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Q.
The most probable velocity (in cm/s) of hydrogen molecule at $27^{\circ} C$ will be
UP CPMTUP CPMT 2008
Solution:
Most probable velocity is $V _{ mps }=\sqrt{\frac{8 RT }{\pi M }}$
$
\begin{array}{l}
R =8.314 \\
T =27+273=300 K
\end{array}
$
$M =2 g / mol$ molecular mass of hydrogen
Most probable velocity will be $\sqrt{\frac{8 \times 8.314 \times 300}{3.14 \times 2 \times 10^{-3}}}=1782.50 m / s =17.8 \times 10^{4} cm / s$