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The average kinetic energy of one molecule of an ideal gas at $27^{\circ}C$ and 1 atm. pressure is :

\[\begin {array} {1 1} (1)\;900\;cals\;k^{-1}\;mol^{-1} & \quad (2)\;6.21 \times 10^{-21}\;Jk^{-1} molecule^{-1}\\ (3)\;336.7\;J.k^{-1}\;molecule {-1} & \quad (4)\;3741.3 \;J.k \;mol^{-1} \end {array}\]

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$ (2)\;6.21 \times 10^{-21}\;Jk^{-1}$
answered Nov 7, 2013 by pady_1

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