Heat / Kinetic Theory of Gases

Pressure Exerted by an Ideal Gas

Question

Derive the expression for the pressure exerted by an ideal gas using the kinetic theory of gases.

Show that the pressure of an ideal gas is given by

\boxed{p=\frac{1}{3}\rho \overline{C^2}}

Given

Consider a quantity of gas enclosed in a cubical vessel of side \(l\). The walls of the vessel are perfectly elastic.

Let \(N\) be the very large number of molecules in the vessel. Consider one molecule of mass \(m\), whose velocity has three components \(u_1,\ v_1,\ w_1\) along the three coordinate axes.

Concept Used

According to the kinetic theory of gases, gas molecules are continuously moving in random directions and collide elastically with the walls of the container.

The pressure of the gas is produced because the moving molecules continuously strike the walls and change their momentum.

For a molecule whose velocity is \(C_1\),

\boxed{ C_1^2=u_1^2+v_1^2+w_1^2 }

Solution

Step 1: Velocity components of a molecule

Let a molecule of mass \(m\) move with velocity \(C_1\). Resolve this velocity into three components:

u_1,\qquad v_1,\qquad w_1

Therefore,

\boxed{ C_1^2=u_1^2+v_1^2+w_1^2 }

Step 2: Collision with the wall

Suppose the molecule strikes the wall \(ABCD\) with velocity component \(u_1\) normal to the wall.

Since the collision is perfectly elastic, the molecule rebounds with the same speed in the opposite direction.

Therefore, momentum before impact is

\[ mu_1 \]

Momentum after impact is

\[ -mu_1 \]

Hence, the change in momentum is

\Delta p = mu_1-(-mu_1) \] \[ \boxed{\Delta p=2mu_1}

Step 3: Time between two successive impacts

After striking the wall \(ABCD\), the molecule travels to the opposite wall and then comes back to \(ABCD\).

Therefore, the total distance travelled before striking the same wall again is

\[ 2l \]

Since the velocity component normal to the wall is \(u_1\), the time taken is

t=\frac{2l}{u_1}

Step 4: Force due to one molecule

Force is the rate of change of momentum. Therefore,

F_1= \frac{\text{change in momentum}}{\text{time}} \] \[ F_1= \frac{2mu_1}{2l/u_1} \] \[ \boxed{ F_1=\frac{mu_1^2}{l} }

Step 5: Force due to all \(N\) molecules

Suppose the velocity components of the molecules in the \(X\)-direction are

u_1,\ u_2,\ u_3,\ldots,u_N

The total force exerted on the wall due to all \(N\) molecules is therefore

F= \frac{m}{l} \left( u_1^2+u_2^2+u_3^2+\cdots+u_N^2 \right)

The area of the wall is

\[ A=l^2 \]

Hence, pressure is

p=\frac{F}{A} \] \[ p= \frac{m}{l^3} \left( u_1^2+u_2^2+\cdots+u_N^2 \right)

Step 6: Use the volume of the vessel

The volume of the cubical vessel is

\[ V=l^3 \]

Therefore,

\boxed{ p= \frac{m}{V} \left( u_1^2+u_2^2+\cdots+u_N^2 \right) }

Step 7: Consider all three directions

In the same way, pressure can be written using the velocity components in the \(Y\)- and \(Z\)-directions.

p_x=\frac{m}{V} \sum_{i=1}^{N}u_i^2 \] \[ p_y=\frac{m}{V} \sum_{i=1}^{N}v_i^2 \] \[ p_z=\frac{m}{V} \sum_{i=1}^{N}w_i^2

Since the gas is assumed to be moving randomly, the pressure exerted by the gas is the same in all directions:

\[ p_x=p_y=p_z=p \]

Therefore, the pressure can be obtained by taking the average of the three expressions:

p= \frac{p_x+p_y+p_z}{3}

Step 8: Add the three velocity components

Adding the three expressions gives

p= \frac{m}{3V} \sum_{i=1}^{N} \left( u_i^2+v_i^2+w_i^2 \right)

But for each molecule,

\[ C_i^2=u_i^2+v_i^2+w_i^2 \]

Hence,

\boxed{ p= \frac{m}{3V} \left( C_1^2+C_2^2+C_3^2+\cdots+C_N^2 \right) }

Step 9: Introduce mean square velocity

The mean square velocity is defined as

\overline{C^2} = \frac{ C_1^2+C_2^2+C_3^2+\cdots+C_N^2 }{N}

Therefore,

C_1^2+C_2^2+\cdots+C_N^2 = N\overline{C^2}

Substituting this into the pressure equation:

p= \frac{mN}{3V}\overline{C^2}

Step 10: Final expression for pressure

The mass density of the gas is

\rho=\frac{mN}{V}

Substituting this in the previous equation:

p= \frac{1}{3}\rho\overline{C^2}

\boxed{ p=\frac{1}{3}\rho\overline{C^2} }

Final Answer

The pressure exerted by an ideal gas according to the kinetic theory of gases is

\boxed{ p=\frac{1}{3}\rho\overline{C^2} }

where \(p\) is the pressure, \(\rho\) is the density of the gas, and \(\overline{C^2}\) is the mean square velocity of the molecules.

Important Relations to Remember

1. Molecular velocity

\boxed{ C^2=u^2+v^2+w^2 }

2. Mean square velocity

\boxed{ \overline{C^2} = \frac{C_1^2+C_2^2+\cdots+C_N^2}{N} }

3. Density of gas

\boxed{ \rho=\frac{mN}{V} }

4. Kinetic theory pressure equation

\boxed{ p=\frac{1}{3}\rho\overline{C^2} }

Exam Tip

The most important result from this derivation is

\boxed{ p=\frac{1}{3}\rho\overline{C^2} }

Remember the factor \(\frac{1}{3}\). It appears because the molecular motion is resolved into three mutually perpendicular directions \(X\), \(Y\), and \(Z\), and the pressure is the same in all three directions.

In short:

\boxed{ \text{Pressure} \propto \text{density} \times \text{mean square molecular velocity} }

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