A cylinder with insulating walls holds a mixture of 2.0 mol of helium and 3.0 mol of nitrogen, at…

Step 1. Gamma (a). In units of $R$: $C_V = 2.0 \times \tfrac32 + 3.0 \times \tfrac52 = 3.0 + 7.5 = 10.5$ and $C_P = 2.0 \times \tfrac52 + 3.0 \times \tfrac72 = 5.0 + 10.5 = 15.5$. So

Step 2. Temperature (b). $TV^{\gamma - 1}$ is fixed, so $T_2 = T_1 \times 4^{\gamma - 1} = 300\ \text{K} \times 4^{0.476} = 300\ \text{K} \times 1.935 = 581\ \text{K}$.

Step 3. Work (c). No heat flows, so the work done on the gas is the rise in internal energy:

Step 4. Pressure (d). $PV^{\gamma}$ is fixed, so $P_2 = P_1 \times 4^{\gamma} = 1.0 \times 10^{5}\ \text{Pa} \times 7.74 = 7.74 \times 10^{5}\ \text{Pa}$.

Step 5. Helium speed (e). Every gas in the mixture has the same temperature. For helium $v_{\text{rms}} = \sqrt{\dfrac{3RT_2}{M}}$. The top line is $3 \times 8.314\ \text{J/(mol K)} \times 580.5\ \text{K} = 14479\ \text{J/mol}$, so

It was $1368\ \text{m/s}$ at the start.