A vessel of 2.0 L holds 4.0 × 10²² helium atoms…

Step 1. The pressure. We use $P = \dfrac{Nmu_{\text{rms}}^2}{3V}$. The mass of all the atoms is $Nm = 4.0 \times 10^{22} \times 6.65 \times 10^{-27} = 2.66 \times 10^{-4}\ \text{kg}$. Then

Step 2. The temperature from the pressure. The gas is ideal, so $PV = NkT$. We want $T$, so $T = \dfrac{PV}{Nk} = \dfrac{7.49 \times 10^4 \times 2.0 \times 10^{-3}}{4.0 \times 10^{22} \times 1.381 \times 10^{-23}} = 271\ \text{K}$.

Step 3. The temperature from the energy. The kinetic energy of one atom is $\tfrac{1}{2}mu_{\text{rms}}^2 = \tfrac{1}{2} \times 6.65 \times 10^{-27} \times 1300^2 = 5.62 \times 10^{-21}\ \text{J}$. This equals $\tfrac{3}{2}kT$, so $T = \dfrac{2 \times 5.62 \times 10^{-21}}{3 \times 1.381 \times 10^{-23}} = 271\ \text{K}$.