Question

The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

Use R = 8.314 JK^-1 mol^-1

Answer 1

Mean velocity is given by \[V_{avg}  = \sqrt{\frac{8RT}{\pi M}}\] 

Let temperature for H and He respectively be  T_​1 and T₂, respectively.
For hydrogen:
M_H = 2g = 2 \[\times\] 10^-3 kg
For helium:
M_He= 4 g = 4 \[\times\] 10^-3 kg
Now,
A/q \[\sqrt{\frac{8R T_1}{\pi M_H}} = \sqrt{\frac{8R T_2}{\pi M_{He}}}\] 

\[ \Rightarrow \sqrt{\frac{8R T_1}{2 \times {10}^{- 3} \pi}} = \sqrt{\frac{8R T_2}{\pi \times 4 \times {10}^{- 3}}}\] 

\[ \Rightarrow \sqrt{\frac{T_1}{2}} = \sqrt{\frac{T_2}{4}}\] 

\[ \Rightarrow \frac{T_1}{T_2} = \frac{1}{2}\] 

\[ \Rightarrow  T_1 :  T_2  = 1: 2\]