1.0 mol of an ideal gas (γ = 1.40) at 320 K and 1.0 atm expands to three times its volume, once…

Step 1. Pressures. Isothermal: $\tfrac13 = 0.333\ \text{atm}$. Adiabatic: $3^{-1.40} = 0.215\ \text{atm}$, lower because the gas also cools.

Step 2. Temperatures. Isothermal: $320\ \text{K}$. Adiabatic: $320\ \text{K} \times 3^{-0.40} = 206\ \text{K}$, because the work comes from $U$.

Step 3. Work. Isothermal: $1.0\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 320\ \text{K} \times \ln 3 = 2921\ \text{J}$. Adiabatic: $1.0\ \text{mol} \times 20.775\ \text{J mol}^{-1}\text{K}^{-1} \times (320\ \text{K} - 206.2\ \text{K}) = 2364\ \text{J}$.