Chapter 14: The Classical Statistical Treatment of an Ideal Gas

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. Internal Energy: We have …

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

6. 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 · )

7. Pressure

8. 14.2 Partition function for a gas assuming g1 = g2= g3 = … gn = 1, ε1 = 0 Z = 1 + e-ε2/kT + e-ε3/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. γs = 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/NA) • 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

16. 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 nQ has the dimension of the reciprocal of the volume, and is called the quantum concentration. • As a result, Z = nQV

18. For MB distribution, • For helium gas under standard conditions. m = 6.65 x 10-27 kg and T = 273 K. nQ ≈ 7 x 1030 m-3 • The number density, the number of particles per unit volume, can be found from N/V ≈ 3 x 1025 m-3 . • Since e(-εj/kT) is of the order of unity, Nj/gj ≈ 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)mv2, 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.