C

To find the temperature at which the rms (root mean square) speed of hydrogen ($H_2$) equals that of nitrogen ($N_2$), we need to use the equation for rms speed, which is given by the formula:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where:
- $v_{rms}$ is the root mean square speed
- $k$ is the Boltzmann constant
- $T$ is the absolute temperature (in Kelvin)
- $m$ is the mass of one molecule in kilograms.

For nitrogen, the molar mass is 28 g/mol, so the mass $m_{N_2}$ of one nitrogen molecule is:
\[ m_{N_2} = \frac{28 g/mol}{N_A} = \frac{28 \times 10^{-3} kg}{6.022 \times 10^{23}} \approx 4.65 \times 10^{-26} kg \]
For hydrogen, the molar mass is 2 g/mol, so the mass $m_{H_2}$ of one hydrogen molecule is:
\[ m_{H_2} = \frac{2 g/mol}{N_A} = \frac{2 \times 10^{-3} kg}{6.022 \times 10^{23}} \approx 3.32 \times 10^{-27} kg \]
Using the rms speed equation, we set the speeds equal:
\[ \sqrt{\frac{3kT_{N_2}}{m_{N_2}}} = \sqrt{\frac{3kT_{H_2}}{m_{H_2}}} \]
As the factor of \( 3k \) cancels, it simplifies to:
\[ \frac{T_{N_2}}{m_{N_2}} = \frac{T_{H_2}}{m_{H_2}} \]
Rearranging gives:
\[ T_{H_2} = T_{N_2} \cdot \frac{m_{H_2}}{m_{N_2}} \]
Plugging in values:
\( T_{N_2} = 300 K \),
\( m_{H_2} \approx 3.32 \times 10^{-27} kg \),
\( m_{N_2} \approx 4.65 \times 10^{-26} kg \)
So,
\[ T_{H_2} = 300 \cdot \frac{3.32 \times 10^{-27}}{4.65 \times 10^{-26}} \approx 21.43 K \]
Therefore, the equivalent temperature at which the rms speed of $H_2$ equals that of $N_2$ is approximately 21.43 K.
In the context of options presented, this is indicated as option C (not explicitly listed in visible output, but inferred). Therefore, C.