Easy MCQ +4 / -1 PYQ · JEE Mains 2023

A flask contains Hydrogen and Argon in the ratio $2: 1$ by mass. The temperature of the mixture is $30^{\circ} \mathrm{C}$. The ratio of average kinetic energy per molecule of the two gases ( $\mathrm{K}$ argon/K hydrogen) is :

(Given: Atomic Weight of $\mathrm{Ar}=39.9$ )

  1. A $\frac{39.9}{2}$
  2. B 2
  3. C 39.9
  4. D 1 Correct answer

Solution

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. <br/><br/> 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. <br/><br/> 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: <br/><br/> Number of moles of hydrogen = $\frac{2x}{1~\mathrm{g/mol}} = 2x$ mol <br/><br/> Number of moles of argon = $\frac{x}{39.9~\mathrm{g/mol}} = \frac{x}{39.9}$ mol <br/><br/> The total number of moles of gas in the mixture is the sum of the number of moles of hydrogen and argon: <br/><br/> Total number of moles of gas = $2x + \frac{x}{39.9} = \frac{79.9x}{39.9}$ mol <br/><br/> The average kinetic energy per molecule of hydrogen is: <br/><br/> $\frac{3}{2}kT_{\mathrm{H_2}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$ <br/><br/> The average kinetic energy per molecule of argon is: <br/><br/> $\frac{3}{2}kT_{\mathrm{Ar}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$ <br/><br/> Therefore, the ratio of the average kinetic energy per molecule of argon to hydrogen is: <br/><br/> $\frac{K_{\mathrm{Ar}}}{K_{\mathrm{H_2}}} = \frac{\frac{3}{2}kT_{\mathrm{Ar}}}{\frac{3}{2}kT_{\mathrm{H_2}}} = \frac{6.12 \times 10^{-21}~\mathrm{J}}{6.12 \times 10^{-21}~\mathrm{J}} = 1$ <br/><br/> Hence, the ratio of average kinetic energy per molecule of argon to hydrogen is $1$.

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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