Ratio of Isothermal to Adiabatic Work for Expansion of Helium Gas

To find the value of $f$, we analyze the work done during each of the two thermodynamic stages for one mole ($n = 1$) of helium gas.

1. Properties of Helium Gas

Helium is a monoatomic ideal gas, which has $3$ degrees of freedom. Therefore, its adiabatic index $\gamma$ is given by: $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$

Let the initial state of the gas be $(P_1, V_1, T_1)$.

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2. Isothermal Expansion Stage

During the first stage, the gas undergoes an isothermal expansion at temperature $T_1$ from volume $V_1$ to volume $V_2 = 4V_1$.

The work done during this isothermal process, $W_{\text{iso}}$, is: $W_{\text{iso}} = n R T_1 \ln\left(\frac{V_2}{V_1}\right)$

Substituting $n = 1$ and $V_2 = 4V_1$: $W_{\text{iso}} = R T_1 \ln\left(\frac{4V_1}{V_1}\right) = R T_1 \ln(4) = 2 R T_1 \ln 2$

At the end of this stage, the temperature of the gas remains $T_2 = T_1$, and its volume is $V_2 = 4V_1$.

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3. Adiabatic Expansion Stage

During the second stage, the gas expands adiabatically from volume $V_2 = 4V_1$ to $V_3 = 32V_1$.

Using the relation $T V^{\gamma - 1} = \text{constant}$ for an adiabatic process: $T_2 V_2^{\gamma - 1} = T_3 V_3^{\gamma - 1}$

Substituting $T_2 = T_1$, $V_2 = 4V_1$, $V_3 = 32V_1$, and $\gamma = \frac{5}{3}$: $T_1 (4V_1)^{\frac{5}{3} - 1} = T_3 (32V_1)^{\frac{5}{3} - 1}$ $T_3 = T_1 \left(\frac{4V_1}{32V_1}\right)^{2/3} = T_1 \left(\frac{1}{8}\right)^{2/3}$

Since $\left(\frac{1}{8}\right)^{2/3} = (2^{-3})^{2/3} = 2^{-2} = \frac{1}{4}$, the final temperature $T_3$ is: $T_3 = \frac{T_1}{4}$

The work done during the adiabatic process, $W_{\text{adia}}$, is given by: $W_{\text{adia}} = \frac{n R (T_2 - T_3)}{\gamma - 1}$

Substituting $n = 1$, $T_2 = T_1$, $T_3 = \frac{T_1}{4}$, and $\gamma - 1 = \frac{2}{3}$: $W_{\text{adia}} = \frac{R \left(T_1 - \frac{T_1}{4}\right)}{\frac{2}{3}} = \frac{R \cdot \frac{3}{4} T_1}{\frac{2}{3}} = \frac{9}{8} R T_1$

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4. Calculation of the Ratio $\frac{W_{\text{iso}}}{W_{\text{adia}}}$

Taking the ratio of $W_{\text{iso}}$ to $W_{\text{adia}}$: $\frac{W_{\text{iso}}}{W_{\text{adia}}} = \frac{2 R T_1 \ln 2}{\frac{9}{8} R T_1} = \frac{2}{\frac{9}{8}} \ln 2 = \frac{16}{9} \ln 2$

Given that $\frac{W_{\text{iso}}}{W_{\text{adia}}} = f \ln 2$, we get: $f = \frac{16}{9} \approx 1.78$