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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2023 · 13 April · Morning Shift · Physics · Kinetic Theory of Gases

The rms speed of oxygen molecule in a vessel at particular...

The rmsrms speed of oxygen molecule in a vessel at particular temperature is (1+5x)12v,\left(1 + \dfrac{5}{x}\right)\dfrac{1}{2}v, when vv is the average speed of the molecule. The value of xx will be:

(take π=227\pi = \dfrac{22}{7})

Options

  1. A2727
  2. B88
  3. C2828
  4. D44

Correct answer

C. 2828

Step-by-step solution

The RMS speed is given by

vrms=3RTM{v}_{rms} = \sqrt{\dfrac{3RT}{M}}

The average speed is given by

vavg=8RT\piM{v}_{avg} = \sqrt{\dfrac{8RT}{\piM}}

The ratio is

⇒vrmsvavg=3RTM8RT\piM=3π8=38×227⇒vrmsvavg=3328⇒ \dfrac{{v}_{rms}}{{v}_{\text{avg}}} = \sqrt{\dfrac{\dfrac{3RT}{M}}{\dfrac{8RT}{\piM}} =}\sqrt{\dfrac{3\pi}{8}} = \sqrt{\dfrac{3}{8} \times \dfrac{22}{7}} ⇒ \dfrac{{v}_{rms}}{{v}_{\text{avg}}} = \sqrt{\dfrac{33}{28}}

Comparing it with the given value \left(1 + \dfrac{5}{x}\right)\dfrac{1}{2},</p><p>,</p><p>\dfrac{{v}_{rms}}{{v}_{\text{avg}}} = \sqrt{1 + \dfrac{5}{28}} ⇒ x = 28

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