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d=\frac{1}{\sqrt{2}}\; \frac{kT}{\pi d^{2}p} 

 

In the given diagram means free path for a gas molecule is along Y- axis then what physical quantity could not be on X-axis.

Option: 1

m (mass each molecule)


Option: 2

T ( Temperature of gas)


Option: 3

\frac{1}{\rho}  ( \rho density of gas )


Option: 4

P ( Pressure of gas)


Answers (1)

best_answer

As we learn

Formula for mean free path -

Y= \frac{KT}{\sqrt{2}\pi \sigma ^{2}p}
 

- wherein

\sigma = Diameter of the molecule

\rho = pressure of the gas

T = temperature

K = Boltzmann's Constant

 

 d=\frac{m}{\sqrt{2}\pi d^{2}\rho }

 

d \; \alpha \; m and 

d \; \alpha \; \frac{1}{p }

and also 

 

d=\frac{1}{\sqrt{2}}\; \frac{kT}{\pi d^{2}p}

d \; \alpha \; T and

  d \; \alpha \; \frac{1}{p }

 

 

Posted by

manish painkra

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