Question

Let (Cms)H2 be the r.m.s speed of H2 at 150 K. At what temperature, the most probable speed of helium [(Cmp)He] will be half of (Crms)H2?

• A. 75K
• B. 112.5K
• C. 225K
• D. 900K

Answer 1

The Correct Option is B

The correct temperature is 112.5 K.

Here's how we can derive this:

• RMS Speed Formula: The root-mean-square (rms) speed is given by: $C_{rms} = \sqrt{\frac{3RT}{M}}$ where:
  • R is the ideal gas constant
  • T is the temperature in Kelvin
  • M is the molar mass (in kg/mol)
• Most Probable Speed Formula: The most probable speed is given by: $C_{mp} = \sqrt{\frac{2RT}{M}}$
• Given Condition: (C_mp)_He = (1/2) * (C_rms)_H2
• Plugging in Values:
  • $M_{He}$ = 4 g/mol
  • $M_{H2}$ = 2 g/mol
  • $T_{H2}$ = 150 K
• Simplifying the Equation: Square both sides to remove the square roots. $\frac{2RT_{He}}{4} = \frac{1}{4} \frac{3R(150)}{2}$
• Solving for T_He $T_{He} = \frac{3(150)}{4}$ $T_{He} = \frac{450}{4}$ $T_{He} = 112.5K$

Therefore, the most probable speed of helium will be half of the RMS speed of hydrogen at 150 K when the temperature of helium is 112.5 K.

Correct Answer: 112.5 K