Derive an expression for pressure exterted by an ideal gas?

Answer

To derive the expression for the pressure exerted by an ideal gas, we will follow a systematic approach based on the kinetic theory of gases.

### Step-by-Step Derivation:

1. **Assumptions of Ideal Gas:**
- The molecules of the gas do not exert any forces on each other (negligible intermolecular forces).
- The molecules are in constant random motion and collide elastically with each other and the walls of the container.
- The size of the molecules is negligible compared to the distance between them.

2. **Consider a Cubical Container:**
- Let’s consider a cube of side length \( l \). The volume \( V \) of the cube is given by:
\[ V = l^3 \]

3. **Velocity Components:**
- The velocity \( v \) of a gas molecule can be resolved into three components: \( v_x \), \( v_y \), and \( v_z \).
- Due to symmetry, we can assume that the average speeds in each direction are equal:
\[ v_x = v_y = v_z = \frac{v}{\sqrt{3}} \]
- Therefore, we can express the total speed squared as:
\[ v^2 = v_x^2 + v_y^2 + v_z^2 = 3v_x^2 \Rightarrow v_x^2 = \frac{v^2}{3} \]

4. **Momentum Change on Collision:**
- When a molecule collides with the wall, it reverses its momentum. The change in momentum \( \Delta p \) for one molecule is:
\[ \Delta p = m(v_x - (-v_x)) = 2mv_x \]
- The time \( t \) taken for the molecule to travel to the wall and back is:
\[ t = \frac{2l}{v_x} \]

5. **Force Calculation:**
- The force \( F \) exerted by one molecule on the wall can be calculated using the rate of change of momentum:
\[ F = \frac{\Delta p}{t} = \frac{2mv_x}{\frac{2l}{v_x}} = \frac{mv_x^2}{l} \]

6. **Total Force from All Molecules:**
- If there are \( n \) molecules, the total force \( F_{total} \) exerted on the wall is:
\[ F_{total} = n \cdot \frac{mv_x^2}{l} \]
- Substituting \( v_x = \frac{v}{\sqrt{3}} \):
\[ F_{total} = n \cdot \frac{m \left(\frac{v^2}{3}\right)}{l} = \frac{n mv^2}{3l} \]

7. **Pressure Calculation:**
- Pressure \( P \) is defined as force per unit area. The area \( A \) of the wall is \( l^2 \):
\[ P = \frac{F_{total}}{A} = \frac{\frac{n mv^2}{3l}}{l^2} = \frac{n mv^2}{3l^3} \]
- Since \( V = l^3 \), we can replace \( l^3 \) with \( V \):
\[ P = \frac{n mv^2}{3V} \]

8. **Final Expression:**
- The final expression for the pressure exerted by an ideal gas can be written as:
\[ P = \frac{1}{3} \frac{n m v_{rms}^2}{V} \]
- Here, \( v_{rms} \) is the root mean square velocity of the gas molecules.