At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at 47°C

The correct option is 20 K.

Step-by-Step Explanation:

1. Recall the formula for Root Mean Square (RMS) Velocity:
The root mean square velocity ($v_{\mathrm{rms}}$) of a gaseous molecule is given by the formula:
$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$
where:
- $R$ is the universal gas constant,
- $T$ is the absolute temperature (in Kelvin), and
- $M$ is the molar mass of the gas.

2. Set up the condition given in the problem:
We are given that the RMS velocity of hydrogen molecules ($H_2$) at an unknown temperature $T_{H_2}$ is equal to the RMS velocity of oxygen molecules ($O_2$) at $47°C$.
$v_{\mathrm{rms}\left(H_2\right)}=v_{\mathrm{rms}\left(O_2\right)}$

3. Convert temperatures and gather molar masses:
- Temperature of oxygen, $T_{O_2}=47°C=47+273=320\text{K}$
- Molar mass of hydrogen gas ($H_2$), $M_{H_2}=2\text{g/mol}$
- Molar mass of oxygen gas ($O_2$), $M_{O_2}=32\text{g/mol}$

4. Equate the two velocities:
$\sqrt{\frac{3R{T}_{H_2}}{M_{H_2}}}=\sqrt{\frac{3R{T}_{O_2}}{M_{O_2}}}$

Squaring both sides and simplifying by canceling the common factor $3R$:
$\frac{T_{H_2}}{M_{H_2}}=\frac{T_{O_2}}{M_{O_2}}$

5. Calculate the unknown temperature ($T_{H_2}$):
Substitute the known values into the equation:
$\frac{T_{H_2}}{2}=\frac{320}{32}$
$\frac{T_{H_2}}{2}=10$
$T_{H_2}=10\times 2=20\text{K}$

Thus, the root mean square velocity of hydrogen molecules at 20 K is equal to that of oxygen molecules at 47°C.