Chapter 18
Thermal Properties of Matter
Dr. Armen Kocharian
Molecular Model of an Ideal Gas
The model shows that the pressure that a gas exerts on the walls of its container is a consequence of the collisions of the gas with the wallsIt is consistent with the macroscopic description developed earlier
Ideal Gas AssumptionsThe number of molecules in the gas is large, and the average separation between the molecules is large compared with their dimensions
The molecules occupy a negligible volume within the containerThis is consistent with the macroscopic model where we assumed the molecules were point-like
Ideal Gas Assumptions, 2The molecules obey Newton’s laws of motion, but as a whole they move randomly
Any molecule can move in any direction with any speedAt any given moment, a certain percentage of molecules move at high speedsAlso, a certain percentage move at low speeds
Ideal Gas Assumptions, 3The molecules interact only by short-range forces during elastic collisions
This is consistent with the macroscopic model, in which the molecules exert no long-range forces on each other
The molecules make elastic collisions with the wallsThe gas under consideration is a pure substance
All molecules are identical
Ideal Gas Notes
An ideal gas is often pictured as consisting of single atomsHowever, the behavior of molecular gases approximate that of ideal gases quite well
Molecular rotations and vibrations have no effect, on average, on the motions considered
Molecular Properties of MatterForces between molecules
Cubic crystal structure of sodium chloride
Pressure and Kinetic EnergyAssume a container is a cube
Edges are length dLook at the motion of the molecule in terms of its velocity componentsLook at its momentum and the average force
Pressure and Kinetic Energy, 2Assume perfectly elastic collisions with the walls of the containerThe relationship between the pressure and the molecular kinetic energy comes from momentum and Newton’s Laws
Pressure and Kinetic Energy, 3The relationship is
This tells us that pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules
___22 1
3 2NP m vV
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Pressure and Kinetic Energy, final
This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speedOne way to increase the pressure is to increase the number of molecules per unit volumeThe pressure can also be increased by increasing the speed (kinetic energy) of the molecules
Molecular Interpretation of Temperature
We can take the pressure as it relates to the kinetic energy and compare it to the pressure from the equation of state for an ideal gas
Therefore, the temperature is a direct measure of the average molecular kinetic energy
___2
B2 13 2
NP m v Nk TV
⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Molecular Interpretation of Temperature, contSimplifying the equation relating temperature and kinetic energy gives
This can be applied to each direction,
with similar expressions for vy and vz
___2
B1 32 2
m v k T=
___2
B1 12 2xm v k T=
A Microscopic Description of Temperature, final
Each translational degree of freedom contributes an equal amount to the energy of the gas
In general, a degree of freedom refers to an independent means by which a molecule can possess energy
A generalization of this result is called the theorem of equipartition of energy
Theorem of Equipartition of Energy
Each degree of freedom contributes ½kBT to the energy of a system, where possible degrees of freedom in addition to those associated with translation arise from rotation and vibration of molecules
Total Kinetic EnergyThe total kinetic energy is just N times the kinetic energy of each molecule
If we have a gas with only translational energy, this is the internal energy of the gasThis tells us that the internal energy of an ideal gas depends only on the temperature
___2
tot trans B1 3 32 2 2
K N m v Nk T nRT⎛ ⎞= = =⎜ ⎟
⎝ ⎠
Root Mean Square SpeedThe root mean square (rms) speed is the square root of the average of the squares of the speeds
Square, average, take the square rootSolving for vrms we find
M is the molar mass and M = mNA
Brms
3 3k T RTvm M
= =
Some Example vrms Values
At a given temperature, lighter molecules move faster, on the average, than heavier molecules
Molar Specific Heat, 2
We define specific heats for two processes that frequently occur:
Changes with constant pressureChanges with constant volume
Using the number of moles, n, we can define molar specific heats for these processes
Specific Heat, 3
Molar specific heats:Q = nCV ΔT for constant-volume processesQ = nCP ΔT for constant-pressure processes
Q (constant pressure) must account for both the increase in internal energy and the transfer of energy out of the system by workQconstant P > Qconstant V for given values of n and ΔT
Ideal Monatomic Gas
A monatomic gas contains one atom per moleculeWhen energy is added to a monatomic gas in a container with a fixed volume, all of the energy goes into increasing the translational kinetic energy of the gas
There is no other way to store energy in such a gas
Ideal Monatomic Gas, contTherefore, ΔΚint = 3/2 nR ΔT
ΔE is a function of T onlyIn general, the internal energy of an ideal gas is a function of T only
The exact relationship depends on the type of gas
At constant volume, Q = ΔΚint = nCV ΔT
This applies to all ideal gases, not just monatomic ones
Monatomic Gases, finalSolving for CV gives CV = 3/2 R = 12.5 J/mol . K
For all monatomic gasesThis is in good agreement with experimental results for monatomic gases
In a constant-pressure process, ΔKint = Q + W and CP – CV = R
This also applies to any ideal gasCo = 5/2 R = 20.8 J/mol . K
Ratio of Molar Specific HeatsWe can also define
Theoretical values of CV , CP , and γ are in excellent agreement for monatomic gasesBut they are in serious disagreement with the values for more complex molecules
Not surprising since the analysis was for monatomic gases
5 / 2 1.673 / 2
P
V
C RC R
γ = = =
Sample Values of Molar Specific Heats
Molar Specific Heats of Other Materials
The internal energy of more complex gases must include contributions from the rotational and vibrational motions of the moleculesIn the cases of solids and liquids heated at constant pressure, very little work is done since the thermal expansion is small and CP and CV are approximately equal
Adiabatic Processes for an Ideal Gas
Assume an ideal gas is in an equilibrium state and so PV = nRT is validThe pressure and volume of an ideal gas at any time during an adiabatic process are related by PV γ = constantγ = CP / CV is assumed to be constant during the processAll three variables in the ideal gas law (P, V, T ) can change during an adiabatic process
Equipartition of EnergyWith complex molecules, other contributions to internal energy must be taken into accountOne possible energy is the translational motion of the center of mass
Equipartition of Energy, 2Rotational motion about the various axes also contributes
We can neglect the rotation around the yaxis since it is negligible compared to the x and z axes
Equipartition of Energy, 3The molecule can also vibrateThere is kinetic energy and potential energy associated with the vibrations
Equipartition of Energy, 4The translational motion adds three degrees of freedomThe rotational motion adds two degrees of freedomThe vibrational motion adds two more degrees of freedomTherefore, Eint = 7/2 nRT and CV = 7/2 RThis is inconsistent with experimental results
Agreement with ExperimentMolar specific heat is a function of temperatureAt low temperatures, a diatomic gas acts like a monatomic gas
CV = 3/2 R
Agreement with Experiment, cont
At about room temperature, the value increases to CV = 5/2 R
This is consistent with adding rotational energy but not vibrational energy
At high temperatures, the value increases to CV = 7/2 R
This includes vibrational energy as well as rotational and translational
Complex MoleculesFor molecules with more than two atoms, the vibrations are more complexThe number of degrees of freedom is largerThe more degrees of freedom available to a molecule, the more “ways” there are to store energy
This results in a higher molar specific heat
Quantization of EnergyTo explain the results of the various molar specific heats, we must use some quantum mechanics
Classical mechanics is not sufficientIn quantum mechanics, the energy is proportional to the frequency of the wave representing the frequencyThe energies of atoms and molecules are quantized
Quantization of Energy, 2This energy level diagram shows the rotational and vibrational states of a diatomic moleculeThe lowest allowed state is the ground state
Quantization of Energy, 3
The vibrational states are separated by larger energy gaps than are rotational statesAt low temperatures, the energy gained during collisions is generally not enough to raise it to the first excited state of either rotation or vibration
Quantization of Energy, 4Even though rotation and vibration are classically allowed, they do not occurAs the temperature increases, the energy of the molecules increasesIn some collisions, the molecules have enough energy to excite to the first excited stateAs the temperature continues to increase, more molecules are in excited states
Quantization of Energy, final
At about room temperature, rotational energy is contributing fullyAt about 1000 K, vibrational energy levels are reachedAt about 10 000 K, vibration is contributing fully to the internal energy
Molar Specific Heat of SolidsMolar specific heats in solids also demonstrate a marked temperature dependenceSolids have molar specific heats that generally decrease in a nonlinear manner with decreasing temperatureIt approaches zero as the temperature approaches absolute zero
DuLong-Petit Law
At high temperatures, the molar specific heats approach the value of 3R
This occurs above 300 K
The molar specific heat of a solid at high temperature can be explained by the equipartition theorem
Each atom of the solid has six degrees of freedomThe internal energy is 3 nRT and Cv = 3 R
Molar Specific Heat of Solids, Graph
As T approaches 0, the molar specific heat approaches 0At high temperatures, CVbecomes a constant at ~3R
Boltzmann Distribution LawThe motion of molecules is extremely chaoticAny individual molecule is colliding with others at an enormous rate
Typically at a rate of a billion times per second
We add the number density nV (E )This is called a distribution functionIt is defined so that nV (E ) dE is the number of molecules per unit volume with energy between Eand E + dE
Number Density and Boltzmann Distribution Law
From statistical mechanics, the number density is nV (E ) = noe –E /kBT
This equation is known as the Boltzmann distribution lawIt states that the probability of finding the molecule in a particular energy state varies exponentially as the energy divided by kBT
Distribution of Molecular Speeds
The observed speed distribution of gas molecules in thermal equilibrium is shown at rightNV is called the Maxwell-Boltzmann speed distribution function
Distribution FunctionThe fundamental expression that describes the distribution of speeds in N gas molecules is
m is the mass of a gas molecule, kB is Boltzmann’s constant and T is the absolute temperature
23/ 2
/ 22
B
( ) 42
Bmv k Tv
mf v N N v ek T
ππ
−⎛ ⎞= = ⎜ ⎟
⎝ ⎠
Most Probable Speed
The average speed is somewhat lower than the rms speedThe most probable speed, vmp is the speed at which the distribution curve reaches a peak
B Bmp
2 1.41k T k Tvm m
= =
Speed DistributionThe peak shifts to the right as Tincreases
This shows that the average speed increases with increasing temperature
The asymmetric shape occurs because the lowest possible speed is 0 and the highest is infinity
Speed Distribution, final
The distribution of molecular speeds depends both on the mass and on temperatureThe speed distribution for liquids is similar to that of gases
EvaporationSome molecules in the liquid are more energetic than othersSome of the faster moving molecules penetrate the surface and leave the liquid
This occurs even before the boiling point is reached
The molecules that escape are those that have enough energy to overcome the attractive forces of the molecules in the liquid phaseThe molecules left behind have lower kinetic energiesTherefore, evaporation is a cooling process
Phases of MatterPhase equilibriumTriple pointCritical pointSublimation
pVT-surfaces
18.7: From Eq. (18.6), C.503K776)cmPa)(4991001.1()cmPa)(46.210(2.821K)15.300( 35
36
11
2212 °==⎟⎟
⎠
⎞⎜⎜⎝
⎛××
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
VpVpTT
18.15: a) C.70.3K343K)molatmL0.08206mol)(0.11(
L)atm)(3.10100(22 °==
⋅⋅==
nRVpT
b) This is a very small temperature increase and the thermalexpansion of the tank may be neglected; in this case, neglecting the expansion means not including expansion in finding the highest safe temperature, and including the expansion would tend to relax safe standards.
18.16: (a) The force of any side of the cube is ,)()( LnRTAVnRTpAF ===
since the ratio of area to volume is .K15.293C20.0For .1 =°== TLVA
N. 103.66 m 200.0
K) (293.15 K)molJ (8.3145 mol) 3( 4×=⋅
==L
nRTF
b) For K, 373.15 C00.100 =°=T
.N1065.4m200.0
)KK)(373.15molJ5mol)(8.3143( 4×=⋅
==L
nRTF