1. Chapter 14: The Classical Statistical Treatment of an Ideal Gas

2. 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!

3. From chapter 13, we know that for dilute gas system, MB 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.

4. 1.Internal Energy: We have …

5. differentiating the above Z equation with respect to T, we have … = (keeping V constant means ε j is constant) Therefore, or

6. 2.Gibbs Function since G = μ N G = - NkT (ln Z – ln N) 3.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 = NkT 2 + NkT = NkT (1 + T · )

7. 4.Pressure

8. 14.2 Partition function for a gas assuming g 1 = g 2 = g 3 = … g n = 1, ε 1 = 0 Z = 1 + e -ε 2 /kT + e -ε 3 /kT + …

9. 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. γ s = 1. For a continuum system

10. 14.3 Properties of a monatomic ideal gas Previous sections illustrate that many of the thermodynamic properties is a function of lnZ, where Z itself is a function of T and V. ln Z = ln V + 3/2 ln ( ) = lnV + (3/2) ln + 3/2 lnT

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

12. 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

13. 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:

15. Continued

17. 14.4 Applicability of the M-B distribution MB statistics is valid for dilute gas system. How good is such an assumption for gases at normal conditions? Recall For a simple notation, assuming that where n Q has the dimension of the reciprocal of the volume, and is called the quantum concentration. As a result, Z = n Q V

18. For MB distribution, For helium gas under standard conditions. m = 6.65 x 10 -27 kg and T = 273 K. n Q ≈ 7 x 10 30 m -3 The number density, the number of particles per unit volume, can be found from N/V ≈ 3 x 10 25 m -3. Since e (-ε j /kT) is of the order of unity, N j /g j ≈ 4 x 10 -6

19. 14.5 Distribution of molecular speeds For a continuum of energy levels, where and

20. Combining the above equations, one has ε is a kinetics energy calculated through (1/2)mv 2, thus dε = mvdv The above equation can be transformed into (in class demonstration)

21. 14.6 Equipartition of energy From Kinetics theory of gases showing that the average energy of a molecule is the number of degrees of freedom (f) of its motion. For a monatomic gas, there are three degrees of freedom, one for each direction of the molecule’s translational motion. The average energy for a single monatomic gas molecule is (3/2)kT (in class derivation). The principle of the equipartition of energy states that for every degree of freedom for which the energy is a quadratic function, the mean energy per particle of a system in equilibrium at temperature T is (1/2)kT.