Lecture 2: Heat Capacities/State functions

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

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110 90 70

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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.