JEE Advanced 2024 Paper 1 Q01 Chemistry Physical Chemistry Kinetic Theory of Gases Medium

JEE Advanced 2024 Paper 1 · Q01 · Kinetic Theory of Gases

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

  1. A. $2\sqrt{2} : \sqrt{3}$
  2. B. $2\sqrt{2} : 1$
  3. C. $1 : 2$
  4. D. $2 : 1$
Reveal answer + step-by-step solution

Correct answer:D

Solution

At constant V and T, $P \propto n$. For X: $P_X V = n_X R T \Rightarrow 2V = (10/M_X) R T$. After adding Y, total pressure is 6 atm so $P_Y = 4$ atm: $4V = (80/M_Y) R T$. Dividing: $\frac{2}{4} = \frac{10/M_X}{80/M_Y} \Rightarrow \frac{M_Y}{M_X} = \frac{80 \times 2}{10 \times 4} = 4 \Rightarrow M_Y = 4 M_X$. RMS velocity $u_{rms} = \sqrt{3RT/M}$, so $\frac{(u_{rms})_X}{(u_{rms})_Y} = \sqrt{M_Y/M_X} = \sqrt{4} = 2$. Hence ratio is $2:1$.

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