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?
- 75K
- 112.5K
- 225K
- 900K
The Correct Option is B
Solution and Explanation
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: (Cmp)He = (1/2) * (Crms)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 THe $T_{He} = \frac{3(150)}{4}$ $T_{He} = \frac{450}{4}$ $T_{He} = 112.5K$
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
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