1. Fundamental Concepts in Thermodynamics
Doba Jackson, Ph.D.
Associate Professor of Chemistry & Biochemistry
Huntingdon College

2. Outline of Chapter What is thermodynamics and what is useful about it?
Macroscopic variables: Volume, Temperature, Pressure
Basic definitions of Thermodynamics
Equations of State

3. 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)

4. 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.

5. 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.

6. What is Thermodynamics?
Macroscopic scale- (also called bulk scale)- visible scale and properties.
Macroscopic 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
Microscopic properties:
- 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)
Thermodynamics
(PCHEM 1)

7. Properties of Matter (Definitions to review)
Chapter 1: Atoms: Matter and Measurement
4/25/2017
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
Copyright © 2010 Pearson Prentice Hall, Inc.

8. Macroscopic Variables: Volume, Temperature, and Pressure

9. Chapter 1: Atoms: Matter and Measurement
4/25/2017
Temperature
Temperature: A measurement of direction and
magnitude of energy flow in the form of heat.
°C + 32 °F
9 °F
5 °C
°F =
The offset from Kelvin to Celsius is exact.
(°F - 32 °F)
5 °C
9 °F
°C =
K = °C
Copyright © 2010 Pearson Prentice Hall, Inc.

10. Derived Units: Quantities based on other quantities
Chapter 1: Atoms: Matter and Measurement
4/25/2017
Derived Units: Quantities based on other quantities
Copyright © 2010 Pearson Prentice Hall, Inc.

11. Chapter 1: Atoms: Matter and Measurement
Volume measurement:
one liter is one cubic decimeter
4/25/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

12. 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

14. Mercury Barometer

16. Gases and Gas Pressure (exact) Conversions 1 torr = 1 mm Hg
1 atm = Pa
1 atm = 760 mm Hg
1 bar = 1 x 105 Pa
Pressure:
Unit area
Force

17. Basic Thermodynamic definitions

18. 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

20. 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.”

21. Equations of state and Ideal gas law

22. 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

23. Volume is a decreasing function
of pressure
Boyle’s Law: PV = const (T,n)
PinitialVinitial = PfinalVfinal
y = 1/x

24. Volume is an increasing function
of temperature
Charles’ Law: V/T = const (P, n) =
Tfinal
Vfinal
Tinitial
Vinitial
Absolute zero ( ºC)

25. Avogadro’s Law (constant T and P) V ∞ n = k n V = nfinal Vfinal
ninitial
Vinitial

26. 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

27. 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?
2Na(s) + 3N2(g)
2NaN3(s)

28. Stoichiometric Relationships with Gases
2Na(s) + 3N2(g)
2NaN3(s)
Moles of N2 produced:
45.0 g NaN3
65.0 g NaN3
1 mol NaN3
2 mol NaN3
3 mol N2 x x
= 1.04 mol N2
Volume of N2 produced:
(1.15 atm)
(1.04 mol)
K mol
L atm
(303.2 K) P
nRT
V = =
= 22.5 L

29. Problem
Calculate the volume that .65 moles of ammonia gas occupies at 37*C and 600 torr.

30. 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.

31. 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.

32. 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

33. Multi-state problems: consider a plot of all states

34. Multi-state problems: Changes in state occur usually some conditions are constant
Isobar- Constant pressure
Isotherm- Constant temperature
Isochore- Constant volume

35. Muti-state problems: often deviations occur from a standard state
Standard Ambient Temperature
& Pressure (SATP)
T = 25ºC or ( K) sig figs
P = 1.0 bar (exactly)
Vm = dm3/mol sig figs
Standard Temperature
and Pressure (STP)
T = 0ºC or ( K) sig figs
P = 1.0 atm (exactly)
Vm = dm3/mol sig figs

36. 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.

37. 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.

38. Dalton’s Law of Partial PressuresB
A + B = +
PB= 20 atm
PA= 5 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.

39. Partial Pressures A + B PA and PB are considered partial pressures
PA + PB = PT
NA + NB = NT
PA and PB are considered
partial pressures
PA= XAPT
PB= XBPT

40. 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 g/mol
Ar g/mol

41. 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
P = .998 atm
T = K (25.0 *C)
V = 1 dm3
n = ?
Mass = 1.21 g
MMO2 = g/mol
MMN2 = g/mol

42. =

43. Section 1.5: Introduction to Real Gases

44. 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

45. Problems with Boyle’s Law
Problem 1: At high pressures
and low molar volumes, inter-
molecular 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

46. Other Equations of state for gases

47. 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

48. Van Der Waals Equation Videal = V – nb “nb- corrects volume”
Discovered by Dutch physicist Johannes Diderik van der Waals ( ) 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”

49. Van Der Waals Equation Ideal Gas Van Der Waals terms
Using Molar Volume
Standard Form

50. The excluded volume term “b” of the van der Waals equation
“b” terms can be calculated by
taking the volume of the molecule
and multiplying by 8

51. Van der Waals Equation
Pressure-Volume (CO2)

52. 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 bar*dm6/mol2, and dm3/mol. Is the attractive or repulsive portion of the potential dominant.

53. 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: K, 100 L. (The van der Waals constants are: a=18.57 atm*L2/mol2 and b= L/mol), R is L*atm/mol*K.

54. Virial Equation Virial Expansion (Expanded Molar Volumes)
Virial Expansion (Expanded Pressures)
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.

55. The Compression Factor

56. Real gases show deviations from the ideal gas
law mainly because of molecular interactions
Z>1
Compression Factor
No forces
Ideal
Z≈1
Z<1

57. Real gases show deviations from the ideal gas
law mainly because of molecular interactions
Compression Factor
Z>1
No forces
Ideal
Z≈1
Z<1

58. Van der Waals constants are used to solve the Second Virial Coeficient
Geometric series
Van der Waals Equation
Virial Expansion
The van der Waals equation can be expanded to form a series
Second Virial Coeficient

59. 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.

60. The Law of Corresponding States

61. Critical behavior of certain substances
Super-
critical
Fluid
Gas
Gas
Pressure
Liquid
Liquid
Temperature
Critical Point- temperature at which
a liquid phase no longer exists and
a phase intermediate of a liquid and
gas exist.

62. Critical constants from Van der Waals constants
Van der Waals Equation
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
First
Second

63. 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.
Let BT = 0

64. 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:
Pc, Tc, and Vc are critical pressure, temperature and volumes