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

VITEEEVITEEE 2009

Solution:

Average kinetic energy per molecule
$=\frac{3}{2} k T$
or $=\frac{3}{2} \frac{R}{N_{0}} T$
$=\frac{3}{2} \times \frac{8.314}{6.023 \times 10^{23}} \times 300$
$=6.21 \times 10^{-21} \,JK ^{-1}$ molecule $^{-1}$