Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The rms speed of oxygen molecule in a vessel at particular temperature is <img alt="\left(1+\frac{5}{x}\right)^{\frac{1}{2}} v," src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28

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: \left(\text { Take } \pi=\frac{22}{7}\right)

 

Option: 1

28


Option: 2

27


Option: 3

8

 


Option: 4

4


Answers (1)

best_answer

\begin{aligned} & \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\ \\& \mathrm{V}_{\mathrm{avg}}=v=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}} \\ \\& \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \pi}{8}} \mathrm{v} \end{aligned}

\begin{aligned} & \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3}{8} \times \frac{22}{7}} v=\left(\frac{33}{28}\right)^{1 / 2} v \\ \\& \mathrm{~V}_{\text {rms }}=\left(1+\frac{5}{28}\right)^{1 / 2} v \\ \\& \mathrm{x}=28 \end{aligned}

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change
anam-khan

Faridabad: 17-year-old girl preparing for JEE Main 2026 dies by suicide
anam-khan

NTA JEE Mains 2026 application closes at 9 am today; correction facility from December 1

Latest Question