Question

The Maxwell distribution of speeds of a gas at 300 K is given below. The molar mass (in g mol$^{-1}$) of this gas is ......... (Round off to one decimal place) (R = 8.3 J mol$^{-1}$ K$^{-1}$) 

Hint:

In Maxwell distribution, the most probable speed is related to the temperature and molar mass. Always remember to use the appropriate equation for speed distribution.

Answer 1

Correct Answer: 19.8 - 20.2

To find the molar mass of the gas, we use the relation for the most probable speed \(v_p\) in a Maxwell distribution:

\[v_p = \sqrt{\frac{2kT}{m}}\]

where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of one molecule.

Given \(v_p \approx 500 \text{ m/s}\) (from the graph) and \(T = 300 \text{ K}\).

For molar mass \(M\), using \(R = 8.3 \text{ J mol}^{-1} \text{K}^{-1}\), we have:

\[v_p = \sqrt{\frac{2RT}{M}}\]

Solving for \(M\):

\[M = \frac{2RT}{v_p^2}\]

Substitute values:

\[M = \frac{2 \times 8.3 \times 300}{500^2}\]

\[M \approx 19.92 \text{ g/mol}\]

Rounded to one decimal place, \(M = 19.9 \text{ g/mol}\).