One mole of an ideal diatomic gas (CV=5/2R, γ=1.4) at 300 K is compressed reversibly and adiabatically to one-eighth of its initial volume. It is…

Step 1 is a reversible adiabatic compression, so $TV^{\gamma-1}$ is constant: $T_2=300\times8^{0.4}=689.2\ \mathrm{K}$. Work done on the gas in this step equals the rise in internal energy, $W_1=C_V\Delta T=2.5\times8.314\times(689.2-300)=8090\ \mathrm{J}$, so the work done by the gas is $-8.09\ \mathrm{kJ}$. Step 2 is an isothermal reversible expansion at $689.2\ \mathrm{K}$ from $V_0/8$ to $V_0$, a volume ratio of $8$. The work done by the gas is $W_2=RT_2\ln8=8.314\times689.2\times2.0794=11916\ \mathrm{J}=11.92\ \mathrm{kJ}$. Net work done by the gas is $11.92-8.09=3.83\ \mathrm{kJ}$. The positive sign means the isothermal expansion at the higher temperature delivers more work than the adiabatic compression consumed; the extra energy comes from the heat absorbed during the isothermal step. The gas ends at its original volume but at $689.2\ \mathrm{K}$.