Mean Pressure of an Ideal Gas

The average force exerted by the gas molecules per unit area of the wall is called the mean pressure or average pressure of the gas.

Consider a rectangular container of volume \(V\) containing \(N\) molecules of an ideal gas, each of mass \(m\). Let the dimensions of the container be \(L_x\), \(L_y\), and \(L_z\). Therefore,

\[ \boxed{V=L_xL_yL_z} \]

Consider a molecule in the \(i\)-th state, having energy \(\varepsilon_i\), and moving in the \(X\)-direction. Let it exert a force \(F_i\) on the wall perpendicular to the \(X\)-axis.

Suppose that the wall is displaced through a small distance \(dL_x\) by this force. If the system is isolated, the work done by the force is at the expense of the energy of the molecule. Therefore,

\[ F_i\,dL_x=-d\varepsilon_i \]

(1)

Hence,

\[ \boxed{ F_i=-\frac{\partial\varepsilon_i}{\partial L_x} } \]

The average force exerted by the molecules in all possible states on the wall is

\[ \boxed{ \overline{F}_x=\sum_iF_iP_i } \]

(2)

where \(P_i\) is the probability of finding a molecule in the \(i\)-th state.

According to the Boltzmann distribution,

\[ P_i= \frac{e^{-\beta\varepsilon_i}} {\displaystyle\sum_j e^{-\beta\varepsilon_j}} \]

where

\[ \boxed{ \beta=\frac{1}{kT} } \]

Therefore,

\[ \overline{F}_x = \frac{ \displaystyle\sum_i F_i e^{-\beta\varepsilon_i} }{ \displaystyle\sum_i e^{-\beta\varepsilon_i} } \]

(3)

Using Eq. (1),

\[ \overline{F}_x = - \frac{ \displaystyle\sum_i \frac{\partial\varepsilon_i}{\partial L_x} e^{-\beta\varepsilon_i} }{ \displaystyle\sum_i e^{-\beta\varepsilon_i} } \]

(4)

The partition function of the system is defined as

\[ \boxed{ Z=\sum_i e^{-\beta\varepsilon_i} } \]

(5)

Differentiating Eq. (5) with respect to \(L_x\), we obtain

\[ \frac{\partial Z}{\partial L_x} = -\beta \sum_i \frac{\partial\varepsilon_i}{\partial L_x} e^{-\beta\varepsilon_i} \]

(6)

Thus,

\[ \sum_i \frac{\partial\varepsilon_i}{\partial L_x} e^{-\beta\varepsilon_i} = - \frac{1}{\beta} \frac{\partial Z}{\partial L_x}. \]

Substituting this result in Eq. (4), we obtain

\[ \boxed{ \overline{F}_x = \frac{1}{\beta Z} \frac{\partial Z}{\partial L_x} } \]

(7)

For a monoatomic ideal gas, the single-particle partition function is

\[ \boxed{ Z= V \left( \frac{m} {2\pi\beta\hbar^2} \right)^{3/2} } \]

(8)

where

\[ V=L_xL_yL_z. \]

Since \(L_y\) and \(L_z\) are independent of \(L_x\),

\[ \frac{\partial Z}{\partial L_x} = \frac{Z}{L_x}. \]

Substituting this result in Eq. (7),

\[ \overline{F}_x = \frac{1}{\beta Z} \frac{Z}{L_x}. \]

Therefore,

\[ \boxed{ \overline{F}_x = \frac{1}{\beta L_x} = \frac{kT}{L_x} } \]

(9)

The area of the wall perpendicular to the \(X\)-axis is

\[ A_x=L_yL_z. \]

Therefore, the mean pressure along the \(X\)-direction is

\[ P_x= \frac{\overline{F}_x}{A_x}. \]

Using Eq. (9),

\[ P_x = \frac{kT/L_x}{L_yL_z}. \]

Hence,

\[ P_x = \frac{kT} {L_xL_yL_z}. \]

Since \(V=L_xL_yL_z\),

\[ \boxed{ P_x=\frac{kT}{V} } \]

By symmetry, the pressure along the \(Y\)- and \(Z\)-directions is also the same. Therefore,

\[ \boxed{ P_x=P_y=P_z=\frac{kT}{V} } \]

Thus, the mean pressure due to one molecule is

\[ \boxed{ P=\frac{kT}{V} } \]

If the system contains \(N\) molecules, the total pressure is obtained by adding the contributions of all the molecules:

\[ P = N\left(\frac{kT}{V}\right). \]

Therefore,

\[ \boxed{ P=\frac{NkT}{V} } \]

(11)

If \(n\) is the number of molecules per unit volume, then

\[ \boxed{ n=\frac{N}{V} } \]

Hence, Eq. (11) can be written as

\[ \boxed{ P=nkT } \]