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 $\frac{200}{\sqrt{3}}m{s}^{-1}$ . Now, find the root-mean-square speed of the gas(in $m{s}^{-1}$ ) at $27^{\circ }C$ and $1atm$ :

Answer 1

Answer: 100

Here, $v_{rm{s}_1}=\frac{200}{\sqrt{3}}m{s}^{-1},T_1={\left(127\right)}^{o}C=\left(127+273\right)K=400K$
$P_1=2atm,v_{rm{s}_2}=?,T_2={\left(27\right)}^{o}C=\left(27+273\right)K=300K$
$P_2=1atm$
RMS velocity of a gas, $v_{rms}=\sqrt{\frac{3RT}{M}}$
Here, $M$ is the molar mass which will remains same.
Therefore, $v_{rms}^2\propto T$
$\therefore \frac{v_{rm{s}_2}^2}{v_{rm{s}_1}^2}=\frac{T_2}{T_1}$
$v_{rm{s}_2}^2=v_{rm{s}_1}^2\times \frac{T_2}{T_1}={\left(\frac{200}{\sqrt{3}}\right)}^2\times \frac{300}{400}$
$\Rightarrow v_{rm{s}_2}=100m{s}^{-1}$