Question

The rms velocity of gas molecules of a given amount of a gas at ${27}^{o}C$ and $1.0\times {10}^{5}N{m}^{-2}$ pressure is $200m{\sec }^{-1}$. If temperature and pressure are respectively ${127}^{o}C$ and $0.5\times {10}^{5}N{m}^{-2}$ the rms velocity will be :-

• A. $\frac{400}{\sqrt{3}}m{s}^{-1}$
• B. $100\sqrt{2}m{s}^{-1}$
• C. $100\frac{\sqrt{2}}{3}m{s}^{-1}$
• D. $50\sqrt{\frac{2}{3}}m{s}^{-1}$

Answer 1

The correct option is A $\frac{400}{\sqrt{3}}m{s}^{-1}$

We know that the rms velocity is given as $V_{rms}=\sqrt[2]{2\frac{3RT}{M}}$

We also know that increasing temperature will increase pressure so we just need to see the formula for temperature dependence and just apply above formula.

From above formula $\frac{v_1}{v_2}=\sqrt[2]{2\frac{T_1}{T_2}}=\sqrt[2]{2\frac{27+273}{127+273}}=\sqrt[2]{2\frac{3}{4}}$

so $v_2=\frac{2v_1}{\sqrt[2]{23}}=\frac{400}{\sqrt{3}}m/s$