Question

At what temperature will hydrogen molecules have the same root mean square speed as nitrogen molecules have at ${35}^{o}C$?

Answer 1

Solution:-

As we know that the root mean square velocity $\left(V_{rms}\right)$ of a gas is given by-

$V_{rms}=\sqrt{\frac{3RT}{M}}$

Therefore, root mean square velocity of nitrogen at $35℃$-

Molar mass of nitrogen $\left(M\right)=28g$

Temperature $\left(T\right)=35℃=(35+273)K=308K$

$\therefore V_{N_2}=\sqrt{\frac{3R\times 308}{28}}$

Let root mean square velocity of hydrogen at temperature $T$ will hve the same as of nitrogen at $35℃$

Molar mass of hydrogen $=2g$

$\therefore V_{H_2}=V_{N_2}=\sqrt{\frac{3R\times 308}{28}}.....\left(1\right)$

From the formula-

$V_{H_2}=\sqrt{\frac{3RT}{2}}.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\sqrt{\frac{3RT}{2}}=\sqrt{\frac{3R\times 308}{28}}$

Squaring both sides, we have

$\frac{3RT}{2}=\frac{3R\times 308}{28}$

$\Rightarrow T=\frac{308}{14}=22K$

Hence, at $22K$, the hydrogen molecules will have the same root mean square speed as of nitrogen molecules have at $35℃$.

Hence the correct answer is $22K$.