1 Lecture 1: State Variables Syllabus is on the web. My web address is in my email signature. If you did not receive an email message from me, contact me at [email protected]immediately. Office hours every day Monday through Thursday. Let me know if these are insufficient or inconvenient. Second session on Friday will deal mainly with problem solving, but new material may also be presented. Attendance is required. The first few Fridays will end early, and in those weeks there will not be any break between sessions. The course will cover the first nine chapters of Atkins. See the syllabus and calendar for further details. Physical chemistry is difficult because it draws from many areas of chemistry and physics. It requires a knowledge of classical mechanics, thermodynamics, electricity and magnetism, quantum mechanics, statistical mechanics, as well as chemistry. It is also very problem oriented. Tips on how to solve problems: 1. Do not use a calculator or computer until the very end. Keep the solution as analytical as possible. 2. Keep track of the dimensions of the variables. Try to use dimensionless variables as much as possible. 3. When you do get a numerical answer, judge whether it has a physically realistic size and proper units. Scope of the Course Chemistry 342: A central problem of 19 th century physics: understanding the macroscopic states of matter and their transformations (e.g., phase transitions, chemical reactions, equilibrium, kinetics). The link to the microscopic world is through the kinetic theory of gases. Chemistry 344: A central problem of 20 th century physics: understanding the microscopic properties of matter and deriving from them the macroscopic observables; relating chemical structure, spectroscopy, and kinetics to quantum mechanics.
Transcript
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Lecture 1: State Variables
Syllabus is on the web. My web address is in my email signature. If you did not receivean email message from me, contact me at [email protected] immediately.
Office hours every day Monday through Thursday. Let me know if these are insufficientor inconvenient.
Second session on Friday will deal mainly with problem solving, but new material mayalso be presented. Attendance is required. The first few Fridays will end early, and inthose weeks there will not be any break between sessions.
The course will cover the first nine chapters of Atkins. See the syllabus and calendar forfurther details.
Physical chemistry is difficult because it draws from many areas of chemistry andphysics. It requires a knowledge of classical mechanics, thermodynamics, electricity andmagnetism, quantum mechanics, statistical mechanics, as well as chemistry. It is alsovery problem oriented.
Tips on how to solve problems:
1. Do not use a calculator or computer until the very end. Keep the solution asanalytical as possible.
2. Keep track of the dimensions of the variables. Try to use dimensionlessvariables as much as possible.
3. When you do get a numerical answer, judge whether it has a physically realisticsize and proper units.
Scope of the Course
Chemistry 342: A central problem of 19th century physics: understanding themacroscopic states of matter and their transformations (e.g., phase transitions, chemicalreactions, equilibrium, kinetics). The link to the microscopic world is through the kinetictheory of gases.
Chemistry 344: A central problem of 20th century physics: understanding themicroscopic properties of matter and deriving from them the macroscopic observables;relating chemical structure, spectroscopy, and kinetics to quantum mechanics.
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Q. How do we describe the macroscopic state of matter?
A. By a list of state variables. By definition, such variables do not depend on theprevious history of the system. Compared with the microscopic ~1025 variables neededto describe the system, only a few are needed to describe a macroscopic object. Our goalis to identify these variables and the physical laws that relate them.
There are two types of state variables: extensive ones and intensive ones.
Think of a swimming pool.Examples of Extensive Variables
Number of moles of each chemical component: n1, n2, …,nr. These are dimensionlessquantities. Knowing themis equivalent to knowing the mass of the system.
Energy of the system: U (kg m2 s-2; joules)There are different sources of energy. Kinetic energy can come from internal motions ofa molecule or crystal, such as vibration, bending, and torsional motion. Kinetic energycan also come for external motion, such as translation and rotation. Potential energy cancome from electronic energy, and contributes to the chemical energy of a system. Theyall add up to the total energy, U. Understanding the different types of energy and theirinterconversions is a major topic of this course.
Volume: V (m3)
Magnetic dipole moment: I (joules/Tesla)
Electric dipole moment: m (C m)
Length (of a rubber band, polymer): L (m)
Surface area (bubble, droplet): A (m2)
Entropy: S, a measure of disorder or chaos. It is dimensionless.
Examples of Intensive Variables
Pressure: P. Defined as the force per unit area: Newtons/m2 = pascal
Temperature: T. Definition is complicated. Think of it for now as proportional thetranslational energy per particle. (Hotter particles move faster.) It has the dimensions ofenergy.
Chemical Potential: mi, chemical energy per mole of substance i (joules)
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Magnetic field, B (Tesla)
Electric field: E (Volts/m)
Tension, TL (joules/m)
Surface tension: g (Joules/m2)
Relation between extensive and intensive variables:
P and V: Applying pressure to an object reduces its volume.PV = (newtons/m2) m3 = force x distance = joules
mi ni =(joules/mole) number of moles = joules
IB = (joules/tesla) tesla = joules
m E = (Cm) V/m = CV = Joules
TL L = (J/m) m = Joules
gA = (Joules/m2) m2 = Joules
The tricky case is temperature. The extensive variable associated with temperature isentropy.TS = joules
The Euler Relation
U = TS - PV + Simini + IB + mE + TLL + gA +….
What we have done is replace the myriads of variables needed to describe themicroscopic system with just a handful of macroscopic state variables. For example, wedo not care about the charge on each particle, just the overall dipole moment of the bulksample. If we need greater precision, we might specify also the quadrupole moment, aswell as some higher multipole moments, but the list is still short.
But not all macroscopic variables are state variables. Example of a swimming pool thatcan be filled with either rain water or water from the faucet. There is no way of tellingfrom the present state of the pool how much of each was used to fill the pool. All we cansay is that the sum of the two sources equals the total water content of the pool:
Water = Waterrain + Wfaucet
Similarly, the energy content of a thermodynamic system is the sum of work done on thesystem (ordered energy input) and heat added to the system (chaotic energy):
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U = w + q
The size of w and q individually depends on the path taken (history of the system) andnot on the state of the system.
Lecture 2: Ideal Gas Law
Assumptions of an ideal gas:1. Particles are point masses2. No forces between the particles (elastic collisions)
We will check these assumptions later.
Relation Between State Variables T,P, n, V
m = mass of one particleV = volume of chamberrn = number density = number of particles/V= n NA/V (NA = Avogodro’s number)A = area of wallv = velocity of particle in one dimensionpin = momentum of particle striking the wall = mvpfinal = momentum of particle bouncing off the wall = -mvDp = momentum change = 2mvIn time Dt, a particle moves a distance vDt.Half of the particles in a volume AvDt strike the wall in time Dt. (The other half aremoving in the wrong direction.)
Dpwall = Total momentum imparted to the wall in time Dt = (2mv) (rn/2) (AvDt)= mv2nNAADt/VForce = Dpwall/Dt = mv2nNAA/VPressure on wall = P = Force/Area = mv2nNA/V
PV = nNAmv2
½ mv2 = kinetic energy of one particle½ NA mv2 = kinetic energy of one mole particles = constant times T = RT
PV = nRT
This is the equation of state of an ideal gas.We define Boltzmann’s constant as k=R/NA, so that PV = nNAkT.
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The problem is more complicated because the particles don’t all have the same speed andthey don’t all move in one dimension, but the proportionality PV ~ nT still holds.R is called the gas constant. Its units areL atm mol-1 K-1 from PV=nRTJ mol-1 K-1 from U ∂ RT
How good is assumption 1?
Suppose the radius of a molecule is 3x10-8 cm. For 1 mole at 1 atm at 298 K, the volumeof the molecules isV0 = (4/3) pr3NA=68 cm3 =0.068 l.If the gas pressure is 1 atm,1 its volume isV=nRT/P = 1 x 0.08206x298/1 = 24.4 l.The error fractional introduced by assuming point masses=V0/V=0.0028.Clearly, we can make the error much smaller by reducing P.
One way of defining the temperature is by the equation of state of a very dilute gas.
Suppose a fixed volume of gas, V, is in thermal equilibrium with some standard system,such as water at its triple point.2
Let the pressure of the gas be P3.We define the absolute temperature of the triple point to be some number, q3. (In thiscase, 273.16 K.)
Under some other condition (for example, the gas thermometer might be in thermalequilibrium with boiling water), the same mass and volume of gas has a pressure P. Theabsolute temperature of the gas in that case is then given by the ideal gas law:
q(P) = (P/P3)q3.
q(P) means that q is a function of P.
To guarantee that the gas has ideal behavior, we take a limit,q(P) = lim (P,P3Ø0) (P/P3)q3.
We can also define the absolute temperature in terms of two fixed points, such as themelting and boiling points of water. We arbitrarily divide the temperature interval into100 parts. Again assume a fixed mass and volume of gas, and define the two absolutetemperatures to be q0 and q100.
The absolute temperature at some arbitrary condition is given by:
(q(P)-q0)/100 = (P-P0)/(P100-P0).Again, take the limit of P0Ø0.We define the Celsius temperature, T, byT=q-q0.
Note: In the water example, it is an arbitrary choice to define q0 = 273.15 K and q100 =373.15 K. These choices are unique to the Kelvin scale. An alternative is given inproblem 1.3. What all absolute temperature scales have in common is that qØ0 in thelimit that PVØ0.
Note: An absolute temperature scale could also be devised by using a fixed pressure andallowing the volume to change when the thermometer comes into contact with thevarious heat baths. In this case we would use the notation q(V). In all of these examples,we use one or two large heat baths as references to define the temperature scale, and weuse the gas sample as a thermometer.
How is pressure measured?
Principle of the manometer: Compare the pressure of the sample with the force ofgravity.
Let P be the pressure of the unknown gas.Let r be the mass density of the fluid in the manometer.
Let h be the height difference of the fluid in the two arms of the manometer.
Let A be the cross sectional area of the arms.The force exerted by the gas on the fluid in one arm of the manometer is PA.
Additional force exerted by the fluid in theevacuated arm of the manometer is Dmg, where Dm is the mass difference of the fluid inthe two arms.
But Dm=rAh, giving a force of rAhg.Therefore PA=rAgh fl P = rgh
Buoyancy
Why does a cork float?
Let h be the height of the cork, and A its cross sectional area. Its volume is V=hA.
The pressure difference is from the top to the bottom of the cork is: DR=rmediumgh.The force difference is:
DF = rmediumghA = rmediumgV.
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Gravitational force on the cork= -mcorkg = -rcorkgV.
Net upward force on the corkF = (rmedium- rcork)gV.
How does a balloon work?
Refer to the accompanying diagram.The radius of the balloon is r.The height above the “equator” is h=r cosqThe pressure difference on an infinitesimal ribbon above and below the equator isDP = rairg(2h)
The vertical force difference on the ribbon isDF = rairg(2h)cosq {(2pr sinq) r dq}
Integrating over the entire surface of the balloon:
= θθθπρπ
dgrair sincos42/
0
23∫
=
gmgVgrdxxgr airairairair ===∫ ρρππρ 31
0
23
3
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Subtracting the force on the gas inside the balloon,DF = (rair -rgas) Vg
Gas Mixtures: Dalton’s Law
Suppose there are n1 moles of substance 1 and n2 moles of substance 2 present.V and T are the same for both substances.
Similarly, for any component i, with mole fraction, xi,
Pi = xi P
Question: A certain gas has a pressure 300 Torr and consists of 1 gram of H2 and 1 gramof He. What are the partial pressures of each component?
θθπθρπ
drgrF air )sin2(cos2 22/
0
2∫=∆
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Question: (Exercise 1.10). The density of air at 740 Torr and 27oC is 1.146g/l.Assuming that air consists only of N2 and O2, what are their mole fractions and partialpressures?
f(v)dv is the probability of observing a speed between v and v+dv.
f(v) is the probability density. What are its dimensions?
Normalization: ∫∞
∞−
= 1)( dvvf
Warning: Be careful to use the correct limits of integration. They are not necessarily≤¶.
Probability of observing a speed between v1 and v2 : ∫=2
1
)(),( 21
v
v
dvvfvvP
Most probable speed: df/dv=0
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Mean speed: ∫∞
∞−
== dvvvfvv )(
RMS speed:
2/1
2 )(
= ∫∞
∞−
dvvfvvrms
One-dimensional Maxwell-Boltzmann distribution
f(v) = C exp{-1/2 mv2/kT} = C exp{-v2/a2}
where ½ ma2 = kT and m is the atomic mass
a º speed sound º 400 m/s
Define the dimensionless variable: x ª v/a
f(x) = C exp(-x2)
Normalization: 12
==∫∞
∞−
− πCdxeC x
f(v) = (1/ap1/2) exp(-v2/a2)Note the limits of integration.
Most probable velocity is zero.
Average velocity is also zero because
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=∫∞
∞−
− dxxe x
Root mean square velocity:
2
1 22
2/22 222 α
πα
παα === −
∞
∞−
∞
∞−
−∫ ∫ dxexdvevv xv
vrms = a/21/2
½ m <v2> = ¼ m a2 = ½ kT = <E>
Lecture 4: Kinetic Theory of Gases, Part II
Three-dimensional distribution
f(vx,vy,vz)dvxdvydvz = f(vx)f(vy)f(vz) dvxdvydvz
= C3 exp{-(vx2 + vy
2 + vz2)/a2} dvxdvydvz
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v2 = vx2 + vy
2 + vz2; x2 = v2/a2
dvxdvydvz = v2 sinq dq df
Take the average over all angles. What remains is
f(x) = C x2 exp(-x2), where 0 § v § ¶
Normalization: ∫∞
− =0
2 4/2
πdxex x
Note the limits of integration.
f(v) = (4/a3p
1/2) v2 exp {-v2/a2}
Note: x2dx = v2dv/a3
Most probable speed:
df/dx = 0 fl xmp=1 fl vmp = a
Average speed:ππ24
0
3 2
∫∞
− == dxexx x
<v> = (2/p1/2)a
Show that ½ m <v2> = 3/2 kT = <E>. This a further example of the EquipartitionTheorem.
Question: What fraction of the molecules have speeds greater than v0?
Answer: dxexdvvfx
x
v∫∫∞
−∞
=0
2
0
24)(
πIntegrate by parts: u=x, dv = x exp(-x2)dx,
du=dx, v = -(1/2) exp(-x2)
Result: )(1222
00 2
0
0
220 xerfedxee
x x
x
xx −+=+ −∞
−−∫ πππ
The error function is defined on page 329.
Question: In two dimensions, the same analysis shows the speed distribution isproportional to v exp(-v2/a2). What is the most probable speed, and what is theprobability of finding a speed greater than this value?
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Kinetic Theory of Collisions
Collision frequency:
Let molecules A and B have radii ra and rb..
Suppose molecule B is stationary, and A moves with an average speed <v>.
Area swept out by A is p(ra+rb)2 = sab.
sab is the collision cross section. In general, it is the effective target area, and may beenergy dependent.
Volume swept out by A in one second is sab<v>.
Number of collisions experienced by one A molecule in one second iszab = sab <v> (nb/V)NA = sab <v>Nb
The total number of collisions between species A and species B per second per unitvolume isZAB = zab (na/V)NA = sab <v> NA
2 (na/V) (nb/V)= sab <v> NA
2 ca cb
The number of moles of A that react per m3 per second is sab <v> NA ca cb
The rate constant is k(T) = sab <v> NA = 2.4x108 m3/mole/sec = 2.4x1014cm3/mole/sec= 4x10-10 cm3/molecule/sec.
This is the rate constant for elastic collisions. Chemical reactions are generally muchslower.
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Proper average over scattering angles introduces a factor of 2 .
Proper average over speeds gives the rate constant ∫= dvvvfvTk )()()( σ .
Mean Free Path
The time between collisions is 1/zab. The distance a molecule travels between collisionsis
l=<v>/zab = (21/2sabNb)
-1
In liquid water, assume sº0.5x10-18 m2
c = 1/18 moles/cm3 = 3.35x1028 molecules/m3
l=0.42x10-10 m = 0.42 C
For 1 Torr N2, 298 K, s=0.42 nm2,c=5.4x10-5 moles/liter, N = 3.24x1022 molecules/m3, l =5.2x10-5 m = 5.2x105
C
Rough rule of thumb:l = 1000 Angstroms at 1000 Torr.
The previous result is within a factor of two of the rule.
How low must the pressure of N2 be to have l = 1 m?N=1.7x1022 molecules/m3
fl P = 5x10-5 Torr.
Effusion
The leak rate through a pinhole is given by the flux of molecules through the hole. Thevolume swept out in one second is A<v>. The number of molecules in that volume is N<v> A. The number of molecules hitting the area A is approximately
Note: volume of the paralellepiped of molecules striking the surface per second is v cosqA.
22 /
0
2/
0
2
0
22/33 cossin απ
θ
π
ϕ
θϕθθπα v
v
evdddvvn −∞
= = =
−∫ ∫ ∫=Γ
Note: Integral over angles is p.
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G= ∫∞
−− =0
32/1 4/2
vndxexn xαπ
Result: rate of effusion = N<v>/4
Number of molecules passing through a hole of area A in one second is N<v>A/4.
Suppose the pressure is 10-5 Torr. How long does it take to form a monolayer coating ofN2 at 298 K, assuming s=0.4 nm2?
Number of molecules per m2 is 1/s = 2.5x1018
N=3.24x1017 m3, <v>=475m/s
3.24x1017 t x475/4 = 2.5x1018
t = 0.065 sec
See the movie Men of Honor.
Question:
A space station has a volume of 1,000 m3. It is filled with 1 atm of N2 at 298 K.Suddenly a hole with an area of 1cm2 is created by a meteor. What is the pressure in thestation after 103 seconds?104 seconds? 105 seconds? 106 seconds?
Homework Problem: A vacuum chamber has a volume of 1 m3 and initially has a perfectvacuum inside. Suddenly a leak develops. The leak turns out to be a pinhole of radius0.1 mm. The chamber is sitting in a lab at one atm external pressure and 298 K. Assumethe lab air is pure nitrogen. How long will it take for the pressure to rise to 1 mTorrinside of the chamber?
Answer: The number of molecules, N, hitting area A in time t is given by N =¼NvAt,Where N=PVNA/RT (at 10-4 and 298 K) = 3.23x1018, N= PNA/RT (at 1 atm and 298 K)= 2.46x1025 m-3, v=475 m/s, A=3.1416x10-8 m2. Solving for t gives t=0.035 seconds.
Lecture 5. Real Gases
The ideal gas law,
Z = PV/nRT = PVm/RT = 1,
is based on the assumption of non-interacting particles.The quantity Vm=V/n is the molar volume.
Real atoms and molecules have an intermolecular potential that is responsible forchemical bonds and van der Waals clusters. The long-range potential between neutral
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molecules varies as -R-6. Short range interaction is always repulsive. Example of achemical bond is Na2:Re=0.308 nm, De = 16.6 kcal/mol, D0=16.4 kcal/mol.Extreme case of a weak van der Waals cluster is He2,Re=0.75 nm, De = 0.1 kcal/mol, D0=2.6 x 10-6 kcal/mol.
Result is a non-ideal equation of state,
Z = 1 + B£ P + C£ P2 + …
Z= 1 + B/Vm + C/Vm2 +…
The “virial coefficients, B£, C£ , … B, C,… can be related to the intermolecular potential.
It is also useful to work with empirical equations of state, just as it is useful to work withempirical potential energy functions. The most famous one is the van der Waals equationof state:
P = nRT/(V - nb) - a(n/V)2
P = RT/(Vm - b) - a/Vm2
The first term contains a correction for repulsive forces. The second term is caused byattractive forces.
Isotherms are plots of P vs Vm for constant T. See Fig. 1.23.
Ideal gas isotherms are hyperbolas, withmV
P
∂∂
< 0 always.
The non-ideal isotherm has regions of
mV
P
∂∂
> 0, which correspond to a phase transition.
At low temperatures the isotherm has a minimum and a maximum. As T increases, themin and max get closer and finally merge at a point of inflection, called the critical point.
This point is defined by the conditions 0=
∂∂
TmV
Pand .0
2
2
=
∂∂
TmV
P
Solving these equations for the van der Waals equation gives
Vm,c = 3bRTc = 8a/27bPc = a/27b3
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Define the dimensionless (reduced) variables:
Pr = P/Pc
Vr = Vm,c/Vc
Tr = T/Tc
Substituting back into the van der Waals equation gives a universal curve:
Pr = 8Tr/(3Vr - 1) - 3/Vr2.
The fact that it is universal is called the law of corresponding states. This implies onlythat the real equation if state can be well described by just two parameters.
We also find that
Zc = PcVc/RTc = 3/8.
Expanding the van der Waals equation of state as a virial series:
P = RT/(Vm - b) - a/Vm2
Z = PVm/RT = Vm/(Vm - b) - a/(RTVm)
= 1/(1- b/Vm) - a/(RTVm) + …
@ 1 + b/Vm - a/(RTVm) + b2/Vm2
.../
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2
++−+=mm V
b
V
RTabZ
Lecture 6: The First Law
1. The internal energy of an isolated system is constant. This is equivalent to thestatement that U is a state variable.
Because U is a state variable, the change in U caused by a change of state does notdepend on the path.
We define w>0 as the work done on the system. It follows that w<0 corresponds to workby the system on its surroundings. Similarly, q>0 is for an endothermic process and q<0is for an exothermic process. If we consider w and q as the only sources of energychange, then
DU = Uf - Ui = w + q
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regardless of the path-dependent values of w and q individually.
Types of paths:
∏ Insulating walls fl q=0This is called an adiabatic path, and the work done is adiabatic work, wad.
w=wad
DU=wad
∏No work done: w=0DU=q
∏ General case: DU= w + q
Important concept: We are free to choose any path we wish connecting a given initialand final state in order to calculate DU. In particular, we may choose an adiabatic path inorder to define energy and heat.
Mechanical definition of energy change: DU=wad
Mechanical definition of heat: q = wad - w
Mechanical equivalent of heat: w=Jq
What is the value of J? In 1849, Joule came up with a value of 1 cal at 15±C = 4.15 J.The value today is 4.1840.
Example of the swimming pool: A pool may be filled either by rain water or by faucetwater. Suppose we wish to measure the amount of rain water, but have no direct way ofdoing so. (It is hard to put a rain gauge on the clouds!) Instead, we use the followingstrategy. First, we cover the pool with a piece of plastic, and fill the pool with faucetwater. We carefully measure the amount of water with a gauge attached to the tap. Next,we drain the pool, remove the sheet, and allow rain water to enter the pool for a period oftime. Finally, cover the pool with plastic and fill the rest of the pool with faucet water,again using the gauge on the tap. The difference between two measurements equals to theamount of rain water that fell into the pool.
In this analogy, the plastic sheet is equivalent to theinsulating walls, the first gauge measurement is wad, the second measurement is w, andthe rain water is q. The definition of heat is therefore wad - w.
2. The work done on an adiabatic system tochange its state from a specified initial state to a specified final state is independent of thetype of work done.
Other types of paths:
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∏ Isothermal path: Tf = Ti
∏ Isochoric path: Vf = Vi
∏ Isobaric path: Pf = Pi
∏ Cyclic path: Initial and final states are the same,
DU=0
3. A system that undergoes cyclic behavior cannot deliver any work without some otherchange occurring to its surroundings. In other words, a perpetual motion machine “of thefirst kind” cannot be built. Example: water running down a hill, turning a turbine, andthen running back up hill.
One of the most useful concepts is that of a∏ reversible path: For such a path, the system is in thermal and mechanical equilibriumthroughout.
Lecture 7: Applications of the First Law
Expansion of a gas:
dw = Fdz = - Fexdz = - PexAdz = - PexdV
Consider the special case of a free expansion(or sudden compression): Pex is constant.
VPVVPdVPw exifex
V
V
ex
fin
in
∆−=−−=−= ∫ )(
We cannot say anything about DU for this process unless additional information is given(i.e., unless additional constraints are imposed).
Suppose V=Vf is fixed. Then PfVf= nRTf, but Pf and Tf are not determined. We haveinsufficient information to calculate q and DU.
Suppose Pf=Pex is fixed. Then PfVf= nRTf, but Vf and Tf are not determined.
What type of additional information do we need?
1) If the process is isothermal, Tf = Ti. Then knowing either Pf or Vf is sufficient todetermine the final state.
2) If the process is adiabatic, q=0, DU = w, and Uf = Ui + w.
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Isothermal Processes
Ti, Vi, Pi fl Tf, Vf, Pf
Ti = Tf
1) Sudden isothermal expansion: Pex = Pf < Pi
)( iff
V
V
f
V
V
ex VVPdVPdVPwf
i
f
i
−−=−=−= ∫∫
But PiVi = PfVf or Vf = Pi Vi /Pf
w = -Vi (Pi - Pf)
q = Vi (Pi - Pf)
Numerical example:
Pi = 2 atm, Pf = 1 atm, Vi = 0.1 liter
w = -0.1 liter atm = -0.1[10-3 m3 x 101,325 pa]=-10.13 J
2) Reversible isothermal expansion
f
iii
i
fV
V
V
V P
PVP
V
VnRT
V
dVnRTdVPw
f
i
f
i
lnln −=−=−=−= ∫∫
= -0.2 ln 2 liter atm =-14.04 J
Look at the indicator diagrams.
Conclusion: More work is extracted in the reversibleprocess.
3) Sudden compression: Pex = Pf > Pi
Numerical example:
Pi = 1 atm, Pf = 2 atm, Vi = 0.2 liter
w = 0.2 liter atm = 20.27 J
4) Reversible compression:
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w = = -0.2 ln 2 liter atm =-14.04 J
Look at the indicator diagrams.
Conclusion: More work is required in the irreversible process.
Claim: Reversible isothermal expansions always deliver more work than irreversibleones.
Path A: 1 Ø 2 at constant pressurePath B: 2 Ø 3 at constant volumePath C: 3 Ø 1 at constant temperature
Calculate w, q, and DU along each reversible path and for the entire cycle. Assume thegas consists of atoms.
DU = (3/2)nRDT = 12.47 DT Joules
DUA = 3.74 kJ
DUB = -3.74 kJ
DUC = 0
wA = -P1 DV = -24.62 liter atm = -2.495 kJwB = 0wc =-nRT ln(V2/V1) = 1.729 kJwcycle= -0.766 kJ
q = DU - w for each step.
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Lecture 9: Enthalpy and Heat Capacity
Suppose the heat is not supplied at constant volume. Typically experiments areperformed in the open air, i.e., at constant pressure. But in that case, the system mustalso undergo PV work, and more heat is required to get the same increase in T. This heatis called the enthalpy, H.
Definition: H = U + PV
Clearly, H is a state variable. What other properties does it have?
dH = dU + PdV + VdP
= dq + dw + PdV + VdP
= dq + VdP
At constant pressure, dP = 0 and
dH = dqp
DH = qp
For an ideal gas, PV = nRT fl H = U + nRTFor constant n, DH = DU + nRDT
For constant T, DH = DU + RTDn
For constant n and T, DH = DU
Note: DH is not equal to DU + PDV +VDP.
Heat capacity at constant pressure:
PP T
HC
∂∂=
dTCHT
T
P∫=∆2
1
How are CP and CV related?
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For an ideal gas: H = U + nRT
CP =PP T
U
T
H
∂∂=
∂∂
+ nR
CP = CV + nR, CP,m = CV,m + R
Empirical result: CP,m = a + bT + c/T2
DH = a(T2 -T1) + ½ b(T22- T1
2) - c(T2-1- T1
-1)
Example: Heating of I2 at constant pressure
Phase a b csolid 40.12 0.04979 0liquid 80.33 0 0vapor 37.40 0.00059 -0.71e+5
Melting point = 386.8 KBoiling point = 458.4 K
Enthalpy of fusion = 15.52 kJ/molEnthalpy of vaporization = 41.80 kJ/mol
Calculate the heat necessary to raise one mole of I2 from T1 to T2 at 1 atm.
a) T1 = 100 K, T2 = 200 K
qP = 40.12 (200-100) + ½ x 0.04979 (2002-1002) = 4.76 kJ
b) T1 = 100 K, T2 = 400 K
qP = 40.12 (386.8-100) + ½ x 0.04979 (386.82-1002) + 15,520 + 80.33(400-386.8) =
31.56 kJ
c) T1 = 100 K, T2 = 500 KqP = 40.12 (386.8-100) + ½ x 0.04979 (386.82
-1002) + 15,520 +80.33(458.4-386.8) + 41,800+ 37.40(500-458.4) + ½ x 0.00059(5002
Hess’ Law: The standard enthalpy of a reactionequals the sum of the standard enthalpiesof individual reactions into which the overall reaction can be divided.
