Question

If root mean square speed of ${CH}_4$ methane at $48K$ is same as the most probable speed of $H_2$ at $TK$, then $T$ is:

• A. $4K$
• B. $9K$
• C. $16K$
• D. $25K$

Answer 1

Answer: (B)

$9K$

Given, RMS of methane at 40 K is equal to MPS of hydrogen at $T$ $K$.

Now we know,

$RMS=u_{rms}=\sqrt{\frac{3RT}{M}}$ ------1

$MPS=u_{mps}=\sqrt{\frac{2RT}{M}}$--------2

Where $R$ is gas constant, $M$ is the molar mass of gas and $T$ is the temperature.

From 1,2 we get,

$RMS{u}_{{CH}_4}=\sqrt{\frac{3RT}{16}}$

$MPS{u}_{H_2}=\sqrt{\frac{2RT}{2}}$

$\sqrt{\frac{3RT}{16}}=\sqrt{\frac{2RT}{2}}$

$\sqrt{T}=\sqrt{\frac{3\times 48}{16}}$

$T=9K$

Temperature for hydrogen is $9K$.