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Ideal Gas

Main Assumption and Partition Function

Ideal Gas is the system where interaction is absent

U\equiv0,\qquad\text{ideal gas}\end{displaymath}

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\end{displaymath}

Free Energy

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


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


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

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

Entropy and Heat Capacity:

 S = -\left(\frac{\partial A}{\partial T}\right)_{N,V} =
 kN\ln\frac{eV}{N} -Nf'(T)\end{displaymath} (4)
or, substituting V by NkT/P  
 S = -kN\ln P + kN\ln(ekT) - Nf'(T)\end{displaymath} (5)
Heat capacity:

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

We obtained:

Cp-Cv = kN


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
 \end{displaymath} (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

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

 \Delta S = 0
 \end{displaymath} (8)

What is wrong? Why we obtained different answers?

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

next up previous
Next: Quiz Up: Systems with Many Particles. Previous: Many-Particle Systems

© 1997 Boris Veytsman and Michael Kotelyanskii
Wed Sep 17 22:58:45 EDT 1997