QJEEM23Ph97d9: JEE Main 2023 Physics - Class 11 JEEM23Ph97d9 Solution: A flask contains Hydrogen and Argon in the ratio by mass. The…

The average kinetic energy per molecule of a gas is given by the expression $\frac{3}{2}kT$ , where $k$ is the Boltzmann constant and $T$ is the temperature of the gas in kelvins. Since the temperature of the mixture is given in Celsius, we need to convert it to kelvins by adding 273.15 to get $303.15$ K.

Let us assume that the total mass of the mixture is $3x$ (where $x$ is a constant), then the mass of hydrogen and argon in the mixture will be $2x$ and $x$ respectively, according to the given ratio.

The number of moles of hydrogen and argon can be calculated using their respective masses and molar masses, which are 1 g/mol and 39.9 g/mol respectively. Therefore:

Number of moles of hydrogen = $\frac{2x}{1~\mathrm{g/mol}} = 2x$ mol

Number of moles of argon = $\frac{x}{39.9~\mathrm{g/mol}} = \frac{x}{39.9}$ mol

The total number of moles of gas in the mixture is the sum of the number of moles of hydrogen and argon:

Total number of moles of gas = $2x + \frac{x}{39.9} = \frac{79.9x}{39.9}$ mol

The average kinetic energy per molecule of hydrogen is:

$\frac{3}{2}kT_{\mathrm{H_2}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$

The average kinetic energy per molecule of argon is:

$\frac{3}{2}kT_{\mathrm{Ar}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$

Therefore, the ratio of the average kinetic energy per molecule of argon to hydrogen is:

$\frac{K_{\mathrm{Ar}}}{K_{\mathrm{H_2}}} = \frac{\frac{3}{2}kT_{\mathrm{Ar}}}{\frac{3}{2}kT_{\mathrm{H_2}}} = \frac{6.12 \times 10^{-21}~\mathrm{J}}{6.12 \times 10^{-21}~\mathrm{J}} = 1$

Hence, the ratio of average kinetic energy per molecule of argon to hydrogen is $1$ .