Question

Consider a certain mass of a gas at $127^\circ C$ and $2atm$ . The root-mean-square speed of the gas at this state is $\frac{200}{\sqrt{3}} \, ms^{- 1}$ . Now, find the root-mean-square speed of the gas(in $ms^{- 1}$ ) at $27^\circ C$ and $1atm$ :

Answer 1

Here, $v_{r m s_{1}}=\frac{200}{\sqrt{3}} \, m \, s^{- 1},T_{1}=\left(127\right)^{o}C=\left(127 + 273\right) \, K=400 \, K$
$P_{1}=2 \, atm, \, v_{r m s_{2}}=?, \, T_{2}=\left(27\right)^{o}C=\left(27 + 273\right) \, K=300 \, K$
$P_{2}=1 \, atm$
RMS velocity of a gas, $v_{r m s}=\sqrt{\frac{3 R T}{M}}$
Here, $M$ is the molar mass which will remains same.
Therefore, $v_{r m s}^{2} \propto T$
$\therefore \frac{v_{r m s_{2}}^{2}}{v_{r m s_{1}}^{2}}=\frac{T_{2}}{T_{1}}$
$v_{r m s_{2}}^{2}=v_{r m s_{1}}^{2}\times \frac{T_{2}}{T_{1}}=\left(\frac{200}{\sqrt{3}}\right)^{2}\times \frac{300}{400}$
$\Rightarrow v_{r m s_{2}}=100ms^{- 1}$