Ideal Gas

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 $$l_Q = \sqrt{\frac{h^2}{2\pi mkT}}$$