$$)
For an isothermal process the work done is
$$W_{\text{iso}} = nRT\ln\!\left(\frac{V_2}{V_1}\right)$$
Here $$n=1,\;V_2=4V_1$$, hence
$$W_{\text{iso}} = R\,T\;\ln 4 = R\,T\;\ln(2^2)=2RT\ln 2 \qquad -(1)$$

Adiabatic expansion (state-2 : $$4V_1,\,T$$ → state-3 : $$32V_1$$)

For an adiabatic process of an ideal gas
$$T\,V^{\gamma-1}=\text{constant}$$

With $$V_2=4V_1,\;T_2=T,\;V_3=32V_1$$ we get
$$T_3 = T_2\left(\frac{V_2}{V_3}\right)^{\gamma-1} = T\left(\frac{4V_1}{32V_1}\right)^{\frac{5}{3}-1} = T\left(\frac{1}{8}\right)^{\frac{2}{3}}$$

Since $$8^{2/3} = (2^3)^{2/3}=2^2=4$$,
$$T_3=\frac{T}{4}$$

The work done in an adiabatic process equals the change in internal energy:
$$W_{\text{adia}} = nC_v\,(T_2-T_3) = 1\cdot\frac{3}{2}R\left(T-\frac{T}{4}\right) = \frac{3}{2}R\left(\frac{3T}{4}\right) = \frac{9}{8}RT \qquad -(2)$$

Required ratio
$$\frac{W_{\text{iso}}}{W_{\text{adia}}} = \frac{2RT\ln 2}{\frac{9}{8}RT} = 2\ln 2 \times \frac{8}{9} = \frac{16}{9}\ln 2$$

Comparing with the given form $$\dfrac{W_{\text{iso}}}{W_{\text{adia}}}=f\ln 2$$,
$$f = \frac{16}{9}$$

Final answer: $$\displaystyle f=\frac{16}{9}$$