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

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Heat & Thermodynamics

The rms speed of oxygen molecule in a vessel at particular temperature is

(1 + \frac{5}{x})^{\frac{1}{2}} v

, where

v

is the average speed of the molecule. The value of

x

will be:(Take

π = \frac{22}{7}

)

Answer: (A)

The rms speed is

v_{rms} = \frac{3 R T}{M}

and the average speed is

v_{avg} = \frac{8 R T}{π M}

. Given that

v_{rms} = (1 + \frac{5}{x})^{\frac{1}{2}} v_{avg}

, we have:

\frac{3 R T}{M} = (1 + \frac{5}{x})^{\frac{1}{2}} \frac{8 R T}{π M}

Squaring both sides:

\frac{3 R T}{M} = (1 + \frac{5}{x}) \frac{8 R T}{π M}

\frac{3 π}{8} = 1 + \frac{5}{x}

Substituting

π = \frac{22}{7}

\frac{3 \times 22}{7 \times 8} = 1 + \frac{5}{x}

\frac{66}{56} = 1 + \frac{5}{x}

\frac{33}{28} = 1 + \frac{5}{x}

\frac{33 - 28}{28} = \frac{5}{x}

\frac{5}{28} = \frac{5}{x}

Therefore,

x = 28

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option