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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 classIn 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: - 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
Thermodynamics (PCHEM 1)
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
(°F - 32 °F)5 °C
9 °F°C =
°C + 32 °F9 °F
5 °C°F =
K = °C + 273.15
Temperature: A measurement of direction and magnitude of energy flow in the form of heat.
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.
Aneroid Barometer
Mercury Barometer
Water Barometer
Mercury Barometer
(exact)
Conversions
1 torr = 1 mm Hg
1 atm = 101 325 Pa
(exact)1 atm = 760 mm Hg
(exact)1 bar = 1 x 105 Pa
Pressure:Unit area
Force
Gases and Gas Pressure
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
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.”
Zeroith Law of Thermodynamics
Equations of state and Ideal gas law
Ideal Gas Law: PV= nRT
PV/nT = constant (R)
Ideal Gas Constant (R)
*R is used in other thermodynamic equations
**
Equations of State are equations that relate the major macroscopic variables of a
physical state
Volume is a decreasing function of pressure
Boyle’s Law: PV = const (T,n)
y = 1/xPinitialVinitial = PfinalVfinal
Volume is an increasing function of temperature
Charles’ Law: V/T = const (P, n)
Absolute zero (-273.15 ºC)
=Tfinal
Vfinal
Tinitial
Vinitial
Avogadro’s Law
(constant T and P)
= kn
V
=nfinal
Vfinal
ninitial
Vinitial
V ∞ n
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?
2Na(s) + 3N2(g)2NaN3(s)
Stoichiometric Relationships with Gases
2Na(s) + 3N2(g)2NaN3(s)
45.0 g NaN3
65.0 g NaN3
1 mol NaN3
2 mol NaN3
3 mol N2
x x
Volume of N2 produced:
= 1.04 mol N2
Moles of N2 produced:
= 22.5 LV =P
nRT=
(1.15 atm)
(1.04 mol) 0.082058K mol
L atm(303.2 K)
Calculate the volume that .65 moles of ammonia gas occupies at 37*C and 600 torr.
Problem
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 thesteam were condensed to a liquid water at25*C? The density of liquid water is 1.00 g/mlat 25*C.
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.
Multistate problem
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
Isobar- Constant pressure
Isotherm- Constant temperature
Isochore- Constant volume
Multi-state problems: Changes in state occur usually some
conditions are constant
Muti-state problems: often deviations occur from a standard state
Standard Ambient Temperature & Pressure (SATP)
T = 25ºC or (298.15 K) 5 sig figsP = 1.0 bar (exactly)Vm = 24.79 dm3/mol 4 sig figs
Standard Temperature and Pressure (STP)
T = 0ºC or (273.15 K) 5 sig figsP = 1.0 atm (exactly)Vm = 22.41 dm3/mol 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.
• 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.
Problem 3
Dalton’s Law of Partial Pressures
• 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.
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
Partial Pressures
A + B
PA + PB = PT
PA= XAPT
PB= XBPT
NA + NB = NT
PA and PB are considered partial pressures
AA
T
NX =
Nmole fractionof A
T
NX =
NB
B mole fractionof B
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/molAr- 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
P = .998 atmT = 298.15 K (25.0 *C)V = 1 dm3
n = ?Mass = 1.21 gMMO2 = 32.00 g/molMMN2 = 28.00 g/mol
3
3
.998 1.0408
.08206 298.15
atm dmPVn
RT dm atm Kmol K
.0408n moles total
2 2 .0408T O Nn n n moles
2 2 1.21T O Nm m m g 2 2 2O O Om n MM
2 2 2 2 1.21T O O N Nm n MM n MM g
2 2 .0408T O Nn n n moles
2 2 1.21T O Nm m m g
2 2 2 2.0408 1.21T N O N Nm n MM n MM g
2 2 2 2 2.0408 1.21T O N O N Nm MM n MM n MM g
2 2 2 2.0408 1.21T O N N Om MM n MM MM g
2 2 2 21.21 .0408N N O On MM MM g MM
2
22 2
1.21 .0408 ON
N O
g MMn
MM MM
0239.00.4
0956.
00.3200.28
00.320408.21.1
=
414.586.1;586.0408.
0239.22
ON atmatmPP
atmatmPP
TOO
TNN
413.998.414.
585.998.586.
22
22
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, 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 becomean 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
• 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”
Discovered by Dutch physicist Johannes Diderik van der Waals (1837-1923) who won the Nobel prize in 1910.
Van Der Waals Equation
2m m
RT aP = -
V - b V
RT b– VV
1aP
nRT nb– VV
naP
nRTVP
m
2
m
2
idealideal
Standard Form
Ideal Gas
Van Der Waals terms
Using Molar Volume
The excluded volume term “b” of the van der Waals equation
343V r
34 23V r
3 34V = 2 πr = 8V3 mol
Excluded Volume
“b” terms can be calculated bytaking the volume of the moleculeand 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 3mT T T
PV = 1 + B P + C P + D P + .....
RT
2 3m
T T Tm m m
PV 1 1 1 = 1 + B + C + D + .....V V VRT
• 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.
The Compression Factor
Real gases show deviations from the ideal gas law mainly because of molecular interactions
m mom
V PVZ= =
V RTomV = Ideal Molar Volume
Compression Factor
Z = 1: Ideal Gas (no forces)
Z < 1: Attractive forces dominate
Z > 1: Repulsive forces dominate
No forcesIdeal
Z≈1
Z>1
Z<1
Real gases show deviations from the ideal gas law mainly because of molecular interactions
m mom
V PVZ= =
V RTomV = Ideal Molar Volume
Compression Factor
Z = 1: Ideal Gas (no forces)
Z < 1: Attractive forces dominate
Z > 1: Repulsive forces dominate
No forcesIdeal
Z≈1
Z>1
Z<1
Van der Waals constants are used to solve the Second Virial
Coeficient
T T T2 3
m m m
B C DPVZ= =1+ + + +.........
RT V V V
2 3 411 .........
1x x x
x
mm
PV 1 aZ= = -
bRT RTV1- V
2m m
RT aP = -
V - b V
Van der Waals Equation
The van der Waals equation can be expanded to form a series
Geometric series
Virial Expansion
m
m m
PV V aZ= = -
RT V - b RTVm
2 3
m m m m
PV b b b aZ = = 1 + + + + ..... -
RT V V V RTV
2 3
m m m
PV 1 b baZ = = 1 + b - + + + .....RTRT V V V
Second Virial Coeficient
TaB = b - RT
Problem 1.15: A gas at 250 K and 15 atm has a molar volume 12% smaller thanthat calculated from the perfect gas law. Calculate (a) the compressionfactor 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.
m mom
V PVZ= = .88
V RT o
m mV =.88 (V )
om
L atm.08206 250KRT mol KV = = =P 15atm
mV =.88 ( 1.37 L ) = 1.20 L
The Law of Corresponding States
Critical behavior of certain substances
Gas Gas
Liquid Liquid
Super-criticalFluid
Pre
ssu
re
Temperature
Critical Point- temperature at whicha liquid phase no longer exists anda phase intermediate of a liquid andgas exist.
Critical constants from Van der Waals constants
2m m
RT aP = -
V - b V
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
2 3m m m
dP -RT 2a = - 0
dV (V -b) V
2
2 3 4m m
d P 2RT 6a = - 0
dV (V -b) Vm
First
Second
cV = 3bc 2
aP =
27bc
8aT =
27Rb
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.
T T T2 3
m m m
B C DPVZ= =1+ + + +.........
RT V V V
TZ 1 as B 0
Let BT = 0
TaB = b - 0RTBoyle
BoyleaT = Rb
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:
rc
TT =
Tr
c
PP =
P rc
VV =
VPc, Tc, and Vc are critical pressure, temperature and volumes