Question

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

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

Hint:

For gases having equal RMS speeds, temperature is directly proportional to molar mass. Lighter gases must be at lower temperature to have the same RMS speed as heavier gases.

Answer 1

The Correct Option is A

To find the temperature of the \( \mathrm{H_2} \) gas when the RMS speed of \( \mathrm{H_2} \) and \( \mathrm{O_2} \) are the same, we can use the formula for Root Mean Square (RMS) speed, which is given by:

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

Where:

• \(v_{\text{rms}}\) is the RMS speed.
• \(k\) is the Boltzmann constant.
• \(T\) is the temperature in Kelvin.
• \(m\) is the mass of one molecule of the gas.

Given that the RMS speeds of \( \mathrm{H_2} \) and \( \mathrm{O_2} \) are the same, we equate the RMS speed formulas for both gases:

\[\sqrt{\frac{3kT_{\mathrm{H_2}}}{m_{\mathrm{H_2}}}} = \sqrt{\frac{3kT_{\mathrm{O_2}}}{m_{\mathrm{O_2}}}}\]

Canceling out common terms, this simplifies to:

\[\frac{T_{\mathrm{H_2}}}{m_{\mathrm{H_2}}} = \frac{T_{\mathrm{O_2}}}{m_{\mathrm{O_2}}}\]

The molecular mass of \( \mathrm{H_2} \) is approximately 2 g/mol, and the molecular mass of \( \mathrm{O_2} \) is approximately 32 g/mol. Therefore, we can write:

\[\frac{T_{\mathrm{H_2}}}{2} = \frac{296.15}{32}\]

Simplifying this gives:

\[\frac{T_{\mathrm{H_2}}}{2} = 9.2547\]

This results in:

\(T_{\mathrm{H_2}} = 2 \times 9.2547 = 18.5094 \text{ K}\]\)

Rounding this value, we find the temperature of the \( \mathrm{H_2} \) gas to be approximately 18.5 K.

Hence, the correct answer is 18.5 K.