lecture 2: heat capacities/state functions reading: zumdahl 9.3 outline –definition of heat...
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Lecture 2: Heat Capacities/State functions
• Reading: Zumdahl 9.3
• Outline– Definition of Heat Capacity (Cv and Cp)
– Calculating E and H using Cv and Cp
– Example of Thermodynamic Pathways – State Functions
Heat Capacity at Constant V• Recall from Chapter 5 (section 5.6):
(KE)ave = 3/2RT (ideal monatomic gas)
• Temperature is a measure of molecular speed
• In thermodynamic terms, an increase in system temperature corresponds to an increase in system kinetic energy ( i.e., q is proportional to <KE>)
Heat Capacity at Constant V• (KE)ave = 3/2RT (ideal monatomic gas)
• How much energy in the form of heat is required to change the gas temperature by an amount T?
Heat required = 3/2R T = 3/2R (for T = 1K)
• Therefore, Cv = 3/2 R is the heat required to raise one mole of an ideal gas by 1K. Cv is called constant volume molar heat capacity.
Heat Capacity at Constant P
• What about at constant pressure? In this case, PV type work can also occur:
PV = nRT = RT (for 1 mole)
= R (for T = 1 K)
• Cp = “heat into translation” + “work to expand the gas” = Cv + R = 5/2R (for ideal monatomic gas)
Cv for Monatomic Gases
• What are the energetic degrees of freedom for a monatomic gas?
• Just translations, which contribute 3/2R to Cv.
z
x
y
1/2 mv2 = 3/2 RT
Cv for Polyatomics
• What are the energetic degrees of freedom for a polyatomic gas?
Translations, rotations, and vibrations. All of which may contribute to Cv (depends on T).
IzIy
IzIy
Ix
N2 Cv = 5/2 R (approx.)
NO2 Cv = 7/2 R (approx.)
Cv for Polyatomics
• When heat is provided , molecules absorb energy and the translational kinetic energy increases
• In polyatomic gases, rotational and vibrational kinetic energies increase as well (depending on T).
IzIy
IzIy
Ix
N2 Cv = 5/2 R (approx.)
NO2 Cv = 7/2 R (approx.)
Cv for Polyatomics
• T measures the average translational kinetic energy
• Increases in rotational and vibrational kinetic energies do not increase T directly
• It takes more heat to increase T by the same amount (Cv/Cp larger)
IzIy
IzIy
Ix
N2 Cv = 5/2 R (approx.)
NO2 Cv = 7/2 R (approx.)
Variation in Cp and Cv
• Monatomics: – Cv = 3/2 R
– Cp = 5/2 R
• Polyatomics: – Cv > 3/2 R
– Cp > 5/2 R
– But….Cp = Cv + R
H2
Ar, He, Ne
Cv Cp
CO2
12.47 20.8
20.54 28.86
28.95 37.27
Units: J/mol.K
Energy and Cv
• Recall from Chapter 5:
Eave = 3/2 nRT (average translational energy)
E = 3/2 nR T
E = n Cv T (since 3/2 R = Cv)
• Why is Cv=E/T
When heating our system at constant volume, all heat goes towards increasing E (no work).
Enthalpy and Cp
• What if we heated our gas at constant pressure? Then, we have a volume change such that work occurs:
q p = n Cp T
= n (Cv + R) T
= E + nRT = E + PV
= H or H = nCpT
Keeping Track
Ideal Monatomic Gas
• Cv = 3/2R
• Cp = Cv + R = 5/2 R
Polyatomic Gas
• Cv > 3/2R
• Cp > 5/2 R
All Ideal Gases
• E = nCvT
• H = nCpT
Example
• What is q, w, E and H for a process in which one mole of an ideal monatomic gas with an initial volume of 5 l and pressure of 2.0 atm is heated until a volume of 10 l is reached with pressure unchanged?
Pinit = 2 atm
Vinit = 5 l
Tinit = ? K
Pfinal = 2 atm
Vfinal = 10 l
Tfinal = ? K
Example (cont.)
• Since PV = nRT (ideal gas law) we can determine T
• ThenV = (10 l - 5 l) = 5 l
• And:
2atm 5l (1mol) .0821l.atmmol.K
121.8K T
Example (cont.)
• Given this:
E nCvT (1mol) 12.5J mol.K 121.8K 1522.5J
H nCpT (1mol) 20.8J mol.K 121.8K 2533.4J
w PextV 2atm 5l 101.3J l.atm 1013.0J
qE w 1522.5J 1013.0J 2535.5J
To Date….
Ideal Monatomic Gas
• Cv = 3/2R
• Cp = Cv + R = 5/2 R
Polyatomic Gas
• Cv > 3/2R
• Cp > 5/2 R
All Ideal Gases
• E = q + w
• w = -PextV (for now)
• E = nCvT = qV
• H = nCpT = qP
• If T = 0, then E = 0
and q = -w
State Functions
• If we start in Seattle and end in Chicago, but you take different paths to get from one place to the other ..
• Will the energy/enthalpy, heat/work we spend be the same along both paths?
United States
L. Superior
OntarioL.Michigan
L.Huron
L.
ErieL.
Gulf of Mexico
Pacific Ocean
AtlanticOcean
WashingtonDC
SanFrancisco
LosAngeles
Chicago
Miami
New York
Detroit Boston
Dallas
Seattle
DenverSt.
Louis
Tampa
Atlanta
Houston
San Diego
Philadelphia
Minneapolis
Cleveland
Portland
Phoenix
OklahomaCity
Omaha
Kansas City
Memphis Raleigh
Cincinnati
NewOrleans
Indianapolis
San Jose
SanAntonio
Salt LakeCity
Orlando
Milwaukee
Jacksonville
Honolulu
Fairbanks
Charlotte
Buffalo
Baltimore
AnchorageCanada
Canada
Cuba
Mexico TheBahamas
200 km
200 miles
1000 km
1000 miles
1000 km
1000 miles
170 160
160
20
70
60
110 90 70
30
40
50
60
Thermodynamic Pathways: an Example
• Example 9.2. We take 2.00 mol of an ideal monatomic gas undergo the following:
– Initial (State A): PA = 2.00 atm, VA = 10.0 L
– Final (State B): PB = 1.00 atm, VB = 30.0 L
• We’ll do this two ways:Path 1: Expansion then CoolingPath 2: Cooling then Expansion
Thermodynamic Jargon
• When doing pathways, we usually keep one variable constant. The language used to indicate what is held constant is:
– Isobaric: Constant Pressure– Isothermal: Constant Temperature– Isochoric: Constant Volume
Thermodynamic Path: A series of manipulations of a system that takes the system from an initial state to
a final state
isochoric
isobaric
Pathway 1
• Step 1. Constant pressure expansion (P = 2 atm) from 10.0 l to 30.0 l.
– PV = (2.00 atm)(30.0 l - 10.0 l) = 40.0 l.atm
= (40.0 l.atm)(101.3 J/l.atm) = 4.0 x 103 J
= -w (the system does work)
– AndT = PV/nR = 4.05 x 103 J/(2 mol)(8.314 J/mol.K)
= 243.6 K (from the ideal gas law)
Pathway 1 (cont.)• Step 1 is isobaric (constant P); therefore,
– q1 = qP = nCpT = (2mol)(5/2R)(243.6 K) = 1.0 x 104 J
= H1
– AndE1 = nCvT = (2mol)(3/2R)(243.6 K) = 6.0 x 103 J
(check: E1 = q1 + w1 = (1.0 x 104 J) -(4.0 x 103 J) = 6.0 x 103 J )
Pathway 1 (cont.)
• Step 2: Isochoric (const. V) cooling until pressure is reduced from 2.00 atm to 1.00 atm.
• First, calculate T:– NowT = PV/nR (note: P changes, not V)
= (-1.00 atm)(30.0 l)/ (2 mol)(.0821 l.atm/mol K)
= -182.7 K
Pathway 1 (cont.)
• q2 = qv = nCvT = (2 mol)(3/2R)(-182.7 K)
= - 4.6 x 103 J
• and E2 = nCvT = -4.6 x 103 J
• and H2 = nCpT = -7.6 x 103 J
• Finally w2 = 0 (isochoric…no V change, no PV-type work)
Pathway 1 (end)
• Thermodynamic totals for this pathway are the sum of values for step 1 and step 2
q = q1 + q2 = 5.5 x 103 J
w = w1 + w2 = -4.0 x 103 J
E = E1 + E2 = 1.5 x 103 J
H = H1 + H2 = 2.5 x 103 J
Next Pathway
• Now we will do the same calculations for the green path.
Pathway 2
• Step 1: Isochoric cooling from P = 2.00 atm to P = 1.00 atm.
• First, calculate T: T = PV/nR = (-1.00 atm)(10.0 l)/ (2 mol)R
= -60.9 K
Pathway 2 (cont.)
• Then, calculate the rest for Step 1:
q1 = qv = nCvT = (2 mol)(3/2 R)(-60.9 K)
= -1.5 x 103 J = E1
H1 = nCPT = (2 mol)(5/2 R)(-60.9 K)
= -2.5 x 103 J
w1 = 0 (constant volume)
Pathway 2 (cont.)
• Step 2: Isobaric (constant P) expansion at 1.0 atm from 10.0 l to 30.0 l.
T = PV/nR = (1 atm)(20.0 l)/(2 mol)R
= 121.8 K
Pathway 2 (cont.)
• Then, calculate the rest:
q2 = qp = nCPT = (2 mol)(5/2 R)(121.8 K)
= 5.1 x 103 J = H2
E2 = nCvT = (2 mol)(3/2 R)(121.8 K)
= 3.1 x 103 J
w1 = -PV = -20 l.atm = -2.0 x 103 J
• Thermodynamic totals for this pathway are again the sum of values for step 1 and step 2:
q = q1 + q2 = 3.6 x 103 J
w = w1 + w2 = -2.0 x 103 J
E = E1 + E2 = 1.5 x 103 J
H = H1 + H2 = 2.5 x 103 J
Pathway 2 (end)
Comparison of Path 1 and 2
• Pathway 1
q = 5.5 x 103 J
w = -4.1 x 103 J
E = 1.5 x 103 J
H = 2.5 x 103 J
• Pathway 2
q = 3.6 x 103 Jw = -2.0 x 103 JE = 1.5 x 103 JH = 2.5 x 103 J
Note: Energy and Enthalpy are the same,but heat and work are not the same!
State Functions
• A State Function is a function in which the value only depends on the initial and final state….NOT on the pathway taken
United States
L. Superior
OntarioL.Michigan
L.Huron
L.
ErieL.
Gulf of Mexico
Pacific Ocean
AtlanticOcean
WashingtonDC
SanFrancisco
LosAngeles
Chicago
Miami
New York
Detroit Boston
Dallas
Seattle
DenverSt.
Louis
Tampa
Atlanta
Houston
San Diego
Philadelphia
Minneapolis
Cleveland
Portland
Phoenix
OklahomaCity
Omaha
Kansas City
Memphis Raleigh
Cincinnati
NewOrleans
Indianapolis
San Jose
SanAntonio
Salt LakeCity
Orlando
Milwaukee
Jacksonville
Honolulu
Fairbanks
Charlotte
Buffalo
Baltimore
AnchorageCanada
Canada
Cuba
Mexico TheBahamas
200 km
200 miles
1000 km
1000 miles
1000 km
1000 miles
170 160
160
20
70
60
110 90 70
30
40
50
60
Thermodynamic State Functions
• Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: E and H)
• Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.