Exercise 2.6
a. Assume that argon is an ideal gas withCV,m12.47 J K−1mol−1. Find∆Uif 1.000 mol of argon gas is heated at constant volume from 298.15 K to 500.0 K. Find the ratio of this energy difference to the rest-mass energy of the system. Find the difference between the observed mass of the system at 298.15 K and at 500.0 K.
b. Explain why it would be difficult to use values of total energies for chemical purposes if the rest-mass energy were included.
For a one-phase simple system containing one component the equilibrium state is specified by the values of three variables, at least one of which must be extensive. Since the internal energy is a state function we can write
UU(T,V,n) (2.3-4)
or
UU(T,P,n) (2.3-5)
The internal energy is a state function, but heat and work are not state functions.
Because heat and work are both means of changing the value of the internal energy, they do not maintain separate identities after a transfer of energy is finished. The following analogy has been used.3Heat transferred to the system is analogous to rain falling on a pond, work done on the system is analogous to the influx of a stream into the pond, and energy is analogous to water in the pond. Evaporation (counted as negative rainfall) is analogous to heat flow to the surroundings, and efflux from the pond into a second stream is analogous to work done on the surroundings. Once rain falls into the pond, it is no longer identifiable as rain, but only as water. Once stream flow is in the pond, it also is identifiable only as water, and not as stream flow. The amount of water in the pond is a well-defined quantity (a state function), but one cannot separately state how much rain and how much stream flow are in the pond. Similarly, there is no such thing as the heat content of a system in a given state and no such thing as the work content of a system in a given state.
wherekBis Boltzmann’s constant,nis the amount of gas in moles,Ris the ideal gas constant, andT is the absolute temperature. Since this energy does not include the gravitational potential energy and the kinetic energy of the entire system, this energy is the internal energy of the model gas:
U 3
2nRT (dilute monotonic gas) (2.3-7)
Real atoms and molecules are not structureless particles. Real atoms and molecules have translational energy and electronic energy, and molecules also have rotational and vibrational energy. The translational energy of all dilute gases is accurately represented by Eq. (2.3-7). The other contributions to the energy of a dilute gas are studied in statistical mechanics, which is discussed in later chapters of this textbook. At ordinary temperatures the electrons of nearly all atoms and molecules are in their lowest possible energy states, and the electronic energy is a constant, which we can set equal to zero. The vibrational contribution to the energy is not quite so small as the electronic contribution at ordinary temperatures, but we will neglect it for now. Statistical mechanics gives the following results for the rotational contributions:
UrotnRT (diatomic gas or linear polyatomic gas) (2.3-8)
Urot 3
2nRT (nonlinear polyatomic gas) (2.3-9) We can now write formulas for the internal energy of dilute gases:
U≈ 3
2nRT +U0 (monatomic gas) (2.3-10)
U≈ 5
2nRT +U0 (diatomic gas or linear polyatomic gas) (2.3-11)
U≈3nRT +U0 (nonlinear polyatomic gas) (2.3-12)
We can set the zero of energy at any convenient energy, so we ordinarily set the constants denoted byU0equal to zero. We will later use experimental heat capacity data to test these equations. Near room temperature, Eq. (2.3-10) is a very good approximation and Eqs. (2.3-11) and (2.3-12) are fairly good approximations.
The atoms or molecules of a solid or liquid have the same average translational energy as the molecules of a gas at the same temperature, although the translational motion is a kind of rattling back instead of motion in a straight line between collisions.
In some liquids containing small molecules the rotational and vibrational contributions will also be nearly the same, although in other cases the rotation is restricted. However, the molecules of a liquid or solid are packed closely together, and the potential energy makes an important contribution to the internal energy. There is no simple representa- tion of this potential energy, and it is not possible to represent the internal energy of a solid or liquid by a general mathematical formula.
P R O B L E M S
Section 2.3: Internal Energy: The First Law of Thermodynamics
2.17 According to special relativity, the total energy of a system is given by
Emc2
wheremis the mass of the system andcis the speed of light.
a. Assuming a substance with a molar mass of 0.100 kg mol−1, calculate the amount of energy necessary to change the molar mass from its rest-mass value to a value larger by 1.0 ppm (part per
million).
b. Compare this energy with a typical chemical bond energy, roughly 400 kJ mol−1.
c. If the energy in part a were all kinetic energy of the center of mass of the system, calculate the speed of the center of mass of the system of 1.000 mol.
2.18 a. Calculateq,w, and∆Uif 2.000 mol of neon (assumed ideal) is heated at a constant pressure of 1.000 atm from a temperature of 0.00◦C to a temperature of 250.00◦C.
b. Calculateq,w, and∆Uif the same sample of neon is heated at a constant volume from the same initial state
to 250.00◦C and is then expanded isothermally to the same final volume as in part a.
2.19 Calculateq,w, and∆Ufor melting 100.0 g of ice at 0.0◦C and a constant pressure of 1.000 atm. The density of ice is 0.916 g mL−1.
2.20 Calculateq,w, and∆Ufor vaporizing 2.000 mol of liquid water at 100.0◦C to steam at 100.0◦C at a constant pressure of 1.000 atm.
2.21 Consider the following three processes: (1) A sample of 2.000 mol of helium gas is isothermally and reversibly expanded from a volume of 10.00 L and a temperature of 400.0 K to a volume of 40.00 L. (2) The same sample is reversibly cooled at a constant volume of 10.00 L from 400.0 K to a temperature of 300.0 K, then expanded reversibly and isothermally to a volume of 40.00 L, and then heated reversibly from 300.0 K to 400.0 K at a constant volume of 40.00 L. (3) The same sample is expanded irreversibly and isothermally at a temperature of 400.0 K from a volume of 10.00 L to a volume of 40.00 L with a constant external pressure of 1.000 atm. Calculate
∆U,q, andwfor each process.
2.22 1.000 kg of water is pressurized isothermally at 298.15 K from a pressure of 1.000 atm to a pressure of 10.00 atm.
Calculatewfor this process. State any assumptions.
2.4 Calculation of Amounts of Heat and Energy Changes
Like work, heat is not a state function. The amount of heat put into a system can depend on the path taken from the initial to the final state, as was the case with work.
Exercise 2.7 Show that if
dUdw+dq
and ifdUis an exact differential and ifdwis an inexact differential thendqmust be an inexact differential.