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The average kinetic energy of an ideal gas per molecule in SI units at $25^{\large\circ}C$ will be

$(a)\;6.2\times10^{-21}KJ\qquad(b)\;6.2\times10^{-21}J\qquad(c)\;6.2\times10^{20}J\qquad(d)\;6.2\times10^{20}KJ$

1 Answer

Average kinetic energy = $\large\frac{3}{2}KT$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\large\frac{3}{2}\times\large\frac{8.314}{6.023\times10^{23}}\times298$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=6.2\times10^{-21}J$
Hence answer is (b)
answered Mar 6, 2014 by sharmaaparna1
 

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