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20B notes: Prof. Daniel Neuhauser
Partially based on notes by Prof. Peter Felker.
Parts written by Christopher Arntsen and Arunima Coomar.
IMPORTANT: ABOUT 20% of these notes are “INSERTS”, derivations which will not be covered in class but you’ll be expected to study and know everything in the notes, including the inserts, for the exam.
Course overview:
This is a course about how molecules behave in aggregate – interact, form solutions and phases, and react with useful results (heat and electricity). Along the way we’ll learn a lot of chemistry and physics, and at times gloss over some proofs, leaving the details to Chem. 110A (thermodynamics)
Brief outline:
- Gases
- Interacting systems (liquids, solids, phases)
- Phases
- Solutions
- Thermodynamics: The 1st^ law
- Thermodynamics: 2nd^ law: disorder increases in world.
- Thermodynamics: Free energy – minimum in reactions.
- From free energy to mass reaction law, chemical equilibrium
- Mass reaction law and LeChatelier’s principle
- Acids and bases; mass reaction law in acids and bases; titrations.
- Electrochemistry
- Kinetics
- (Time permitting:) Raul’s law, physical properties of solutions, colligative properties.
Part 1: Units and Ideal-Gases (Gas-law, Kinetics):
- Units, Avogadro’s number.
- PV=nRT – experiments, temperature scale.
- PV=nRT – theoretical proof (kinetic).
- Maxwell’s velocity distribution
- Boltzmann factor – application to vibrations.
PRELUDE (NOT COVERED IN CLASS): Review from 20A
Units, Avogadro’s numbers and moles
in practice it is difficult to deal with big # of molecules, so we measure it in terms’ of moles. A “mol” of things means it contains
Avogadro’s’ #: NAvogadro = 6.02*10^23
[We call NAvogadro interchangeably NA ]
(Note that the concept of “mol” is analogous to “dozen”; dozen molecules are 12 molecules, and a mole of molecules is 6.02*10^23 molecules).
NA was adjusted to be almost exactly the number of hydrogen atoms in one gram of hydrogen atoms. However, we have to be careful when we say that, since hydrogen is made of separate individual H atoms only at very high temperatures (above, say, 2000-3000 degrees). At room temperature, hydrogen is a diatomic molecule H 2 ; each H 2 molecule weighs twice as much as one individual hydrogen atom, so at room temperature a mole of hydrogen molecules (H 2 ) will weigh two grams. (A dozen hydrogen molecules weigh twice as much as a dozen hydrogen atoms, so a mole of hydrogen molecules weighs twice as much as a mole of hydrogen atoms).
The mass of a hydrogen atom is almost exactly
Mass(H atom)~ 1gram/ NA = 1gram/ (6.0210^23 ) = 1.6610-24^ gram
(I won’t prove this equation; you learned it or will learn in the physics series).
A side note on energy units.
Note: energy = (mass)* (speed)^2 = mass distance^2 / time^2 so energy is measured in units of (mass distance^2 / time^2 )
Energy units reminder: two sets (one called CGS and the other MKS) are commonly used; MKS is more prevalent. The energy units get special name in each unit system:
1 erg = 1 g cm^2 / sec^2 (CGS units)
1 J = 1kg m^2 / sec (MKS units)
End of side note
Say that we have “N” atoms (N will be of course NAvog if we have a mole of atoms, but we take the general case). Let’s consider the “j'th atom, i.e., we consider either the 1st^ atom (in which case j=1), or the 2nd^ (then j=2), or the 3rd, etc.
The K.E. of the “j’th atom will be therefore K.E.(jth^ atom)= mj vj^2 /
So the total kinetic energy will be the sum over all “j” atoms where j will range from “1” (the first atom) to “N” (the Nth atom) will be:
Kinetic energy for N atom: K.E. = ½ ∑^ 〷⢀⡩,…,〕ᡥ〷ᡴ〷⡰
(The strange symbol, Σ^ , means “sum”, it is the capitalized form of the Greek letter sigma; we’ll encounter more Greek letters below).
The total energy, E, has, in addition to the kinetic energy, a 2nd^ part called potential energy, defined as:
Potential energy: energy that can turn to K.E.
E.g., when we raise an iron bar up, we increase its potential energy; as we let it go, it will fall, and while it falls it acquires kinetic energy and its potential energy will decrease, its total energy will be conserved (will actually slightly decrease, due to friction, i.e., transfer of energy from the iron bar to the air molecules ).
Example for Potential energy and Kinetic Energy—a mass falling;
Densities
There are several types of densities we will deal with. Densities in general are defined as something/volume.
The most common type of densities are mass-density and molar density.
Mass density is defined as :
Mass density = Mass/Volume, and has units of g/cm^3 or g/L or kg/L.
Molar density is:
Molar density = # moles/Volume = n/V ; So for example the molar density of water =1mol/(18 cm⡱)
K.E.=
Pot.E.=10 J
Total E=10J
K.E.=3J
Pot.E.=7 J
Total E=10J
Next: pressure (to be covered in class):
P=Pressure = Force/Area , and labeling the force as “F” and the area as “A”,
P =
F
A
P is measured in its own unit in MKS, which has been given a special name:
Pascal =
Newton m⡰
Multiply by m (meter) both numerator and denom.:
Pascal =
Newton ∗ m m⡱^ =^
J
m⡱ (i.e., PRESSURE = ENERGY/VOLUME!)
However, 1 Newton is really a weak force in daily life (it is about the force exerted by a quarter-pound object – e.g., a small pear – on your hand when you hold it so it does not fall) , and a meter squared is a large area, so 1 Pascal is actually a very weak pressure.
Ambient atmospheric pressure, i.e., pressure near sea level, is much higher than 1 Pascal
1 atmosphere ~ 1.02 * 10^5 Pascal.
One atmosphere is denoted as P^0
It is a coincidence of nature that 1 Atmosphere is so close to a power of ten (here 100,000) times Pascal; so to use this fact, scientists introduced another pressure, called a bar, defined as
1bar = exactly 10^5 Pascal
So
P^0 = 1.02 bar~1bar.
A secret: since we are a little higher than sea level, the pressure at UCLA is actually lower than P^0 and is closer to a bar (and fluctuates daily with weather). So for me, in this course, we’ll just approximate that ᡂ⡨^ " = " 1ᡔᡓᡰ
and we'll ignore the 1atm vs. 1bar difference (don’t do it in other courses or labs or you will lose points there…)
Useful: Note that
1bar = 100,000 J/m^3 , and m^3 =1000L (L means Liter),
1bar = 100,000J/(1000L)
1bar = 100 J/L
Useful later.
So: mathematically we write these observations as:
PV=nfunction(T)
However, so far this “function(T)” could depend on the material; the next experiment proves it does not depend on the material, i.e., “function(T)” has universal characteristics!
2 nd^ experimental observation: for same T, and P, the ratios of the volumes between 2 different gases = ratio of the numbers of moles;
E.g., consider making hydrogen and oxygen gases
H⡰O䙦l䙧 → H⡰䙦g䙧 ㎗ ½ O⡰䙦g䙧
The outcome of this reaction is twice as many hydrogen moles as oxygen moles
nH2 = 2 nO
V=50L V=25L^ P=2ba (4mol) P=2ba, (2mol)
Example for PV=n*const. at fixed T for ideal gases: Double volume and n while P and T fixed.
The observation is that if we collect the hydrogen in one container and the oxygen in the other, and both containers have the same P and T, then
VH2 = 2VO2 so: ᡂᡈ〉⡰ ᡦ〉⡰^ =
And we’ll find this for any reaction, i.e., ‘function(T)’ above is universal, i.e.,
ᡂᡈ = ᡦᡘ䙦ᡆ䙧
where ᡘ䙦ᡆ䙧 is a function of temperature.
Note on raising volumes without reactions: The gas laws tells us that if we were to increase the temperature at a fixed pressure, then two gases that have initially the same volume will continue having the same volume.
(Example:, say we have a gas of hydrogen and a gas of bromine, all in 1 bar, and that initially, when each is at the same initial temperature T each occupies 100L then if at another temperature, T’, one gas occupies, say, 160L , the other gas will also occupy 160L at that temperature).
Next Defining the temperature in Kelvin, based on the Gas Law.
So far we avoided the question of what temperature scale to actually use. An example: we intuitively “know” that the temperature of boiling water is higher than the temperature of ice-water; but we can label the two as being 100 and 0 (as in the Celsius scale) or 212 and 32 (as in the Fahrenheit scale), or other arbitrary designations.
Clearly, it will be best to use a temperature definition that is based on a physical law, rather than being completely arbitrary. The simplest way for that is to use directly the gas law, i.e., we define:
R T= PV/n
R: constant used to match T to more historical definitions.
1bar ∗ 50.00L 1.61 mol ∗ 373.16 K
0.0831 bar ∗ L mol K
But recall that we learned that
1bar * L = 100 J (where J=Joule),
So
R = 8.
J
K mol
You don’t need to memorize this number (I’ll give it to you in the exams), but with time you’ll remember it naturally.
Room temperature, and R T(room):
Officially people define, for reference, room temperature as
T(room) = 25Celisus = 298 K ;
I’ll round in this course T(room)~300K
Note: RT(room) = 8.3 J/(K mol) * 300K ~2500J/mol
WORD OF CAUTION ON THE IDEAL GAS LAW,
PV=nRT is only valid for “ideal” gases, i.e., ones in which the distances are so large that the molecules barely meet each other once in a while. At high pressures (e.g., bigger than say, 30bar and definitely for more than100 bar) we have to start worrying about corrections to the ideal gas law.
But for most gases the ideal gas law is extremely accurate at room pressures.
We’ll study shortly deviations from ideal gas laws and how they teach us about the forces that hold the atoms together in solids and liquids.
IMPORTANT: never apply the ideal gas law to liquids and solids!
Now to the pressure calculation: we imagine a container with unit cross sectional area in the left wall (e.g., if we use cgs units, the area will be A=1 cm^2 , etc.); the force on that area will be the pressure.
ᡂ = ᠲ䙦ᡧᡦ ᡳᡦᡡᡲ ᡓᡰᡗᡓ䙧
But the force exerted by the molecules on the left walls the same as the force exerted on the molecules; the latter is, as mentioned, simply the rate of change in momentum of the molecules that hit the wall,
ᡂ = ᠲᡧᡰᡕᡗ䙦ᡳᡦᡡᡲ ᡓᡰᡗᡓ䙧 = ᡕℎᡓᡦᡙᡗ^ ᡡᡦ^ ᡥᡧᡥᡗᡦᡲᡳᡥ
where “ " is a short time; and since the momentum equals mass*velocity, we can write the pressure as
ᡂ = 䙦ᡥᡓᡱᡱ^ ᡧᡘ^ ᡨᡓᡰᡲᡡᡕᡤᡗᡱ^ ᡲℎᡓᡲ^ ℎᡡᡲ^ ᡵᡓᡤᡤ^ ᡡᡦ^ ᡲᡡᡥᡗ^ 䙧^ ∗^ 䙦ᡕℎᡓᡦᡙᡗ^ ᡡᡦ^ ᡴᡗᡤᡧᡕᡡᡲᡷ^ ᡵℎᡗᡦ^ ᡥᡧᡤᡗᡕ. ℎᡡᡲ^ ᡵᡓᡤᡤ^ 䙧
Let’s then calculate the term.
The velocity will change because when a particle hits the left wall it will bounce back, so its velocity change will be (see figure)
ᡕℎᡓᡦᡙᡗ ᡡᡦ ᡴᡗᡧᡤᡕᡡᡲᡷ = 2ᡳ~ᡳ
where we ignored a factor of 2, as promised.
So we are only left with the need to calculate what’s the mass that hits the wall in a short time.
initially: velocity = - u
Finally: u
WALL
Now we know that in a time , only particles that are within a distance ᡳ from the wall can hit the wall; particles that are further way simply wont make it in time. (See figure)
Figure: Only molecules that are within a distance ∃≔ from the wall will hit it in time ≔ (molecules in black, that start to the left of the imaginary “red line” at a distance of ∃≔ from the wall); molecules which are further away (blue in figure) will not hit the wall in time ≔.
Therefore, the volume from which particles that hit the wall come is its cross- section area (1) times its length (u 䙧, i.e., it will be ᡳ
Therefore, the mass of particles (within a unit area) that hit the wall will be the mass density times the volume from which the particles come.
i.e.,
䙦ᡥᡓᡱᡱ ᡧᡘ ᡨᡓᡰᡲᡡᡕᡤᡗᡱ ᡲℎᡓᡲ ℎᡡᡲ ᡲℎᡗ ᡳᡦᡡᡲ − ᡓᡰᡗᡓ ᡵᡓᡤᡤ ᡡᡦ ᡲᡡᡥᡗ 䙧
= ᡥᡓᡱᡱ ᡖᡗᡦᡱᡡᡲᡷ ∗ 䙦ᡴᡧᡤᡳᡥᡗ ᡘᡰᡧᡥ ᡵℎᡡᡕℎ ᡲℎᡗ ᡨᡓᡰᡲᡡᡕᡤᡗᡱ ᡲℎᡓᡲ ℎᡡᡲ ᡡᡦ ᡓᡰᡰᡡᡴᡗ䙧
=
ᡈ^ ∗ ᡳ
So collecting it all we get, using the eq. we derived:
Wall, (^) A=
distance=u≔
3 m
RT ≡ E
Then by plugging the last equation into the one before it, we get the ideal gas law,
PVm=RT!
(Note that as a bonus, we get from the definition of RT the “result” – really just a definition – Em= 3 RT/2 , which we will need later)
Note: this derivation makes it clear that T cannot be negative, since it is proportional to kinetic energy, which is always positive.
ALSO Note: this is a mathematically involved derivation, but I'll expect you to be able to redeliver it!
Finally, note that Em is NOT THE TOTAL ENERGY, but is instead the kinetic energy associated with center of mass motion. For atoms that are far from each other, there is no other energy, so Em=3RT/2 is the energy of atomic gases (e.g., He, Ar, , etc.); for molecules, we have to supplement this energy by intermolecular kinetic and potential energy, so
Molar Energy(molecules)>Em=3RT/2.
Maxwell’s Velocity Distribution:
The derivation above was approximate as it assumed that all molecules have the same speed in the x direction (with ½ going left, ½ right). Typically such approximate derivations give results that are accurate within an order of magnitude, i.e., have mistakes of order of 0.3 - 3.0. However, in this case we were lucky and even the final factors were correct (so PV=nRT is correct exactly for an ideal gas).
One issue we ignored in our derivation is that molecules do not have a fixed speed; if we want to really know what molecules behave like, they have distribution of speeds.
Let’s use “u” to denote the total speed. This is different from the previous section, where u denoted the speed along x; here it denotes the total speed.
Further, u will be variable, so different molecules (or even the same molecule at different times) will have a different u.
Since u is a continuous variable, we define the fraction of particles within a range between u and u+du, i.e., with speeds around u but within a range du,
F䙦ᡳ, ᡳ ㎗ ᡖᡳ䙧 = FRACTION OF ᡥᡧᡤᡗᡕ. ᡵᡡᡲℎ ᡳ < ᡱᡨᡗᡗᡖ < ᡳ ㎗ ᡖᡳ
For example, a gas can have a temperature where, say, 0.03% of its molecules have speeds between 50 and 50.1 m/s; in that case,
F(50 m/s, 50.1 m/s ) = 0.03% = 0.
Next, note that if du is small, F(u,u+du) is proportional to du:
F䙦ᡳ, ᡳ ㎗ ᡖᡳ䙧 ∝ ᡖᡳ
(where the symbol ∝^ means: proportional to).
E.g., if u is, say, 50m/s, then the number of particles that have speeds between say 50 and 50.2 m/s will be about twice as large as the number of particles between 50 and 50.1 m/s
