A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is...

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JEE Advanced 2024 Paper 1 Chemistry Question 1 Solution

From the ideal gas equation $PV = nRT$ , at constant volume and temperature, $P \propto n$ .

Given: $P_X = 2$ atm and $P_{total} = 6$ atm.

So, partial pressure of gas Y, $P_Y = P_{total} - P_X = 6 - 2 = 4$ atm.

Using the ratio of pressures:

$\frac{P_X}{P_Y} = \frac{n_X}{n_Y}$

$\frac{2}{4} = \frac{W_X / M_X}{W_Y / M_Y}$

$\frac{1}{2} = \frac{10 / M_X}{80 / M_Y} = \frac{10}{80} \times \frac{M_Y}{M_X} = \frac{1}{8} \times \frac{M_Y}{M_X}$

$\frac{M_Y}{M_X} = \frac{8}{2} = 4$

The root mean square velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$ .

At the same temperature:

$\frac{v_X}{v_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{4} = 2$

Therefore, the ratio is $2 : 1$ .