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The mean kinetic energy of a molecule at $0^{\large\circ}C$ is $5.62\times10^{-14} erg$ . Calculate Boltzmann's Constant . If the value of $R = 8.314\times10^7 erg$ . then also calculate the no. of molecules present in one mole of gas.

$\begin {array} {1 1}(a)\;1.372\times10^{-16}erg\;molecule^{-1}K , 6.059\times10^{23}\\(b)\;1.372\times10^{-19}erg\;molecule^{-1}K , 6.059\times10^{20}\\(c)\;1.372\times10^{-14}erg\;molecule^{-1}K , 6.059\times10^{14}\\(d)\;1.372\times10^{-15}erg\;molecule^{-1}K , 6.059\times10^{19}\end {array}$

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Average Kinetic energy = $\large\frac{K.E / mol}{Av. No.}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \large\frac{3RT}{2\times N}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \large\frac{3}{2}KT$
$ K = \large\frac{5.621\times10^{-14}\times2}{3\times273}$
(Since T = 273 K)
$ = 1.372\times10^{-16}erg\;molecule^{-1}K$
Now Avogadro No. = $\large\frac{R}{K} = \large\frac{8.314\times10^7}{1.372\times10^{-16}}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 6.059\times10^{23}$
Hence answer is (a)
answered Apr 2, 2014 by sharmaaparna1

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