Class 12 > Unit # 15:Molecular Theory of Gases > Effect of Pressure on Gases

Q.5 Derive an expression for the pressure of an ideal gas on the basis of Kinetic Theory of Gases.

INTERPRETATION OF PRESSURE ON KINETIC THEORY OF GASES:-

Let "$N$" number of gas molecules be enclosed in a cubical container whose length, breadth, and height are each equal to $L$. The total internal volume of the container is $V = L^3$. Let the mass of a single molecule be "$m$".

Consider a molecule moving parallel to the x-axis with a velocity component $v_x$. When this molecule collides elastically with the opposite boundary wall of the container, it rebounds back with a velocity of $-v_x$.

$$\text{Initial Momentum of gas molecule} = m v_x$$ $$\text{Final Momentum of gas molecule} = -(m v_x)$$ $$\text{Change in Momentum } (\Delta p) = m v_x - (-m v_x) = m v_x + m v_x = 2m v_x$$

The momentum transferred to the wall during this single impact is $+2m v_x$. The time interval ($\Delta t$) required for this molecule to make a round-trip distance of $2L$ across the container and strike the exact same wall face again is given by the definition of velocity:

$$v_x = \frac{2L}{\Delta t} \implies \Delta t = \frac{2L}{v_x}$$

According to Newton's Second Law of Motion, force is defined as the rate of change of momentum. Substituting our values gives the force ($f_1$) exerted by this single molecule:

$$f_1 = \frac{\text{Change in Momentum}}{\Delta t} = \frac{2m v_x}{\left(\frac{2L}{v_x}\right)} = \frac{m v_x^2}{L} \quad \text{--- (i)}$$

Inside the container, there are multiple molecules possessing various x-component velocities designated as $v_{x1}, v_{x2}, v_{x3}, \dots, v_{xN}$. The total collective force ($F_x$) acting upon this container wall face is the sum of their individual contributions:

$$\text{Total Force } (F_x) = \frac{m v_{x1}^2}{L} + \frac{m v_{x2}^2}{L} + \dots + \frac{m v_{xN}^2}{L}$$ $$F_x = \frac{m}{L} \left( v_{x1}^2 + v_{x2}^2 + v_{x3}^2 + \dots + v_{xN}^2 \right)$$

By definition, macroscopic pressure is given by force per unit area ($\text{Area } A = L^2$). Therefore, the pressure ($P$) on this face wall is written as:

$$P = \frac{\text{Force}}{\text{Area}} = \frac{F_x}{L^2} = \frac{m}{L^3} \left( v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2 \right) \quad \text{--- (ii)}$$

Let us express this in terms of the bulk physical properties of the gas. We know that:

$$\text{Density } (\rho) = \frac{\text{Total Mass}}{\text{Volume}}$$ $$\text{Mass of one molecule} = m \implies \text{Mass of } N \text{ molecules} = mN$$ $$\text{Density } (\rho) = \frac{mN}{L^3} \implies \frac{m}{L^3} = \frac{\rho}{N}$$

Substituting $\frac{m}{L^3} = \frac{\rho}{N}$ into equation (ii) yields:

$$P = \rho \left[ \frac{v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2}{N} \right]$$

The expression inside the brackets is defined as the mean of the square of velocity along the x-axis, denoted as $\overline{v_x^2}$:

$$\overline{v_x^2} = \frac{v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2}{N}$$ $$P = \rho \overline{v_x^2} \quad \text{--- (iii)}$$

The term $\overline{v_x^2}$ represents only a single dimensional component of the total molecular velocity vector. For any given molecule, the total velocity magnitude squared is:

$$\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}$$

Due to the completely chaotic and random nature of molecular motion, there is no preferred path orientation inside the box. On average, the mean square components along all three dimensions are equal:

$$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$$ $$\overline{v^2} = 3\overline{v_x^2} \implies \overline{v_x^2} = \frac{1}{3}\overline{v^2}$$

Substituting $\overline{v_x^2} = \frac{1}{3}\overline{v^2}$ back into equation (iii) gives the final general kinetic expression for gas pressure:

$$P = \frac{1}{3}\rho\overline{v^2}$$

Where $\sqrt{\overline{v^2}}$ is known as the Root Mean Square Velocity ($v_{\text{rms}}$), which can be written as:

$$v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}$$