Fundamental Concepts in
Thermodynamics
Doba Jackson, Ph.D.
Associate Professor of Chemistry & Biochemistry
Huntingdon College
Outline of Chapter
• What is thermodynamics and what is useful
about it?
• Macroscopic variables: Volume, Temperature,
Pressure
• Basic definitions of Thermodynamics
• Equations of State
What is Thermodynamics?
Chemistry- the study of matter and changes that it
may undergo.
Physical Chemistry- the branch of chemistry
that establishes and develops the principles of the
subject using the underlying concepts of physics and
mathematics.
Thermodynamics- The branch of science that
describes the behavior of matter on the macroscopic
scale (visible scale)
How to Study for this class
Time:
For every 1 hour class, you should spend 2 hrs of
studying.
Taking Notes:
Unlike most classes, you don’t need to write down my
conversation. Occasionally I will give you something
you need to remember but not often. Also the
PowerPoint's will be made available.
How to Study for this class
In Class Problems:
You should pay particular attention to problems we work
on in class. These problems will be very similar to test
problems. Try to follow in class and review them after
class and prior to the test.
MasteringChemistry problems:
No problems this semester
End of Chapter Problems:
You must work on the end of chapter problems outlined
in the syllabus. The selected problems are similar to the
problem we discuss in class.
What is Thermodynamics?
Macroscopic scale- (also called bulk scale)visible scale and properties.
Macroscopic properties: Microscopic properties:
- Physical state (gas, liquid, solid)
- Volume
- Temperature
- Pressure
- Amount (mass, moles)
- Heat Capacity
- Color
- Boiling, melting points
- Density (mass/volume)
- Bulk Energies (KE, PE, H, S, G, U)
- Electrical conductor, producer
Thermodynamics
(PCHEM 1)
- Absorption/emission of energy
- Dipole moment/ charges present
- Atoms present (type, amount)
- Bonds (covalent, noncovalent)
- Orbitals occupied
- Molecular movements (T, R, V)
- Atomic movements (spin, orbits)
- Atomic energies
- Solubility/Miscibility
Quantum Mechanics
(PCHEM 2)
Properties of Matter
(Definitions to review)
Property: Any characteristic that can be used to
describe or identify matter.
Intensive Properties: Does not depend on the amount
of sample. Ex: Temperature, Melting point, Density
Extensive Properties: Does depend on the amount of
sample. Ex: Length, Volume, Mass, Moles
Extensive properties can be converted to intensive
properties
Density: is mass divided by volume (extensive) to
produce density
Molar Volume: is volume divided by moles
Macroscopic Variables:
Volume, Temperature, and
Pressure
Temperature
Temperature: A measurement of direction and
magnitude of energy flow in the form of heat.
°F =
°C =
9 °F
5 °C
5 °C
9 °F
°C + 32 °F
(°F - 32 °F)
K = °C + 273.15
Derived Units: Quantities based on
other quantities
Volume measurement:
one liter is one cubic decimeter
Measurement of Pressure
• Evangelista Torricelli, in 1863 first
devised a method for measuring the
pressure of an atmosphere using a
mercury barometer.
Mercury Barometer
Aneroid Barometer
Water Barometer
Mercury Barometer
Gases and Gas Pressure
Force
Pressure:
Unit area
Conversions
1 atm = 760 mm Hg
(exact)
1 torr = 1 mm Hg
(exact)
1 bar = 1 x 105 Pa
1 atm = 101 325 Pa
(exact)
Basic Thermodynamic
definitions
Basic Thermodynamic Definitions
• System- Part of the world of interest
• Surroundings- Region outside the
system
• Open system- Allows matter and energy
to pass
• Closed system- Cannot allow matter to
pass
• Isolated system- Cannot allow matter or
energy to pass
Zeroith Law of Thermodynamics
If “A” is in thermal equilibrium with “B” and “B” is
in thermal equilibrium with “C”, then “A” should be
in thermal equilibrium with “C.”
Equations of state and Ideal
gas law
Equations of State are equations that relate
the major macroscopic variables of a
physical state
Ideal Gas Law: PV= nRT
PV/nT = constant (R)
Ideal Gas Constant (R)
*R is used in other
thermodynamic equations
*
*
Volume is a decreasing function
of pressure
Boyle’s Law: PV = const (T,n)
PinitialVinitial = PfinalVfinal
y = 1/x
Volume is an increasing function
of temperature
Charles’ Law: V/T = const (P, n)
Vinitial
Tinitial
=
Vfinal
Tfinal
Absolute zero (-273.15 ºC)
Avogadro’s Law
(constant T and P)
V∞n
V
=k
n
Vinitial
ninitial
=
Vfinal
nfinal
Two major types of problems that can
be solved using gas laws
ONE STATE PROBLEM:
- Using known variables in one state, find the
unknown variable in that same state.
- Given T, P, n; find V
MULTI-STATE PROBLEM:
- Using known variables in one state, find an
unknown variable in another state assuming
some variables remain constant.
P1V1 = P2V2 ; assumes n,T are constant
Example of a Single-State Problem
The reaction used in the deployment of automobile
airbags is the high-temperature decomposition of
sodium azide, NaN3, to produce N2 gas. How many liters
of N2 at 1.15 atm and 30.0 °C are produced by
decomposition of 45.0 g NaN3?
2NaN3(s)
2Na(s) + 3N2(g)
Stoichiometric Relationships with
Gases
2NaN3(s)
2Na(s) + 3N2(g)
Moles of N2 produced:
45.0 g NaN3
x
1 mol NaN3
65.0 g NaN3
x
3 mol N2
= 1.04 mol N2
2 mol NaN3
Volume of N2 produced:
(1.04 mol)
V=
nRT
P
0.082058
L atm
(303.2 K)
K mol
=
= 22.5 L
(1.15 atm)
Problem
Calculate the volume that .65 moles of
ammonia gas occupies at 37*C and 600
torr.
Problem 2
• Calculate the pressure exerted by 18 g of steam
(H2O) confined to a volume of 18 L at 100*C.
What volume would the water occupy if the
steam were condensed to a liquid water at
25*C? The density of liquid water is 1.00 g/ml
at 25*C.
Multistate problem
Show the approximate level of the movable piston in
drawings (a) and (b) after the indicated changes have been
made to the initial gas sample.
Multi-State problems use Boundaries:
ex. Temperature Boundaries
Diathermic Boundary: A boundary
that allows energy in the form of
heat to transfer from one object to
the next.
Adiabatic Boundary: (insulating)Will not allow energy to transfer as
heat between two objects in contact
Multi-state problems: consider a
plot of all states
Multi-state problems: Changes
in state occur usually some
conditions are constant
Isobar- Constant pressure
Isotherm- Constant temperature
Isochore- Constant volume
Muti-state problems: often deviations
occur from a standard state
Standard Ambient Temperature
& Pressure (SATP)
T = 25ºC or (298.15 K)
P = 1.0 bar
Vm = 24.79 dm3/mol
5 sig figs
(exactly)
4 sig figs
Standard Temperature
and Pressure (STP)
T = 0ºC or (273.15 K)
P = 1.0 atm
Vm = 22.41 dm3/mol
5 sig figs
(exactly)
4 sig figs
Example of a Multi-State Problem
In an industrial process, nitrogen is heated
to 500 K in a vessel of isochoric conditions
(constant volume). If it enters the vessel at
100 atm and 300 K, what pressure would it
exert at the working temperature if it
behaved as a perfect gas.
Problem 3
• A weather balloon is partially filled with
helium at 20*C to a volume of 43.7 L and a
pressure of 1.16 atm. The balloon rises to the
stratosphere where the temperature and
pressure are -23.0*C and 6.00 x 10-3 atm.
Calculate the volume of the balloon in the
stratosphere.
Dalton’s Law of Partial
Pressures
A
+
B
=
A+B
PA= 5 atm
PB= 20 atm
PA + PB = PT= 25 atm
NA= 5 moles
NB= 20 moles
NA + NB = NT= 25 moles
• Dalton’s law- Pressure exerted by a mixture of perfect
gases is the sum of the pressure the gases would exert
is they were alone in a container at the same
temperature.
Partial Pressures
PA + PB = PT
A+B
PA= XAPT
PB= XBPT
NA + NB = NT
PA and PB are considered
partial pressures
NA
XA =
 mole fraction of A
NT
NB
XB =
 mole fraction of B
NT
Problem 1.10b
• A gas mixture consists of 320 mg of
methane, 175 mg of argon, and 225 mg of
neon. The partial pressure of neon at 300
K is 8.87 kPa. Calculate (a) the volume and
(b) the total pressure of the mixture.
Ne- 20.18g/mol
Ar- 39.95g/mol
Question 1:
Air at 25.0*C and .998 atm has a density
of 1.21 g/dm3. Assuming air consists of only N2
and O2. Calculate the partial pressure of N2
and O2
3
.998
atm
1
dm


 
PV
P = .998 atm
T = 298.15 K (25.0 *C)
V = 1 dm3
n=?
Mass = 1.21 g
MMO2 = 32.00 g/mol
MMN2 = 28.00 g/mol
n
RT


.08206 dm  atm
3
mol  K
 298.15K 
 .0408
n  .0408 moles total
mO 2  nO 2  MM O 2
nT  nO 2  nN 2  .0408 moles
mT  mO 2  mN 2  1.21 g
mT  nO 2 MM O 2  nN 2 MM N 2  1.21 g
nT  nO 2  nN 2  .0408 moles
mT  mO 2  mN 2  1.21 g
mT  .0408  nN 2  MM O 2  nN 2 MM N 2  1.21 g
mT  .0408MM O 2  nN 2 MM O 2  nN 2 MM N 2  1.21 g
mT  .0408MM O 2  nN 2  MM N 2  MM O 2   1.21 g
nN 2  MM N 2  MM O 2   1.21 g  .0408MM O 2
nN 2
1.21 g  .0408MM O 2

 MM N 2  MM O 2 
.0239
 N2 
 .586 ;
.0408
=
1.21  .040832.00  .0956

 .0239
28.00  32.00
 4.00
 O  1  .586  .414
2
PN 2   N 2 PT  .586.998 atm  .585 atm
PO2   O2 PT  .414.998 atm  .413 atm
Section 1.5: Introduction to
Real Gases
Why are real gases not Ideal?
Ideal Gas Model Assumptions
• The size of the molecules is negligible
because the diameters are much smaller
than the distance traveled between
collisions.
• The molecules do not interact with each
other outside of brief, infrequent and elastic
collisions
Problems with Boyle’s Law
Problem 1: At high pressures
and low molar volumes, intermolecular forces between
molecules become an important
factor to consider.
Problem 2: At high pressures
and low molar volumes, the
volume occupied by the
molecules themselves become
an important factor to consider.
Deviations
Other Equations of state for gases
Outline of Chapter
• 1) The Difference between Real and Ideal gases
• 2) Equations of State for Real gases
– Van Der Waals Equation
– Virial Equation
• 3) Compression Factor
• 4) Law of Corresponding States
Van Der Waals Equation
Discovered by Dutch physicist
Johannes Diderik van der Waals (18371923) who won the Nobel prize in 1910.
• Molecules do occupy space and their volume must be
excluded from the ideal volume. This reduces any
repulsive interactions. The “b” term.
• Attractive interactions between molecules are
proportional to the square of the density (or molar
concentration) of the gas. The “a” term.
Videal = V – nb
“nb- corrects volume”
Pideal = P + a(n/V)2 “a(n/V)2- corrects pressure”
Van Der Waals Equation
PidealVideal  nRT
2


n


 P  a   V – nb   nRT


V





 P  a  1
V

 m




2

V – b   RT
 m

Ideal Gas
Van Der Waals terms
Using Molar Volume
Standard Form
RT
a
P=
2
Vm - b
Vm
The excluded volume term “b”
of the van der Waals equation
Excluded Volume
V  4  r3
3
3
4
V    2r 
3


V = 23 4 πr3 = 8Vmol
3
“b” terms can be calculated by
taking the volume of the molecule
and multiplying by 8
Van der Waals Equation
Pressure-Volume (CO2)
Problem 1.3
• Calculate the pressure exerted by Ar for a molar volume
of 1.31 L/mol at 426 K using the van der Waals equation
of state. The van der Waals parameters are 1.355
bar*dm6/mol2, and .0320 dm3/mol. Is the attractive or
repulsive portion of the potential dominant.
Problem 2.1
• Calculate the pressure exerted by 1.0 mol of C2H6
behaving as a perfect gas and a van der Waals gas. The
gas is confined under the following conditions:
Condition1: 1000 K, 100 L. (The van der Waals constants
are: a=18.57 atm*L2/mol2 and b= .1193 L/mol), R is .08206
L*atm/mol*K.
Virial Equation
• Virial Expansion (Expanded Molar Volumes)
2
3
PVm
= 1 + BT  1  + CT  1  + DT  1  + .....
 Vm 
 Vm 
 Vm 
RT
• Virial Expansion (Expanded Pressures)
PVm
= 1 + BT P + CT P 2 + DT P3 + .....
RT
• Each term in the Virial equation becomes
successively smaller.
• The series does not converge at very high
pressures which make molar volumes less than 1.
• Usually the equation is trunicated after the
second or third term.
The Compression Factor
Real gases show deviations from the ideal gas
law mainly because of molecular interactions
Z>1
Compression Factor
Vm PVm
Z= o =
Vm RT
Vmo = Ideal Molar Volume
Z = 1: Ideal Gas (no forces)
Z < 1: Attractive forces dominate
Z > 1: Repulsive forces dominate
No forces
Ideal
Z≈1
Z<1
Real gases show deviations from the ideal gas
law mainly because of molecular interactions
Compression Factor
Vm PVm
Z= o =
Vm RT
Vmo = Ideal Molar Volume
Z = 1: Ideal Gas (no forces)
Z < 1: Attractive forces dominate
Z > 1: Repulsive forces dominate
Z>1
No forces
Ideal
Z<1
Z≈1
Van der Waals constants are
used to solve the Second Virial
Coeficient Geometric series
Van der Waals Equation
1
 1  x 2  x 3  x 4  .........
1 x
RT
a
P=
- 2
Vm - b Vm
Virial Expansion
The van der Waals equation can be
expanded to form a series
Z=
Z=
BT CT DT
PV
Z=
=1+
+ 2 + 3 +.........
RT
Vm Vm Vm
PVm
Vm
a
=
RT
Vm - b RTVm
PV
1
=
RT
1- b
a
RTVm
Vm
2
 b 
PV 
b
Z=
= 1 +
+
 +
RT 
Vm  Vm 

3

 b 
a
+
.....



 Vm 
 RTVm
2




PV 
1
b
Z=
= 1 + b - a
+
+
RT  V   V 
RT 
 m  m



3

 b 

 + .....
V
 m

Second Virial Coeficient
BT = b - a
RT
Problem 1.15:
A gas at 250 K and 15 atm has a molar volume 12% smaller than
that calculated from the perfect gas law. Calculate (a) the compression
factor under these conditions and (b) the molar volume of the gas.
Which are dominating in the sample, the attractive or repulsive forces?
12% smaller volume means the real gas is 88% of the actual gas.
Vm PVm
Z= o =
 .88
Vm RT
Vm =.88 (Vmo )
L  atm
.08206
 250K 


RT
mol

K
V =
=
=
o
m
P
15atm 
Vm =.88 ( 1.37 L ) = 1.20 L
The Law of Corresponding
States
Gas
Gas
Liquid
Liquid
Supercritical
Fluid
Critical Point- temperature at which
a liquid phase no longer exists and
a phase intermediate of a liquid and
gas exist.
Pressure
Critical behavior of certain substances
Temperature
Critical constants from Van der
Waals constants
Van der Waals Equation
P=
RT
a
- 2
Vm - b Vm
The critical Temp, Pressure and Volume
will be the point at which the first and
second derivative both equal zero. The
inflection point of a cubic equation
First and second derivative
dP
-RT
2a
=
0
First
2
3
dVm
(Vm -b)
Vm
d2P
2RT
6a
=
0
Second
2
3
4
dVm
(Vm -b)
Vm
Vc = 3b
8a
Tc =
27Rb
a
Pc =
27b 2
Van der Waals constants are used
to find the Boyle Temperature
Boyle Temperature- the temperature
at which the compression factor is
the most ideal (Z ≈ 1) over a broad
range of pressures and volumes.
Z=
B
C
D
PV
=1+ T + T2 + T3 +.........
RT
Vm Vm Vm
Z  1 as BT  0
Let BT = 0
BT = b - a
RTBoyle
TBoyle = a
Rb
0
Principle of corresponding
states
• Ideal gas law is independent of the molecular
substance.
• Real gases depend on each individual gas.
• Find a relative scale for each substance and
use it for all substances.
• Use the critical point as a relative scale:
P
Pr =
Pc
T
Tr =
Tc
V
Vr =
Vc
Pc, Tc, and Vc are critical pressure, temperature and volumes