logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
0 votes

At what temperature the molecule of nitrogen will have the same rms velocity as the molecule of oxygen at $\;127^{0}C\;$ ?

$(a)\;457^{0}C\qquad(b)\;273^{0}C\qquad(c)\;350^{0}C\qquad(d)\;77^{0}C$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : $\;457^{0}C$
Explanation :
$V_{m}=\sqrt{\large\frac{3RT}{M}}$
$\large\frac{V_{N_{2}}}{V_{O_{2}}} = 1$
$\sqrt{\large\frac{3RT_{1}}{M_{N_{2}}}} = \sqrt{\large\frac{3RT_{2}}{M_{O_{2}}}}$
$T_{1} = T_{2} \times \large\frac{M_{N_{2}}}{M_{O_{2}}}$
$= \large\frac{400 \times 28}{32} = 350 K$
Therefore , $\;t=350 - 273 =77^{0}C $
answered Mar 31, 2014 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...