Question

A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to_. If it were to go up straight without colliding with any other molecules, how high would it rise? Assume that the height attained is much less than the radius of the earth. (${k_B}$ is Boltzmann constant)
$\left( {\text{A}} \right){\text{zero}}$
$\left( B \right)\dfrac{{273{k_B}}}{{2Mg}}$
$\left( C \right)\dfrac{{546{k_B}}}{{3Mg}}$
$\left( D \right)\dfrac{{819{k_B}}}{{2Mg}}$

Answer 1

Hint:Firstly, determine the degree of freedom of the gas. The degree of freedom of a system refers to the possible independent motion a system can have. Then apply it to the translational kinetic energy of the molecules. Here the kinetic energy is equal to potential energy. Equate kinetic energy and potential energy to obtain the height.

Formula used:
$KE = \dfrac{f}{2}RT = \dfrac{f}{2}{k_B}T$
Where $KE$ is the kinetic energy, $f$ is the degree of freedom, ${k_B}$ is the Boltzmann constant, $R$ is universal gas constant.

Complete step by step solution:
Degree of freedom is the total number of independent modes in which a system can possess energy. The monatomic gas, diatomic gas and triatomic gas have three, five and six degrees of freedom respectively.
In the kinetic energy per degree of freedom is proportional to Boltzmann constant .
Here the degree of freedom for the above question is $3$.
In any system at thermal equilibrium, the total energy is equally distributed among its degree of freedom and each degree $\dfrac{1}{2}{k_B}T$ per each molecule of the gas.
Translational kinetic energy of molecules is given by the formula
$KE = \dfrac{f}{2}RT = \dfrac{f}{2}{k_B}T$
$KE = \dfrac{3}{2}{k_B}T$
Kinetic theory of gas refers to the macroscopic properties of gas to the microscopic properties of the gas molecule. The kinetic energy is an ideal gas in the internal energy
From conservation of energy concept, we can equate kinetic energy and potential energy.
$\dfrac{3}{2}{k_B}T = mgh$
$h = \dfrac{{3{k_B}\left( {273} \right)}}{{2mg}}$
$h = \dfrac{{819{k_B}}}{{2mg}}$

Hence option $\left( D \right)$ is the right option.

Note:Degree of freedom is the total number of independent modes in which a system can possess energy. Triatomic gas molecules have six degrees of freedom; Diatomic gases have five degrees of freedom. Monatomic gases have three degrees of freedom. Absolute temperature of the gas is proportional to the average kinetic energy. Kinetic energy is independent of pressure, volume, nature of the ideal gas.