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What is the relation between pressure and kinetic energy of gas?

seo-qna
Last updated date: 20th Sep 2024
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Hint: In order to solve this question, we are going to firstly give the equation for the calculation of pressure. After that, the equation for kinetic energy is given and is simplified in a manner to the simplest form to find the relation and then the two equations are divided to get the final relation.

Formula used: The pressure of the gas molecules is given as:
\[P = \dfrac{1}{3}\rho {v^2}\]
Where, \[\rho \]is the density of the gas molecules and \[v\]is the R.M.S speed of the gas molecules
The kinetic energy of any molecules in the motion is given as:
\[K = \dfrac{1}{2}m{v^2}\]
Where, \[m\]is the mass of the gas molecules and \[v\]is the R.M.S velocity of the gas molecules.

Complete step-by-step solution:
According to the Kinetic theory of gases, the pressure of the gas molecules is given as:
\[P = \dfrac{1}{3}\rho {v^2}\]
Where, \[\rho \]is the density of the gas molecules and \[v\]is the R.M.S speed of the gas molecules.
The kinetic energy of any molecules in the motion is given as:
\[K = \dfrac{1}{2}m{v^2}\]
Where, \[m\]is the mass of the gas molecules and \[v\]is the R.M.S velocity of the gas molecules.
Putting the value of the mass in the kinetic energy equation,
\[K = \dfrac{1}{2}\rho V{v^2}\]
Now, for unit volume of gas,
\[K = \dfrac{1}{2}\rho {v^2}\]
Dividing the pressure and kinetic energy equations, we get
\[\dfrac{P}{K} = \dfrac{{\dfrac{1}{3}\rho {v^2}}}{{\dfrac{1}{2}\rho {v^2}}} = \dfrac{2}{3}\]
Thus, the relation between pressure and kinetic energy is
\[P = \dfrac{2}{3}K\]
This implies that the pressure is \[\dfrac{2}{3}\]times of the kinetic energy per unit volume for a given gas.

Note:It is to be noted that the pressure of a gas is defined as the force that the molecules exert on each other inside the gas per unit area. The kinetic energy of a gas is defined as the energy associated with the motion of the gas molecules. In terms of temperature, the kinetic energy is given as:\[E = \dfrac{3}{2}kT\]