Question

The RMS speeds of \( \mathrm{H_2} \) and \( \mathrm{O_2} \) gases are the same. If the temperature of \( \mathrm{O_2} \) gas is \(23^\circ\mathrm{C}\), find the temperature of \( \mathrm{H_2} \) gas.

• A. \(18.5\ \mathrm{K}\)
• B. \(2.5^\circ\mathrm{C}\)
• C. \(18^\circ\mathrm{C}\)
• D. \(164\ \mathrm{K}\)

Hint:

If RMS speeds are equal: \[ T \propto M \] Always convert temperature into Kelvin before substituting in gas equations.

Answer 1

The Correct Option is A

To solve the problem, we need to use the concept of Root Mean Square (RMS) speed for gases, which is derived from the kinetic theory of gases.

The RMS speed, \( v_{\text{rms}} \), of a gas is given by the formula: 

\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)

where:

• \(k\) is the Boltzmann constant.
• \(T\) is the absolute temperature in Kelvin.
• \(m\) is the mass of a molecule of the gas.

Given that the RMS speeds of \(\mathrm{H_2}\) and \(\mathrm{O_2}\) gases are the same, we have:

\(\sqrt{\frac{3kT_{\mathrm{H_2}}}{m_{\mathrm{H_2}}}} = \sqrt{\frac{3kT_{\mathrm{O_2}}}{m_{\mathrm{O_2}}}}\)

By canceling out the common terms and squaring both sides, we get:

\(\frac{T_{\mathrm{H_2}}}{m_{\mathrm{H_2}}} = \frac{T_{\mathrm{O_2}}}{m_{\mathrm{O_2}}}\)

Rearranging gives:

\(T_{\mathrm{H_2}} = T_{\mathrm{O_2}} \cdot \frac{m_{\mathrm{H_2}}}{m_{\mathrm{O_2}}}\)

Let's plug in the known values:

• Temperature of \(\mathrm{O_2}\) gas: \(T_{\mathrm{O_2}} = 23^\circ\mathrm{C} = 296\ \mathrm{K}\).
• Molar mass of \(\mathrm{H_2}\): 2 g/mol.
• Molar mass of \(\mathrm{O_2}\): 32 g/mol.

Substitute these values into the equation:

\(T_{\mathrm{H_2}} = 296\ \mathrm{K} \times \frac{2\ \mathrm{g/mol}}{32\ \mathrm{g/mol}}\)

Simplifying, we find:

\(T_{\mathrm{H_2}} = 296\ \mathrm{K} \times \frac{1}{16} = 18.5\ \mathrm{K}\)

Thus, the temperature of the \(\mathrm{H_2}\) gas is \(18.5\ \mathrm{K}\).