QJEEM23Ch4e07: JEE Main 2023 Chemistry - Class 11 JEEM23Ch4e07 Solution: At , the root mean square (rms) speed of gas (molar mass ) is…

The root mean square speed ( $v_{rms}$ ) of a gas can be calculated using the following formula:

$v_{rms} = \sqrt{\frac{3kT}{M}}$

where:

• $k$ is Boltzmann's constant ( $1.38 \times 10^{-23}$ J/K),
• $T$ is the temperature in Kelvin,
• $M$ is the molar mass of the gas in kilograms per mole.

The most probable speed ( $v_{mp}$ ) of a gas can be calculated using the following formula:

$v_{mp} = \sqrt{\frac{2kT}{M}}$

Given that the root mean square speed of gas X at 600 K is equal to the most probable speed of gas Y at 90 K, we can set the two equations equal to each other and solve for the molar mass of gas Y:

$\sqrt{\frac{3kT_X}{M_X}} = \sqrt{\frac{2kT_Y}{M_Y}}$

Square both sides to eliminate the square root:

$\frac{3kT_X}{M_X} = \frac{2kT_Y}{M_Y}$

We are asked to find $M_Y$ , so let's rearrange the equation to solve for $M_Y$ :

$M_Y = \frac{2kT_YM_X}{3kT_X}$

We know that $T_X = 600$ K, $M_X = 40$ g/mol $= 40 \times 10^{-3}$ kg/mol (converted from grams to kilograms), $T_Y = 90$ K, and $k = 1.38 \times 10^{-23}$ J/K. Substituting these values into the equation gives:

$M_Y = \frac{2 \times 1.38 \times 10^{-23} \text{J/K} \times 90 \text{K} \times 40 \times 10^{-3} \text{kg/mol}}{3 \times 1.38 \times 10^{-23} \text{J/K} \times 600 \text{K}}$

Solving this equation gives:

$M_Y = 0.004 \text{kg/mol}$

To convert this value back into grams per mole, multiply by 1000:

$M_Y = 4 \text{g/mol}$

So, the molar mass of gas Y is approximately 4 g/mol.