Question

If the RMS velocity of nitrogen molecules is $5.15{ms}^{-1}$ at 298 K, then a velocity of $10.30{ms}^{-1}$ will be possessed at a temperature of:

• A. 149 K
• B. 172.6 K
• C. 596 K
• D. 1192 K

Answer 1

The correct option is D 1192 K

Solution:- (D) $1192K$

As we know that the r.m.s. velocity is given as-

$V_{r.m.s.}=\sqrt{\frac{3RT}{M}}$

For same gas at different temperature,

$V_{r.m.s.}\propto \sqrt{T}$

$\frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}.....\left(1\right)$

Let the r.m.s. velocity of nitrogen molecules is $10.30m/s$ will be possessed at $TK$.

Given:-

$V_{298K}=5.15m/s$

$V_T=10.30m/s$

$T=?$

Now, from ${eq}^n\left(1\right)$, we have

$\frac{5.15}{10.30}=\sqrt{\frac{298}{T}}$

$\frac{1}{2}=\sqrt{\frac{298}{T}}$

Squaring both sides, we have

${\left(\frac{1}{2}\right)}^2={\left(\sqrt{\frac{298}{T}}\right)}^2$

$\frac{1}{4}=\frac{298}{T}$

$\Rightarrow T=298\times 4=1192K$

Hence the r.m.s. velocity $10.30m/s$ will be possessed at $1192K$.