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Week 1 Wednesday September 26, 2012 page 1
Thermodynamic properties: composition, internal energy, volume, pressure, temperature, entropy, etc
Properties:
1. Extensive
a. Depends on size
b. Larger system means larger value
c. Volume is an example
2. Intensive
a. Density is an example
b. More important than extensive
c. Pressure and temperature are examples
Always think about whether a quantity is intensive or extensive.
Systems:
1. Homogeneous
a. Uniform, 1 phase
b. Refers to mixture (more than one substance)
2. Heterogeneous
a. More than one phase (like oil and water)
State: We need at least two intensive properties to define the state of a pure substance.
Temperature:


Most important intensive quantity
Can’t measure directly (thermometer shows volume, not temperature)
We can skip section 1.3
Gas Laws
Equation of state for ideal gas (perfect gas): PV=nRT
𝐿 𝑎𝑡𝑚
𝑐𝑎𝑙
𝐽
𝑐𝑚3 𝑎𝑡𝑚
𝑅 = .08206
= 1.987
= 8.314
= 82.06
𝑚𝑜𝑙 𝐾
𝑚𝑜𝑙 𝐾
𝑚𝑜𝑙 𝐾
𝑚𝑜𝑙 𝐾
So L atm is a unit of energy.
R is universal gas constant.
PV=nRT is true at low pressure and high temperature only.

Ignores intermolecular forces since molecules are so far apart


Assumes molecules take up no space
Have to memorize
Van der Waals Equation (chapter 8)
𝑃=
𝑛𝑅𝑇
𝑎𝑛2
− 2
𝑉 − 𝑛𝑏 𝑉
a and b are both gas specific constants.
The pressure correction (a term) is subtracted because intermolecular forces are always attractions,
which reduces the pressure on the wall.
The b term is due to the size of the molecules.
(𝑃 +
𝑎𝑛2
) (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇
𝑉2
We don’t have to memorize the Van der Waals equation, but it’s important.
Symbols can have more than one use or meaning to pay attention to context.
We don’t have to memorize every equation.
PVT behavior in condense phases (solid and liquid)
For any substance, V=f(T,P) assuming n is constant.
dV is change in volume
d is differential (infinitesimal change)
𝑑𝑉 = (
𝜕𝑉
𝜕𝑉
) 𝑑𝑇 + ( ) 𝑑𝑃
𝜕𝑇 P
𝜕𝑃 𝑇
dV is a total differential
The P in the first term means constant pressure.
The T in the second term means constant temperature.
Divide that equation by V.
𝑑𝑉 1 𝜕𝑉
1 𝜕𝑉
= ( ) 𝑑𝑇 + ( ) 𝑑𝑃
𝑉
𝑉 𝜕𝑇 𝑃
𝑉 𝜕𝑃 𝑇
The equation is now intensive.
The ratio of two extensive quantities is intensive.
𝜌=
𝑚
𝑣
m is extensive, v is extensive, 𝜌 is intensive
1 𝜕𝑉
𝛼 = 𝑉 (𝜕𝑇 )
𝑃
alpha
1 𝜕𝑉
−𝐾 = 𝑉 (𝜕𝑃)
kappa
𝑇
𝛼 is the coefficient of thermal expansion or thermal expansivity, or cubic expansion coefficient.
K is isothermal compressibility.
Isothermal means constant temperature.
𝐾=
−1 𝜕𝑉
( )
𝑉 𝜕𝑃 𝑇
Constant temperature
1 𝜕𝑉
𝛼 = ( ) constant pressure
𝑉 𝜕𝑇 𝑃
𝑉𝑚 =
𝐾=
𝑉
= 𝑚𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
𝑛
−1 𝜕𝑉𝑚
1 𝜕𝑉𝑚
(
) 𝑎𝑛𝑑 𝛼 =
(
)
𝑉𝑚 𝜕𝑃 𝑇
𝑉𝑚 𝜕𝑇 𝑃
𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑣𝑒
= 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑣𝑒
𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑣𝑒
𝛼, 𝐾 are both intensive.
𝛼:
10-5 to 10-4 K-1 for solids
10-3.5 to 10-3 K-1 for liquids
10-3 to 10-2 K-1 for gasses
K:
10-6 to 10-5 atm-1 for solids
10-4 to ? atm-1 for liquids
.1 to 1 atm-1 for gasses
Liquid is the most complicated phase.
K=f(T,P)
𝛼 = 𝑓(𝑇, 𝑉)
Example: Calculate the volume change that occurs when 50cm3 of argon, treated as an ideal gas, is
heated though 5.0K at 298K.
𝜕𝑉
𝜕𝑉
𝑑𝑉 = ( ) 𝑑𝑇 + ( ) 𝑑𝑃
𝑃
𝜕𝑇
𝜕𝑃 𝑇
𝜕𝑉
𝑇ℎ𝑒 ( ) 𝑑𝑃 𝑡𝑒𝑟𝑚 = 0 𝑠𝑖𝑛𝑐𝑒 𝑃 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
𝜕𝑃 𝑇
∆V is a finite change.
𝜕𝑉
∆𝑉 ≈ ( ) ∆𝑇 ≈ 𝛼𝑉𝑇
𝜕𝑇 𝑃
What is α for an ideal gas?
𝛼=
1 𝜕𝑉
1 𝜕 𝑛𝑅𝑇
1 𝑛𝑅 1
( ) = ( (
)) =
=
𝑃 𝑉 𝑃
𝑉 𝜕𝑇 𝑃 𝑉 𝜕𝑇 𝑃
𝑇
∆𝑉 ≈
End of chapter 1
1
50𝑐𝑚3 ∗ 5𝐾
𝑉∆𝑇 ≈
≈ .84𝑐𝑚3
𝑇
298𝐾