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:

For gases, \[ v_{\text{rms}} \propto \sqrt{\frac{T}{M}} \] If two gases have equal r.m.s velocity, their temperature-to-molar-mass ratios must be equal.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the temperature (\(T_H\)) for hydrogen gas (\(H_2\)) at which its root-mean-square (r.m.s.) velocity is the same as the r.m.s. velocity of oxygen gas (\(O_2\)) at a given temperature (\(T_O = 47^\circ\text{C}\)).
Step 2: Key Formula or Approach:
The r.m.s. velocity (\(v_{\text{rms}}\)) of a gas molecule is given by the formula:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \(R\) is the universal gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas.
Step 3: Detailed Explanation:
First, convert the given temperature of oxygen to Kelvin:
\[ T_O = 47^\circ\text{C} + 273.15 \approx 47 + 273 = 320\,\text{K} \] We are given the condition \(v_{\text{rms}}(H_2) = v_{\text{rms}}(O_2)\). Using the formula, we can write:
\[ \sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}} \] Squaring both sides and canceling the constant term \(3R\), we get:
\[ \frac{T_H}{M_H} = \frac{T_O}{M_O} \] Now we need the molar masses:
Molar mass of hydrogen (\(H_2\)), \(M_H \approx 2\,\text{g/mol}\)
Molar mass of oxygen (\(O_2\)), \(M_O \approx 32\,\text{g/mol}\)
Substitute the known values into the equation:
\[ \frac{T_H}{2} = \frac{320}{32} \] \[ \frac{T_H}{2} = 10 \] Solving for \(T_H\):
\[ T_H = 2 \times 10 = 20\,\text{K} \] Step 4: Final Answer:
The temperature of the hydrogen molecule must be \(20\,\text{K}\).