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

Concept: The root mean square (RMS) speed of a gas is given by: \[ v_{\text{rms}}=\sqrt{\frac{3RT}{M}} \] where \(T\) is the absolute temperature and \(M\) is the molar mass of the gas. If two gases have the same RMS speed, then: \[ \frac{T_1}{M_1}=\frac{T_2}{M_2} \]
Step 1: Write the equality for RMS speeds For hydrogen and oxygen: \[ \frac{T_{\mathrm{H_2}}}{M_{\mathrm{H_2}}} = \frac{T_{\mathrm{O_2}}}{M_{\mathrm{O_2}}} \]
Step 2: Substitute molar masses \[ M_{\mathrm{H_2}} = 2,\qquad M_{\mathrm{O_2}} = 32 \] \[ T_{\mathrm{H_2}} = T_{\mathrm{O_2}} \times \frac{2}{32} \]
Step 3: Convert temperature of \( \mathrm{O_2} \) to Kelvin \[ T_{\mathrm{O_2}} = 23^\circ\mathrm{C} = 23 + 273 = 296\ \mathrm{K} \]
Step 4: Calculate temperature of \( \mathrm{H_2} \) \[ T_{\mathrm{H_2}} = 296 \times \frac{2}{32} = \frac{296}{16} = 18.5\ \mathrm{K} \] Final Answer: \[ \boxed{18.5\ \mathrm{K}} \]