Question

The average thermal energy of an oxygen atom at room temperature \[(27{}^\circ C)\] is
(A) \[4.5\times {{10}^{-21}}J\]
(B) \[6.2\times {{10}^{-21}}J\]
(C) \[3.4\times {{10}^{-21}}J\]
(D) \[1.8\times {{10}^{-21}}J\]

Answer 1

Hint: To solve this question, we need to know the relation between the average thermal energy and the temperature of the gas. Now the kinetic theory of gases describes a gas as a large number of small particles (atoms or molecules), all of which are in constant, random motion. In ideal gases, there is no inter-particle interaction. Therefore, only the kinetic energy contributes to internal energy.

Formula Used:
Average Kinetic Energy \[(E)=\dfrac{3}{2}kT\]

Complete step by step answer:
From the kinetic theory of gases, the average translational kinetic energy of a molecule is equivalent to \[\dfrac{3}{2}kT\] and is called thermal energy, where \[k\] is the Boltzmann’s Constant and \[T\] is the absolute temperature of the gas.
The temperature of the gas has been provided to us on the Celsius scale.
Converting it to Kelvin, we get, absolute temperature \[(T)=27{}^\circ C+273=300K\]
Boltzmann’s constant is a physical constant relating energy at the particle level with temperature observed at the bulk level. It is calculated as the ratio of the gas constant to the Avogadro’s number.
The value of Boltzmann constant \[(k)=1.38\times {{10}^{-23}}{{m}^{2}}kg{{s}^{-2}}{{K}^{-1}}\]
Substituting these values in the expression for thermal energy, we get
Thermal energy of the gas \[=\dfrac{3\times 1.38\times {{10}^{-23}}\times 300}{2}=6.2\times {{10}^{-21}}J\]
Hence we can say that option (B) is the correct answer.

Note: Another explanation for the thermal energy of the gas sample is from the law of equipartition of energy which says that since atomic motion is random and isotropic, each degree of freedom contributes \[\dfrac{1}{2}kT\] per atom to the internal energy or the thermal energy. Since we are given only one atom of oxygen and monatomic species have three degrees of freedom, the average kinetic energy or the thermal energy again comes out to be \[\dfrac{3}{2}kT\], where the symbols have their usual meaning.