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The molecules of an ideal gas at 270 C have a mean velocity  v. at what temperature (in oC) will the mean velocity be \frac{3}{2}\: v

 

 

 

 

Option: 1

350


Option: 2

450


Option: 3

402


Option: 4

315


Answers (1)

best_answer

 

The mean velocity of the gas molecule is

\bar{v}= \sqrt{\frac{8RT}{\pi M}}

for same gas

\bar{v}= \sqrt{T}

\frac{\bar{v_{1}}}{\bar{v_{2}}}= \sqrt{\frac{T_{1}}{T_{2}}}= \sqrt{\frac{\left ( 273+27 \right )}{T_{2}}}

\frac{2v}{3v}= \sqrt{\frac{300}{T_{2}}}\Rightarrow \frac{300}{T_{2}}= \frac{4}{9}

T_{2}= \frac{2700}{4}= 675 K\\*675-373=402^{0}C

Posted by

Sanket Gandhi

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