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Q.
At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at $47^{\circ} C$ ?
Kinetic Theory
Solution:
For oxygen $v_{ O _{2}}=\sqrt{\frac{3 R T_{ O _{2}}}{M_{ O _{2}}}}$ and
For hydrogen $v_{ H _{2}}=\sqrt{3 R \frac{T_{ H _{2}}}{M_{ H _{2}}}}$
According to problem $=\sqrt{\frac{3 R T_{ O _{2}}}{M_{ O _{2}}}}$
$=\sqrt{3 R \frac{T_{ H _{2}}}{M_{ H _{2}}}}$
$\Rightarrow \frac{T_{ O _{2}}}{M_{ O _{2}}}=\frac{T_{ H _{2}}}{M_{ H _{2}}}$
$\Rightarrow \frac{47+273}{32}=\frac{T_{ H _{2}}}{M_{ H _{2}}}$
$\Rightarrow T_{ H _{2}}=\frac{320}{32} \times 2=20\, K$