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 [JEE Advanced 2024, Paper-1]

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YESPYQ · Accepted Answer

Correct answer: d) $2:1$. At constant V and T, P ∝ n. Initial: P1 = 2 atm, moles of X = nX. After adding Y: P2 = 6 atm, total moles = nX + nY. (nX + nY)/nX = P2/P1 = 6/2 = 3 ⇒ nY = 2nX. Given mX = 10 g, mY = 80 g. Mx = 10/nX, My = 80/nY = 80/(2nX) = 40/nX ⇒ My = 4Mx. u(rms) ∝ 1/√M, so uX/uY = √(My/Mx) = √4 = 2. Ratio uX : uY = 2 : 1.

A closed vessel contains \(10\mathrm{g}\) of an ideal gas X at \(300\mathrm{K}\), which exerts \(2\mathrm{atm}\) pressur…

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