Question

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

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

Answer 1

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

Given,
Pressure, $P_1=1\text{atm}{P}_2=2\text{atm}$
Temperature, $T_1={127}^{\circ }\text{C}=400\text{K}{T}_2={227}^{\circ }\text{C}=500\text{K}$
$V_{rms,1}=200\text{m/s}{V}_{rms,2}=?$

For case - 1 $V_{rms,1}=\sqrt{\frac{3R{T}_1}{M}}\therefore V_{rms,1}=200=\sqrt{\frac{3R\times 400}{M}}..\left(1\right)$
For case - 2
$V_{rms,2}=\sqrt{\frac{3R\times 500}{M}}...\left(2\right)$
[Gas is same so molar mass $M$ is same]

On dividing equation $\left(2\right)$ by equation $\left(1\right)$,
$\frac{V_{rms,2}}{200}=\sqrt{\frac{\frac{3R\times 500}{M}}{\frac{3R\times 400}{M}}}\Rightarrow V_{rms,2}=\frac{\sqrt{5}}{2}\times 200\Rightarrow V_{rms,2}=100\sqrt{5}\text{m/s}$