• # question_answer The kinetic energy of one gramme molecule of a gas, at normal temperature and pressure will be $(R=8.31\,J\,mole\,K)$: A)  $3.4\times {{10}^{3}}J$         B)  $2.7\times {{10}^{2}}J$C)  $1.2\times {{10}^{2}}J$          D)  none of these

Solution :

We know that the kinetic energy of $1g$ molecule of an ideal gas is $E=\frac{3}{2}\,RT$ Here     $R=8.31\,J/mol\,K$   and        $~T={{273}^{o}}K$ $\therefore$  $E=\frac{3}{2}\times 8.31\times 273=3.4\times {{10}^{3}}J$

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