Question

At what 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. \(40\,\text{K}\)
• B. \(20\,\text{K}\)
• C. \(10\,\text{K}\)
• D. \(5\,\text{K}\)

Hint:

Always convert the given temperature from Celsius to Kelvin before doing anything else, since this is a common place to lose marks. Then use the fact that $v_{rms}$ depends on temperature divided by molar mass, so equal r.m.s speeds mean equal ratios of temperature to molar mass for the two gases.

Answer 1

The Correct Option is B

Concept: The r.m.s. velocity of a gas molecule is given by: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \(T\) = absolute temperature and \(M\) = molar mass. Thus, \[ v_{\text{rms}} \propto \sqrt{\frac{T}{M}} \]

Step 1: Equate the r.m.s velocities. \[ \sqrt{\frac{T_H}{M_H}} = \sqrt{\frac{T_O}{M_O}} \] Squaring both sides: \[ \frac{T_H}{M_H} = \frac{T_O}{M_O} \]

Step 2: Substitute molar masses. \[ M_H = 2, \quad M_O = 32 \] Temperature of oxygen: \[ 47^\circ\text{C} = 320\,\text{K} \] \[ \frac{T_H}{2} = \frac{320}{32} \]

Step 3: Solve for \(T_H\). \[ \frac{T_H}{2} = 10 \] \[ T_H = 20\,\text{K} \] Thus, the required temperature is \[ \boxed{20\,\text{K}} \]

Answer 2

Concept:

• At the molecular level, the average kinetic energy of a gas molecule depends only on the absolute temperature, not on the identity of the gas: $\dfrac{1}{2}mv_{rms}^2 = \dfrac{3}{2}k_BT$, where $m$ is the mass of a single molecule and $k_B$ is the Boltzmann constant.
• Setting the $v_{rms}$ of the two gases equal at their respective temperatures links temperature and molecular mass directly, since the Boltzmann constant cancels out.

Step 1: Write the per molecule kinetic energy relation for each gas.
For hydrogen: $\dfrac{1}{2}m_Hv_{rms}^2 = \dfrac{3}{2}k_BT_H \Rightarrow v_{rms}^2 = \dfrac{3k_BT_H}{m_H}$
For oxygen: $v_{rms}^2 = \dfrac{3k_BT_O}{m_O}$

Step 2: Equate the two expressions since the r.m.s velocities are equal.
$\dfrac{3k_BT_H}{m_H} = \dfrac{3k_BT_O}{m_O}$
The Boltzmann constant cancels, leaving:
$\dfrac{T_H}{m_H} = \dfrac{T_O}{m_O}$
The ratio of molecular masses equals the ratio of molar masses, so $\dfrac{m_H}{m_O} = \dfrac{M_H}{M_O} = \dfrac{2}{32}$

Step 3: Substitute known values and solve for $T_H$.
$T_O = 47 + 273 = 320\ \text{K}$
$T_H = T_O\times\dfrac{M_H}{M_O} = 320\times\dfrac{2}{32} = 20\ \text{K}$

Final Answer: 20 K