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: $$\boxed{20 \, \text{K}}$$