Question

For a given gas at $1\text{atm}$ pressure, rms speed of the molecules is $200\text{m/s}$ at $127°\text{C}$. At $2\text{atm}$ pressure and at $227°\text{C}$, the rms speed of the molecules will be

• A. $100\text{m/s}$
• B. $80\text{m/s}$
• C. $80\sqrt{5}\text{m/s}$
• D. $100\sqrt{5}\text{m/s}$

Answer 1

The correct option is D $100\sqrt{5}\text{m/s}$

Given:
$P_1=1\text{atm},v_{rm{s}_1}=200\text{m/s},$
$T_1=127°\text{C}$
$P_2=2\text{atm},T_2=227°\text{C}$
The rms speed of the molecule$V_{rms}=\sqrt{\frac{3RT}{M}}$

$\Rightarrow V_{rms}\propto \sqrt{T}$

So,
$\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{(273+127)}{(273+227)}}=\sqrt{\frac{400}{500}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}$

$\therefore v_2=\frac{\sqrt{5}}{2}{v}_1=\frac{\sqrt{5}}{2}\times 200$

$v_2=100\sqrt{5}\text{m/s}$

Hence option $\left(C\right)$ is correct.