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Calculate the average volume available to a molecule in a sample of nitrogen gas at STP. What is the average distance between neighbouring molecules if nitrogen molecules are spherical in nature?

$(a)\;40.4\times10^{-8}cm\qquad(b)\;41.4\times10^{-8}cm\qquad(c)\;42.4\times10^{-8}cm\qquad(d)\;44.4\times10^{-8}cm$

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1 Answer

Since $6.023\times10^23$ molecules of $N_2$ occupy $22400cm^3$
One molecule of $N_2$ occupies $\large\frac{22400}{6.023\times10^23} cm^3$
(or) Volume of one molecule of $N_2 = 3.73\times 10^{-20}cm^3$
Also the average distance between two molecules = 2r
and $\large\frac{4}{3}\pi r^3 = 3.72\times10^{-20}$
$r^3 = \large\frac{3.72\times10^{-20}\times3\times7}{4\times22}$
$r = 20.7\times10^{-8}cm$
Thus average distance = $2\times20.7\times10^{-8}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Rightarrow = 41.4\times10^{-8}cm$
Hence answer is (b)
answered Feb 21, 2014 by sharmaaparna1
 

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