Change in Internal Energy in Adiabatic Expansion of Gas

For a monoatomic ideal gas, the degrees of freedom are $f = 3$, and the adiabatic index (ratio of specific heats) is given by: $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$

Let the initial state of the gas be $(P_1, V_1) = (P, V)$ and the final state be $(P_2, V_2) = (P_2, 27V)$.

For a reversible adiabatic expansion, the state variables follow the relation: $P_1 V_1^\gamma = P_2 V_2^\gamma$

Substituting the given values into the equation: $P V^{5/3} = P_2 (27V)^{5/3}$

Solving for the final pressure $P_2$: $P_2 = P \left(\frac{V}{27V}\right)^{5/3} = P \left(\frac{1}{27}\right)^{5/3} = P \left(3^{-3}\right)^{5/3} = P \cdot 3^{-5} = \frac{P}{243}$

Now, the final product of pressure and volume $P_2 V_2$ is: $P_2 V_2 = \left(\frac{P}{243}\right) (27V) = \frac{27}{243} PV = \frac{1}{9} PV$

The work done $W$ by an ideal gas during an adiabatic expansion is given by: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

Substituting the values of $P_1 V_1$, $P_2 V_2$, and $\gamma$: $W = \frac{PV - \frac{1}{9}PV}{\frac{5}{3} - 1} = \frac{\frac{8}{9}PV}{\frac{2}{3}} = \frac{8}{9} \times \frac{3}{2} PV = \frac{4}{3} PV$

According to the First Law of Thermodynamics: $Q = \Delta U + W$

Since the process is adiabatic, the heat exchange $Q = 0$, which gives: $\Delta U = -W = -\frac{4}{3} PV$

Thus, the change in internal energy of the gas is $-\frac{4}{3} PV$.