2.13 SUMMARY
In this chapter, we introduced the fundamental variables and laws of thermodynamics.
Temperature, pressure, volume, and energy are state variables who value depends only on the state of the system and not the path taken to that state. Two other fundamental variables, work and heat, are not state variables and their value is path dependent in transformations. Relationships between state variables are known as equations of state. Most often we are interested in changes in state variables rather than their absolute values and we often express these in terms of partial differential equations, for example, the dependence of volume on T and P is written as:(2.17)
The first law states the principle of conservation of energy: even though work and heat are path dependent, their sum is the energy change in a transformation and is path independent:(2.22)
We introduced another important state variable, entropy, which is a measure of the randomness of a system and is defined as:(2.47) where Ω is the number of states accessible to the system and k is Boltzmann's constant.
The second law states that in any real transformation the increase in entropy will always exceed the ratio of heat exchanged to temperature:(2.51) In the fictional case of a reversible reaction, entropy change equals the ratio of heat exchanged to temperature.
The third law states that the entropy of a perfectly crystalline substance at the absolute 0 of temperature is 0. Any other substance will have a finite entropy at absolute 0, which is known as the configurational entropy:(2.110)
We then introduced another useful variable, H, the enthalpy, which can be thought of as the heat content of a system and is related to other state variables as:(2.65) The value of enthalpy is in measuring the energy consumed or released in changes of state of a system, including phase changes such as melting.
The heat capacity of a system, C, is the amount of heat required to raise its temperature. This will depend on whether volume or pressure is held constant. For the latter:(2.68)
With these variables, we could then define a particularly useful state function called the Gibbs free energy, G:(2.125) Written in terms of its characteristic variables:(2.124) The Gibbs free energy is the amount of energy available to drive chemical transformations. It has two important properties:Produces and reactants are in equilibrium when the Gibbs free energies are equal and At fixed temperature and pressure, a chemical reaction will proceed in the direction of lower Gibbs free energy.
REFERENCES AND SUGGESTIONS FOR FURTHER READING
1 Anderson, G.M. and Crerar, D.A. 1993. Thermodynamics in Geochemistry. New York, Oxford University Press.
2 Berman, R.G. 1988. Internally-consistent thermodynamic data for minerals in the system Na2O-K2O-CaO-MgO-FeO-Fe2O3-Al2O3-SiO2-TiO2-H2O-CO2. Journal of Petrology 29, 445–552.
3 Einstein, A. 1905. Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig? Annalen der Physik 18, 639–643.
4 Feynman, R., Leighton, R.B. and Sands, M.L. 1989. The Feynman Lectures on Physics Vol. I. Pasadena: California Institute of Technology.
5 Fletcher, P. 1993. Chemical Thermodynamics for Earth Scientists. Essex: Longman Scientific and Technical.
6 Haas, J.L. and Fisher, J.R. 1976. Simultaneous evaluation and correlation of thermodynamic data. American Journal of Science 276: 525–45.
7 Helgeson, H.C., Delany, J.M., Nesbitt, H.W. and Bird, D.K. 1978. Summary and critique of the thermodynamic properties of rock-forming minerals. American Journal of Science 278A: 1–229.
8 Holland, T. J. B. and Powell, R. 1998. An internally consistent thermodynamic data set for phases of petrological interest. Journal of Metamorphic Geology 16(3), 309–343. doi: 10.1111/j.1525-1314.1998.00140.x.
9 Nordstrom, D.K. and Munoz, J.L. 1986. Geochemical Thermodynamics. Palo Alto, Blackwell Scientific.
PROBLEMS
1 For a pure olivine mantle, calculate the adiabatic temperature gradient (∂T/∂P)s at 0.1 MPa (1 atm) and 1000°C. Use the thermodynamic data in Table 2.2 for forsterite (Mg-olivine, Mg2SiO4), and , and .Note that: 1 cc/mol = 1 J/MPa/mol.
2 Complete the proof that V is a state variable by showing that for an ideal gas:
3 A quartz crystal has a volume of 7.5 ml at 298 K and 0.1 MPa. What is the volume of the crystal at 840K and 12.3 MPa ifα = 1.4654 × 10−5 K−1 and β = 2.276 × 10−11 Pa−1 and α and β are independent of T and P.α = 1.4310 × 10−5 K−1 + 1.1587 × 10−9 K−2T β = 1.8553 × 10−11Pa−1 + 7.9453 × 10−8 Pa−1
4 One mole of an ideal gas is allowed to expand against a piston at constant temperature of 0°C. The initial pressure is 1 MPa and the final pressure is 0.04 MPa. Assuming the reaction is reversible,What is the work done by the gas during the expansion?What is the change in the internal energy and enthalpy of the gas?How much heat is gained/lost during the expansion?
5 A typical eruption temperature of basaltic lava is about 1200°C. Assuming that basaltic magma travels from its source region in the mantle quickly enough so that negligible heat is lost to wall rocks, calculate the temperature of the magma at a depth of 40 km. The density of basaltic magma at 1200°C is 2610 kg/m3; the coefficient of thermal expansion is about 1 × 10−4/K. Assume a heat capacity of 850 J/kg-K and that pressure is related to depth as 1 km = 33 MPa (surface pressure is 0.1 MPa.).(HINT: “Negligible heat loss” means the system may be treated as adiabatic.)
6 Show that the Cp of an ideal monatomic gas is 5/2 R.
7 Show that:
8 Show that for a reversible process:(2.73) (Hint: Begin with the statement of the first law (eqn. 2.58), make use of the Maxwell relations, and your proof in problem 7.)
9 Imagine that there are 30 units of energy to distribute among three copper blocks.If the energy is distributed completely randomly, what is the probability of the first block having all the energy?If n1 is the number of units of energy of the first block, construct a graph (a histogram) showing the probability of a given value of n1 occurring as a function of n1.(HINT: Use eqn. 2.37, but modify it for the case where there are three blocks.)
10 Consider a box partitioned into equal volumes, with the left half containing 1 mole of Ne and the right half containing 1 mole of He. When the partition is removed, the gases mix. Show, using a classical thermodynamic approach (i.e., macroscopic), that the entropy change of this process is . Assume that He and Ne are ideal gases and that temperature is constant.
11 Find expressions for Cp and Cv for a van der Waals gas.
12 Show that β (the compressibility, defined in eqn. 2.12) of an ideal gas is equal to 1/P.
13 Show that Hint: Start with equations 2.47 and 2.36a using the approximation that .
14 Show that Hint: Begin with equation 2.63 and express dU as a function of temperature and volume change.
15 Helium at 298K and 1 atm has . Assume He is an ideal gas.Calculate V, H, G, α, β, Cp, Cv, for He at 298K and 1 atm.What are the values for these functions at 600K and 100 atm?What is the entropy at 600 K and 100 atm?
16 Using the enthalpies of formation given in Table 2.02, find ΔH in Joules for the reaction:
17 Using