Question

Given Avagadro number = $6.02\times {10}^{23}$ and Boltzmann's constant =$1.38\times {10}^{-23}$ J/(mol.-K). Calculate (a) the average kinetic energy of translation of an oxygen molecule at ${27}^{\circ }$C, (b) the total average kinetic energy of an oxygen molecule at ${27}^{\circ }$C, (c) the total kinetic energy in Joules of a gram-molecule of oxygen at ${27}^{\circ }$C.

• A. joule/molecule , 6231 joule/molecule.
• B. joule/molecule , 728 joule/molecule.
• C. joule/molecule , 600 joule/molecule.
• D. joule/molecule , 298 joule/molecule.

Answer 1

The correct option is A joule/molecule , 6231 joule/molecule.

Given, Avogadro number
$N=6.02\times {10}^{23}$
Boltzmann's constant
$k=1.38\times {10}^{-23}$ Joule/molecule K
kelvin temperature T = 27 + 273 = 300 K
(a) An oxygen molecule has three degrees of freedom with respect to translation. Hence the average kineticenergy of translation of molecule
$3\times \frac{1}{2}kT=\frac{3}{2}kT$
($\Theta$ Average kinetic energy of a gas molecule Per degree of freedom is $\frac{1}{2}$. kT.)
$3\times \frac{1}{2}(1.38\times {10}^{-23})\times 300$
$6.21\times {10}^{-21}$ joule/molecule
(b) The oxygen molecule has five degrees of freedom (three degrees of freedom with respect to translation as it is a diatomic molecule). Hence the total kinetic energy of the molecule,
$5\times \frac{1}{2}kT=\frac{5}{2}kT$
$\frac{5}{2}(1.38\times {10}^{-23})\times 300$
$=10.35\times {10}^{-21}$ joule/molecule
(c) One gm molecule of oxygen contains N molecules. Hence the total kinetic energy of 1 gm molecule of the gas,
$=N\times \frac{5}{2}kT$
$=(6.02\times {10}^{23})\times 10.35\times {10}^{-21}$
= 6231 joule/molecule.