Question

RMS speed for \( \mathrm{H_2} \) and \( \mathrm{O_2} \) are same. If temperature of \( \mathrm{O_2} \) gas is \( 23^\circ\text{C} \), find the temperature of \( \mathrm{H_2} \) gas.

• A. 18.5 K
• B. \( 2.5^\circ\text{C} \)
• C. \( 18^\circ\text{C} \)
• D. 164 K

Hint:

For gases having equal RMS speeds, temperature is directly proportional to molar mass. Lighter gases must be at lower temperature to have the same RMS speed as heavier gases.

Answer 1

The Correct Option is A

Step 1: Write the formula for RMS speed.
The RMS speed of a gas molecule is given by the formula: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \( R \) is the gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.
Step 2: Use the condition given in the question.
It is given that the RMS speeds of \( \mathrm{H_2} \) and \( \mathrm{O_2} \) are equal. Therefore: \[ \sqrt{\frac{3RT_{\mathrm{H_2}}}{M_{\mathrm{H_2}}}} = \sqrt{\frac{3RT_{\mathrm{O_2}}}{M_{\mathrm{O_2}}}} \]
Step 3: Simplify the equation.
Squaring both sides and cancelling common terms: \[ \frac{T_{\mathrm{H_2}}}{M_{\mathrm{H_2}}} = \frac{T_{\mathrm{O_2}}}{M_{\mathrm{O_2}}} \]
Step 4: Substitute known values.
Molar mass of hydrogen gas: \[ M_{\mathrm{H_2}} = 2 \] Molar mass of oxygen gas: \[ M_{\mathrm{O_2}} = 32 \] Temperature of oxygen gas: \[ T_{\mathrm{O_2}} = 23^\circ\text{C} = 296 \text{ K} \] Substituting: \[ \frac{T_{\mathrm{H_2}}}{2} = \frac{296}{32} \]
Step 5: Calculate the temperature of hydrogen gas.
\[ T_{\mathrm{H_2}} = \frac{2 \times 296}{32} = \frac{296}{16} = 18.5 \text{ K} \]