Question

The pressure exerted by a perfect gas is equal to

• A. $\frac{2}{3}$ of mean kinetic energy.
• B. half of mean kinetic energy per unit volume.
• C. mean kinetic energy per unit volume
• D. $\frac{2}{3}$ of mean kinetic energy per unit volume.

Answer 1

The correct option is D

$\frac{2}{3}$ of mean kinetic energy per unit volume.

Step 1: Given parameters

The given gas is an ideal gas.

Step 2: Formulas Used:

1. $\begin{array}{l}P=\frac{1}{3}\rho {\nu }^2\end{array}$ ; Where $P$ is pressure, $\rho$ is density, and $v$ is r.m.s velocity.
2. $\rho =\frac{M}{V}$ ; Where $\rho$is the density, $M$ is the Mass, and $V$ is the volume.
3. $\begin{array}{l}K.E=\frac{1}{2}M{\nu }^2\end{array}$; Where $K.E.$ is kinetic energy, $M$ is the Mass, and $v$ is r.m.s velocity.

Step 3: Derive the relation between kinetic energy and pressure.

According to the kinetic theory of gasses,

$\begin{array}{l}P=\frac{1}{3}\rho {\nu }^2\end{array}...\left(1\right)$

The density of a gas, $\rho =\frac{M}{V}...\left(2\right)$

Substitute the value of density from equation (2) in equation (1).

$\begin{array}{l}P=\frac{M{\nu }^2}{3V}\end{array}...\left(3\right)$

The kinetic energy of gases is expressed as,

$\begin{array}{l}K.E=\frac{1}{2}M{\nu }^2\end{array}...\left(4\right)$

Divide equation (3) by (4).

$\begin{array}{l}\frac{P}{K.E}=\frac{\frac{M{\nu }^2}{3V}}{\frac{1}{2}M{\nu }^2}\end{array}$

$\begin{array}{l}\frac{P}{K.E}=\frac{2}{3V}\end{array}$

$\begin{array}{l}P=\frac{2}{3}\times \frac{K.E}{V}\end{array}$

Hence, the pressure exerted by a perfect gas is equal to two-thirds of kinetic energy per unit volume.

Hence, option (D) is correct.