Solution:

The root mean square (r.m.s.) speed \( v_{\text{rms}} \) of gas molecules is given by the formula:

\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

where:
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature (assumed to be room temperature here),
- \( m \) is the mass of one molecule of the gas.

Rearranging to find \( m \):

\[
m = \frac{3 k_B T}{v_{\text{rms}}^2}
\]

For a diatomic gas like \( \text{H}_2 \), using the given \( v_{\text{rms}} = 1930 \, \text{m/s} \), we calculate that this speed corresponds to the molecular mass of hydrogen (\( \text{H}_2 \)). Therefore, the gas is identified as \( \text{H}_2 \).