Definition
Ideal gas is a model of gas in which the gas molecules
- are identical, i.e. indistinguishable
- have negligible volume, i.e. are point masses
- do not exert forces on each other except in collision
- all collisions are elastic
Entropy
Ideal gas consists of identical point masses with no interaction with each other. So the total internal energy is their total kinetic energy
$$
\sum^N_{i}\frac{p_i^2}{2m} = U
$$
Expressing the momentum in terms of \(U\)
$$
p_1^2 + p_2^2 + ... +p_N^2 = 2mU
$$
Since each momentum is
$$p_i^2 = p_x^2 + p_y^2 + p_z^2$$
The expression \(\sqrt{2mU}\) is the radius of a \(3N\) dimension sphere. Thus, the number of states of momentum of the ideal gas can have is equivalent to the surface area of this \(3N\) dimension sphere
$$\frac{2\pi^{3N/2}}{\left(\frac{3N}{2}-1\right)!}(2mU)^{\frac{3N-1}{2}}$$
As each particle can be anywhere within the volume \(V\), the number of state of position of each particle is \(V\). Thus, the total number of states of position of the ideal gas is \(V^N\). As each state of position can have each state of momentum, the total multiplicity is
\begin{align}
\Omega &\propto V^N \frac{2\pi^{3N/2}}{\left(\frac{3N}{2}-1\right)!}(2mU)^{\frac{3N-1}{2}} \\
&\approx \frac{(2\pi m)^{3N/2}}{\left(3N/2\right)!}V^NU^{3N/2}\\
\end{align}
Since the particles are indistinguishable, the number of states should be divided by \(n!\). Also, to make the dimension right (multiplicity should be dimensionless),
$$\Omega = \frac{1}{h^{3N}}\frac{1}{N!}\frac{(2\pi m)^{3N/2}}{\left(3N/2\right)!}V^NU^{3N/2}$$
The entropy, applying Stirling approximation, is
\begin{align}
S &= k \ln \Omega = k \ln \left[\frac{1}{h^{3N}}\frac{1}{N!}\frac{(2\pi m)^{3N/2}}{\left(3N/2\right)!}V^NU^{3N/2} \right] \\
&= k \left[-\ln(N!)-\ln\left(\frac{3N}{2}!\right) + \ln \left(\frac{V}{h^3}\pi^{3/2}(2mU)^{3/2} \right)^N \right] \\
&\approx k \left[-N\ln N + N - \frac{3N}{2}\ln\frac{3N}{2} + \frac{3N}{2} + N\ln\left(\frac{V}{h^3}\pi^{3/2}(2mU)^{3/2}\right) \right] \\
&= Nk\left[-\ln N -\frac{3}{2}\ln\frac{3N}{2} + \frac{5}{2} + \ln\left(V\left(\frac{2\pi mU}{h^2}\right)^{3/2}\right)\right] \\
&= Nk\left[\frac{5}{2}+\ln\left(\frac{V}{N}\left(\frac{2}{3N}\frac{2\pi mU}{h^2}\right)^{3/2}\right)\right] \\
\end{align}
$$ \bbox[5px,border:2px solid #666]
{
S = Nk\left[\frac{5}{2} + \ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^2}\right)^{3/2}\right)\right]
}
$$
Ideal Gas Law
From experiments, it was found that the pressure, \(P\), and volume, \(V\), of a gas is related as
$$ PV = \text{constant}$$
when holding temperature, number of molecules constant. This is known as the Boyle's law. From experiments, it was found that the volume and temperature, \(T\), of a gas is related as
$$ \frac{V}{T} = \text{constant}$$
when holding pressure, number of molecules constant. This is known as the Charles's law. Also, from experiments, it was found that the volume and the number of molecules, \(N\) are related as
$$ \frac{V}{N} = \text{constant}$$
when holding pressure, temperature constant. Combining these results,
$$ \bbox[5px,border:2px solid #666]
{
PV = NkT
}$$
where \(k\) is the Boltzmann constant.
Kinetic Energy
Suppose the ideal gas is contained in a box with width \(L\). The change of momentum of an ideal gas molecule when it collides with the wall of the container is
$$\Delta p_x = 2 mv_x$$
Within the duration the molecule collides the spot again
$$\Delta t = 2 \frac{L}{v_x}$$
the average force is
$$F_{\text{avg}} = \frac{\Delta p_x}{\Delta t} = \frac{mv_x^2}{L}$$
The pressure experienced by the wall due to the collisions by the molecules is
\begin{align}
P &= \frac{\sum_i F_{\text{avg}}}{A} \\
&= \frac{1}{AL}\sum_i m_iv_{x\, i}^2 = \frac{N}{V}m\bar{v_x}^2
\end{align}
where
$$\bar{v_x}^2 = \frac{1}{N}\sum_i v_{x\, i}^2$$
Thus,
$$ \bbox[5px,border:2px solid #666]
{
PV = Nm\bar{v_x}^2
}$$
comparing to the ideal gas law
$$PV = NkT$$
the average kinetic energy of a molecule in the x-direction is
$$\frac{1}{2}m\bar{v_x}^2 = \frac{1}{2}kT$$
Similarly for y and z-direction
$$\frac{1}{2}m\bar{v_y}^2 = \frac{1}{2}m\bar{v_z}^2 = \frac{1}{2}kT$$
So, the average translational kinetic energy of the ideal gas is
\begin{align}
KE &= \frac{1}{2}Nm \bar{v} ^2 \\
&= \frac{1}{2}m\bar{v_x}^2 + \frac{1}{2}m\bar{v_y}^2 + \frac{1}{2}m\bar{v_z}^2 \\
&= \frac{3}{2}NkT
\end{align}
Internal Energy
The entropy of an ideal gas is
$$S &= k \left[-\ln(N!)-\ln\left(\frac{3N}{2}!\right) + \ln \left(\frac{V}{h^3}\pi^{3/2}(2mU)^{3/2} \right)^N \right]$$
Applying the definition of temperature
\begin{align}
\frac{1}{T} &= \frac{\partial S}{\partial U} \\
&= k\frac{\partial }{\partial U} \left[\ln (U^{\frac{3N}{2}}) -\ln(N!)-\ln\left(\frac{3N}{2}!\right) + \ln \left(\frac{V}{h^3}\pi^{3/2}(2m)^{3/2} \right)^N \right]\\
&= \frac{3Nk}{2}\frac{1}{U}
\end{align}
So,
$$ \bbox[5px,border:2px solid #666]
{
U = \frac{3}{2}NkT
}$$
Heat Capacity
Heat capacity is defined by
$$C = \lim_{\Delta T \to 0}\frac{\Delta Q}{\Delta T}$$
At constant volume, no work is done by the gas
$$C_V = \frac{\partial}{\partial T} Nf\left(\frac{1}{2}kT\right) = \frac{f}{2}Nk$$
At constant pressure, \(Q = \Delta U - W\),
\begin{align}
C_P &= \frac{\Delta U - W}{\Delta T} \\
&= \frac{\Delta U + P\Delta V}{\Delta T} \\
&= \left(\frac{\partial U}{\partial T}\right)_P + P\left(\frac{\partial V}{\partial T}\right)_P
\end{align}
Isothermal Process
Isothermal process is a process in which temperature of the ideal gas is kept constant.
Work done is
\begin{align}
W &= - \int^{V_f}_{V_i} P dV \\
&= - NkT \int^{V_f}_{V_i} \frac{dV}{V} \\
&= -NkT\ln \frac{V_i}{V_f}
\end{align}
The internal energy, by equipartition theorem, is given by
$$ U = N\left(\frac{f}{2}kT \right)$$
Since \(\Delta T = 0\), the change in internal energy is
$$\Delta U = N\left(\frac{f}{2}k\Delta T = 0$$
$$\Delta U = Q + W$$
the heat flowed into the ideal gas is
$$Q = -W = -NkT\ln \frac{V_i}{V_f}$$
Adiabatic Process
Adiabatic process is a process in which there is no heat flow into or out of the ideal gas
$$\Delta U = Q + W = W$$
Since
$$dU = Nf\left(\frac{1}{2}kdT\right)$$
and
$$dW = -PdV$$
we have
\begin{align}
dU &= dW \\
Nf\left(\frac{1}{2}kdT\right) &= -PdV \\
&= - \frac{NkT}{V} dV \\
\frac{f}{2}\frac{dT}{T} &= -\frac{dV}{V} \\
\frac{f}{2}\ln\frac{T'}{T} = - \ln\frac{V'}{V} \\
\end{align}
Thus,
$$VT^{f/2} = \text{constant}$$
Substitute ideal gas law,
$$PV^{(f+2)/f} = \text{constant}$$
Chemical Potential
The chemical potential of ideal gas is
\begin{align}
\mu &= \left(\frac{\partial S}{\partial N}\right)_{U,V} \\
&= -kT\left[\frac{5}{2} + \frac{3}{2}\ln\left(\frac{4\pi mUV^{2/3}}{3h^2}\right)-\frac{5}{2}\ln N + N\left(-\frac{5}{2}\right)\frac{1}{N}\right] \\
&= -kT\ln\left[\left(\frac{4\pi mUV^{2/3}}{3h^2}\right)^{3/2}N^{-5/2}\right] \\
&= -kT\ln\left[\left(\frac{4\pi m(3NkT/2)}{3h^2}\right)^{3/2}\frac{V}{N^{5/2}}\right]\\
&= -kT\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\frac{V}{N}\right] \\
\end{align}
Define the quantum length as
$$l_Q = \sqrt{\frac{h^2}{2\pi mkT}}$$
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