The rms speed of oxygen molecule in a vessel at particular temperature is (1 + 5/x)1/2 v, where v is the average speed of the molecule. The value of x will be: ( Take π = 22/7) (a) 4 (b) 8 (c) 28 (d) 27

Answer: (c). vrms = √(3RT/M), vav = √((8RT)/(π M)), so vrms²vav² = 3π/8 With π = 22/7: 3π/8 = 66/56 = 33/28 = 1 + 5/28 So x = 28; the answer is (c) 28.

The rms speed of oxygen molecule in a vessel at particular… | JEE Main 2023 (13 Apr, Shift 1)

Q43JEE Main 2023 · 13 APR · SHIFT 1Medium

Molecular Speeds & Maxwell Distribution

The rms speed of oxygen molecule in a vessel at particular temperature is $\left(1 + \frac{5}{x}\right)^{\frac{1}{2}} v$ , where $v$ is the average speed of the molecule. The value of $x$ will be:

$($ Take $\pi = \frac{22}{7})$

Show answer and solution

Answer: (c)

$v_{rms} = \sqrt{\frac{3RT}{M}}$ , $v_{av} = \sqrt{\frac{8RT}{\pi M}}$ , so $\frac{v_{rms}^2}{v_{av}^2} = \frac{3\pi}{8}$
With $\pi = \frac{22}{7}$ : $\frac{3\pi}{8} = \frac{66}{56} = \frac{33}{28} = 1 + \frac{5}{28}$
So $x = 28$ ; the answer is (c) 28.

Report a mistake

Molecular Speeds & Maxwell Distribution question from JEE Main 2023 (13 Apr, Shift 1)

More questions on Molecular Speeds & Maxwell Distribution