Chapter 13, Problem 54P

To determine

To Prove: The ratio of speed is equal to the inverse ratio of the square roots of molecular mass.

Explanation of Solution

Given info:

The masses of mixture of the two gases are ${\text{m}}_{\text{1}}$ and ${\text{m}}_{\text{2}}$ .

Temperature kept constant.

Formula used:

The root mean square formula,

  ${\text{v}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3kT}}{\text{m}}}$

Here, $k$ is the Boltzmann constant,

  $T$ is the temperature of the gas,

  $m$ is mass of the gas.

Calculation:

The rms value for the first case,

  ${\left({\text{v}}_{\text{rms}}\right)}_{\text{1}}\text{=}\sqrt{\frac{\text{3kT}}{{\text{m}}_{\text{1}}}}$......... Equation (1)

And, also for the second gas,

  ${\left({\text{v}}_{\text{rms}}\right)}_{\text{2}}\text{=}\sqrt{\frac{\text{3kT}}{{\text{m}}_{\text{2}}}}$......... Equation (2)

Now, dividing the equation (1) by the equation (2), then the result is,

  $\left(\frac{{\left({\text{v}}_{\text{rms}}\right)}_{\text{1}}}{{\left({\text{v}}_{\text{rms}}\right)}_{\text{2}}}\right)\text{=}\frac{\sqrt{\frac{\text{3kT}}{{\text{m}}_{\text{1}}}}}{\sqrt{\frac{\text{3kT}}{{\text{m}}_{\text{2}}}}}\text{=}\sqrt{\frac{{\text{m}}_{\text{2}}}{{\text{m}}_{\text{1}}}}$

Conclusion:

Hence, the ratio of speed is equal to the inverse ratio of the square roots of molecular mass.