Subsections

• Main Assumption and Partition Function
• Thermodynamic Properties
  • Stirling Formula
  • Free Energy
  • Pressure
  • Entropy and Heat Capacity:
  • Energy
• Gibbs Paradox

Ideal Gas

Main Assumption and Partition Function

Definition:

Ideal Gas is the system where interaction is absent

$$U\equiv0,\qquad\text{ideal gas}$$

Configuration Integral:

This is factorizable!

$$Z_N = \int d\tilde q = \int dq_1\dots dq_N = V^N$$

Partition Function:

$$Q_N = \frac{1}{N!}\left(\frac{q_{\text{internal}}V}{\Lambda^3}\right)^N$$

Thermodynamic Properties

Stirling Formula

Let us calculate $\ln N!$ for $N\gg1$:

$$\ln N! = \ln 1 + \ln 2 + \dots\approx \int_0^N \ln x\,dx = N\ln N - N$$

Free Energy

$$A = - kT\ln Q_N = -kTN\ln\frac{eV}{N} + Nf(T)$$

with

$$f(T)=3kT\ln\Lambda - kT\ln q_{\text{internal}}$$

Pressure

$$P = -\left(\frac{\partial A}{\partial V}\right)_{N,T} = \frac{kTN}{V}$$

Pressure of ideal gas does not depend on the nature of the gas!

Entropy and Heat Capacity:

Entropy:  

$$S = -\left(\frac{\partial A}{\partial T}\right)_{N,V} = kN\ln\frac{eV}{N} -Nf'(T)$$ (4)

or, substituting V by NkT/P  

$$S = -kN\ln P + kN\ln(ekT) - Nf'(T)$$ (5)

Heat capacity:

$$\begin{gathered} C_v = T\left(\frac{\partial S}{\partial T}... ...rtial S}{\partial T}\right)_{N,P} = kN-NTf''(T) \end{gathered}$$

We obtained:

C_p-C_v = kN

Energy

 

E = A+TS = Nf(T)-NTf'(T)

Gibbs Paradox

Suppose we have a container of the volume 2V with 2N particles, divided into two parts. We deleted the divider. What is the entropy change?

First Method:

Let us paint the particles in the left in white, and particle in the right in black. After we deleted the divider, ``white'' and ``black'' particles mix. From equation (4) the change of entropy for ``white particles'' is

$$\Delta S_w = kN\ln(2V) - kN\ln V = kN\ln2$$

For ``black particles'' $\Delta S_b = \Delta S_w$. Total change of entropy is  

$$\Delta S = 2kN\ln2$$ (7)

Second Method:

Entropy is an extensive variable, so the entropy for 2N molecules in the volume 2V is twice the entropy for N molecules in the volume V

$$\begin{gathered} S(N,V)+S(N,V) = 2S(N,V)\ S(2N,2V)=2S(N,V) \end{gathered}$$

and  

$$\Delta S = 0$$ (8)

What is wrong? Why we obtained different answers?

Solution:

The distinguishability is a quantum effect: it is either there, or not (particles cannot be a little bit different!). If they are identical, we cannot speak about entropy of white or black particles separately--and only (8) works. If they are not identical, only (7) works.

Consequence:

There is an entropy change when mixing different isotopes--even while the molecules are quite similar!