JEE Advanced 2024 · Paper 1 · Chemistry Q1

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ChemistrySingle correct+3 / -1Question 35 of 51

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

2\sqrt{2} : \sqrt{3}

2\sqrt{2} : 1

1 : 2

2 : 1

Answer and worked solution

Official answer: D

At fixed $V$ and $T$ , pressure is proportional to moles. Adding Y raises the pressure from 2 to 6 atm, so Y contributes 4 atm and $n_Y = 2n_X$ .

$M_X = \dfrac{10}{n_X}$ and $M_Y = \dfrac{80}{2n_X} = 4M_X$ .

$\frac{u_{rms}(X)}{u_{rms}(Y)} = \sqrt{\frac{M_Y}{M_X}} = 2 \implies 2 : 1.$

Topics:Kinetic theory of gases Gas laws

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JEE Advanced 2024 · Paper 1, Chemistry question 1