Question

At which temperature will the r.m.s. velocity of a hydrogen molecule be equal to that of an oxygen molecule at \( 47^\circ \text{C} \)?

• A. \( 80 \, \text{K} \)
• B. \( -73 \, \text{K} \)
• C. \( 4 \, \text{K} \)
• D. \( 20 \, \text{K} \)

Hint:

The r.m.s. velocity is influenced by both the temperature and the molar mass of the gas. Use the proportionality \( v_{\text{rms}} \propto \sqrt{\frac{T}{M}} \) to connect the properties of different gases.

Answer 1

The Correct Option is D

The root mean square (r.m.s.) velocity \( v_{\text{rms}} \) of gas molecules is determined by the formula: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \] where: - \( R \) is the universal gas constant, - \( T \) is the absolute temperature in Kelvin, - \( M \) is the molar mass of the gas. Step 1: Equating \( v_{\text{rms}} \) for Hydrogen and Oxygen For hydrogen (\( H_2 \)) and oxygen (\( O_2 \)): \[ v_{\text{rms}}(H_2) = v_{\text{rms}}(O_2). \] Substitute the expression for \( v_{\text{rms}} \): \[ \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}. \] Square both sides: \[ \frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}. \] Rearrange to solve for \( T_{H_2} \): \[ T_{H_2} = T_{O_2} \cdot \frac{M_{H_2}}{M_{O_2}}. \] Step 2: Substitute Known Values The temperature of oxygen gas is: \[ T_{O_2} = 47^\circ \text{C} + 273 = 320 \, \text{K}. \] Molar masses are: \[ M_{H_2} = 2, \quad M_{O_2} = 32. \] Substitute into the formula for \( T_{H_2} \): \[ T_{H_2} = 320 \cdot \frac{2}{32}. \] Simplify: \[ T_{H_2} = 320 \cdot \frac{1}{16} = 20 \, \text{K}. \] Final Answer: \[ \boxed{20 \, \text{K}} \]