Question 1[Maximum number: 3]A sealed vessel contains a mass of 0.0424 kg of an ideal gas at 227∘C227^{\circ} \mathrm{C}227∘C. The pressure of the gas is 1.37×105 Pa1.37 \times 10^{5} \mathrm{~Pa}1.37×105 Pa and the volume of the gas is 0.640 m30.640 \mathrm{~m}^{3}0.640 m3.Calculate:the root-mean-square (r.m.s.) speed v of the molecules of the gas.v=ms−1\begin{aligned} & v= \\ & \mathrm{ms}^{-1} \end{aligned}v=ms−1Show Answer12m⟨c2⟩=(3/2)kT\frac{1}{2} m\left\langle c^{2}\right\rangle=(3 / 2) k T21m⟨c2⟩=(3/2)kT3.34×10−27×v2=3×1.38×10−23×5003.34 \times 10^{-27} \times v^{2}=3 \times 1.38 \times 10^{-23} \times 5003.34×10−27×v2=3×1.38×10−23×500v=2490 m s−1v=2490 \mathrm{~m} \mathrm{~s}^{-1}v=2490 m s−1orpV=1/3(Nm)⟨c2⟩p V=1 / 3(N m)\left\langle c^{2}\right\ranglepV=1/3(Nm)⟨c2⟩ and N m= mass of gas0.0424×v2=3×1.37×105×0.6400.0424 \times v^{2}=3 \times 1.37 \times 10^{5} \times 0.6400.0424×v2=3×1.37×105×0.640v=2490 m s−1v=2490 \mathrm{~m} \mathrm{~s}^{-1}v=2490 m s−1Add to Test