Chapter 14: The Classical Statistical Treatment of an Ideal Gas

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Presentation transcript:

Chapter 14: The Classical Statistical Treatment of an Ideal Gas

14.1 Thermodynamic properties from the Partition Function All the thermodynamic properties can be expressed in terms of the logarithm of the partition function and its derivatives. Thus, one only needs to evaluate the partition function to obtain its thermodynamic properties!

From chapter 13, we know that for diluted gas system, M-B statistics is applicable, where S= U/T + Nk (ln Z – ln N +1) F= -NkT (ln Z – ln N +1) μ= = – kT (lnZ – lnN) Now one can derive expressions for other thermodynamic properties based on the above relationship.

Internal Energy: We have …

differentiating the above equation with respect to T, we have … = (keeping V constant means E(V) is constant) Therefore, or

Gibbs Function since G = μ N G = - NkT (ln Z – ln N) Enthalpy G = H – TS → H = G + TS H =-NkT(lnZ - ln N) + T(U/T + NklnZ – NklnN + Nk) = -NkTlnZ + NkTlnN + U + NkTlnZ – NkTlnN + NkT = U + NkT = NkT2 + NkT = NkT [1 + T · ]

Pressure

14.2 Partition function for a gas assuming g1 = g2= g3 = … gn = 1, E1 = 0 Z = 1 + e + e + …

For a gas in a container of macroscopic size, the energy levels are very closely spaced, which can be regarded as a continuum. Recall Since the gas is composed of molecules rather than spin ½ particles, the spin factor does not apply, i.e. γs = 1. For continuum system

14.3 Properties of a monatomic ideal gas Previous sections illustrate that many of the thermodynamic properties is a function of lnZ, which Z itself is a function of T and V. ln Z = ln V + 3/2 ln ( ) = lnV + (3/2) ln + 3/2 lnT

Recall Now we have Thus, PV = NkT (note that k = R/NA) Similarly, we have and Therefore, U = 3/2 NkT

From equation 14.1 S= U/T + Nk (ln Z – ln N +1) and U = 3/2 NkT ; We have The above equation become invalid as T 0

Example (textbook 14.1) (a) Calculate the entropy S and the Helmholtz function for an assembly of distinguishable particles. (b) Show that the total energy U and the pressure P are the same for distinguishable particles as for molecules of an ideal gas while S is different. Explain why this makes sense. Solution: