Lecture 1 - Earth and Atmospheric Sciences

Equilibrium &
Lecture 1
Plan of the Course
• The geochemical toolbox
Aquatic systems
Trace elements & magmatic systems
• Radiogenic
• Stable
• The Big Picture: Cosmochemistry
o Formation of the elements
o Formation of the Earth and the Solar System
• Chemistry of the Earth
• Thermodynamics is the study of energy & its
o Chemical changes involve energy; by “following the energy” we can
predict the ‘equilibrium state of a system and therefore the outcome of
• For example, we can predict the minerals that will crystallize from a
cooling magma.
• We can predict that as the concentration of atmospheric CO2
increases, ocean pH will decrease and calcium carbonate skeletons
of coral will become more soluble.
• Thermodynamics uses a macroscopic approach.
o We can use it without knowledge of atoms or molecules.
o We will occasionally consider the microscopic viewpoint using statistical
mechanics when our understanding can be enhanced by doing so.
Thermodynamics and
• The equilibrium state of a system is independent of
any previous state. So, for example, if we do a
partial melting experiment with rock, it should not
matter is we start with a solid and partially melt it or
with a melt and partially crystallize it.
• Kinetics is the study of rates and mechanisms of
reaction. Kinetics concerns itself with the pathway
to equilibrium; thermodynamics does not. Very
often, equilibrium in the Earth is not achieved, or
achieved only very slowly, which naturally limits the
usefulness of thermodynamics.
‘The System’
• A thermodynamic
system is the part of the
universe we are
considering. Everything
else is referred to as the
o We are free to define the
‘system’ anyway we chose.
However, how we define it
may determine whether we
can successfully apply
• Four kinds of ‘systems’
• The equilibrium state is
the one toward which
a system will change in
the absence of
• It is time invariant on
the macroscopic scale,
but not necessarily on
the microscopic one.
Equilibium &
• Conundrum: strictly speaking, we can apply
thermodynamics only to the equilibrium state. If a
reaction is proceeding, then the system is out of
equilibrium and thermodynamic analysis cannot be
• Solution: we imagine reversible processes in which
systems are only infinitesimally out of equilibrium. In
contrast, natural processes can proceed only in
one direction and are irreversible.
• We might also imagine local equilibrium where
even if the whole system (e.g. the ocean, magma
and crystals) is out of equilibrium, we can imagine a
part of it is (rim of the crystal).
Fundamental Variables
• Pressure: P
o Force/unit area
• Volume; V
• Work: W
o we are mainly concerned with P-V work: pressure integrated over volume
o Work done by a system is negative
Heat: Q
Energy: U
Temperature: T
Entropy: S – more on that in next lecture
State Variables and
Equations of State
• State variables depend only on the state of the system,
not on the path taken. Not all the variables listed above
are state variables.
• Extensive vs. Intensive variables. We can convert
extensive variables to intensive ones.
• Equations of state express the relationship between state
variables, e.g.
• Tells us, for a given number of moles, now temperature,
volume, and pressure of an ideal gas relate to each
other; i.e., if we heat the gas, what happens to volume
and pressure?
• Van der Waals Equation and Ideality
General Equation of State
• Obviously, there is no one solution to the ideal gas
law, but we can imagine a couple of special cases.
• Isothermal compressibility (β): change in volume
with change in pressure at constant temperature
per unit volume: 1/V(∂V/∂P)T
• Coefficient of thermal expansion (α): change in
volume with change in temperature at constant
pressure per unit volume 1/V(∂V/∂T)P.
• We can write a general equation relating V, T & P:
dV = ∂V/∂T)PdT + (∂V/∂P)TdP
dV = VαdT + VβdP
• Temperature is a measure
of the average kinetic
energy of a system.
• How do we measure it?
o Volume of an ideal gas
o We can arbitrarily define a scale
such that
o V = V0(1+γτ)
o If so, we might have negative τ.
o But note V cannot be negative,
so there must be an absolute
minimum to temperature.
o The absolute minimum of T
occurs where the volume of an
ideal gas is 0.
• Use the absolute scale (K)
in all thermodynamic
Zeroth Law
• Two bodies in thermal equilibrium have the same
• Two bodies in equilibrium with a third body are in
equilibrium with each other.
The First Law
• Energy is conserved in all transformations
• Heat and work are equivalent (the sum of the two is
always the same).
• Change in energy of a system is independent of the
path taken.
• Mathematically: ΔU = Q +W
• Energy change is path independent and thus is a state
variable. Heat and work are not.
• State variables have exact differentials (heat and work
do not).
• (we are mainly concerned with energy changes, not
absolute amounts).
• SI units of energy are Joules (kg-m2/s2). Calorie is another
metric unit of energy, but not SI.
State Functions & Path
• State functions are path independent and have exact
• Think about an internal combustion engine. Chemical
energy is released by burning gas. Some of that energy
goes into heat and some to work. There is no fixed rule
about how much goes to each – it depends on your
engine design (engineers work to increase the amount
going to work).
• Therefore, heat and work cannot be state functions.
• However, no matter how you design the engine, the sum
of heat and work for a given amount of (fully)
combusted gasoline is the same.
• Energy is path independent and a state function.
State Functions & Exact
• State functions have exact differentials.
o (These are not new, they are the kind you have learned about in calculus).
o This means we can obtain (in principle anyway) an exact solution if we
differentiate them (or integrate the differentials).
• Exact differentials have the property that the cross
differentials are equal (in other words, if we differentiate
by two separate variables, the order doesn’t matter).
o Again, this is what you learn in calculus.
• Consider
dV = (∂V/∂T)PdT + (∂V/∂P)TdP
• If V is a state function, then (∂V2/∂T∂P)= (∂V2/∂P∂T)
• This is not true of non-state functions like work and heat.
• Work is not a state function
• dW = -PdV
• For an ideal gas:

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