Lecture 2: Heat Capacities/State
functions
• Reading: Zumdahl 9.3
• Outline
–
–
–
–
Definition of Heat Capacity (Cv and Cp)
Calculating DE and DH 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 DT?
Heat required = 3/2R DT
= 3/2R (for DT = 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:
PDV = nRDT = RDT (for 1 mole)
= R (for DT = 1 K)
• Cp = “heat into translation” + “work to expand the
gas” = Cv + R = 5/2R (for ideal monatomic gas)
Cv for Monatomic Gases
z
1/2 mv 2 = 3/2 RT
x
y
• What are the energetic
degrees of freedom for
a monatomic gas?
• Just translations,
which contribute 3/2R
to Cv.
Cv for Polyatomics
Iz
• What are the energetic
degrees of freedom for
a polyatomic gas?
Iy
N2 Cv = 5/2 R (approx.)
Iz
Iy
Ix
NO2 C v = 7/2 R (approx.)
Translations, rotations,
and vibrations. All of
which may contribute
to Cv (depends on T).
Cv for Polyatomics
Iz
Iy
N2 Cv = 5/2 R (approx.)
Iz
Iy
Ix
NO2 C v = 7/2 R (approx.)
• 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).
Cv for Polyatomics
Iz
• T measures the average
translational kinetic energy
Iy
N2 Cv = 5/2 R (approx.)
Iz
Iy
Ix
NO2 C v = 7/2 R (approx.)
• 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)
Variation in Cp and Cv
Cv
Cp
12.47
20.8
H2
20.54
28.86
CO2
28.95
37.27
Ar, He, Ne
Units: J/mol.
• Monatomics:
– Cv = 3/2 R
– Cp = 5/2 R
• Polyatomics:
– Cv > 3/2 R
– Cp > 5/2 R
K
– But….Cp = Cv + R
Energy and Cv
• Recall from Chapter 5:
Eave = 3/2 nRT (average translational energy)
DE = 3/2 nR DT
DE = n Cv DT
(since 3/2 R = Cv)
• Why is Cv=DE/DT
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 DT
= n (Cv + R) DT
= DE + nRDT = DE + PDV
= DH or DH = nCpDT
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
• DE = nCvDT
• DH = nCpDT
Example
• What is q, w, DE and DH 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
Pfinal = 2 atm
Vinit = 5 l
Vfinal = 10 l
Tinit = ? K
Tfinal = ? K
Example (cont.)
• Since PDV = nRDT (ideal gas law) we can
determine DT
• Then DV = (10 l - 5 l) = 5 l
• And:
2atm5l

(1mol) .0821 l.atm mol.K

 121.8K  DT
Example (cont.)
• Given this:

121.8K   1522.5J


mol.K

121.8K   2533.4J


mol.K
DE  nCv DT  (1mol) 12.5 J
DH  nC p DT  (1mol) 20.8 J

w  Pext DV  2atm5l 101.3 J
 1013.0J

l.atm
q  DE  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
•
•
•
•
•
DE = q + w
w = -PextDV (for now)
DE = nCvDT = qV
DH = nCpDT = qP
If DT = 0, then DE = 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?
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 Cooling
Path 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.
– PDV = (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)
– And DT = PDV/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 = nCpDT = (2mol)(5/2R)(243.6 K)
= 1.0 x 104 J
= DH1
– And DE1 = nCvDT = (2mol)(3/2R)(243.6 K)
= 6.0 x 103 J
(check: DE1 = 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 DT:
– Now,DT = DPV/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 = nCvDT = (2 mol)(3/2R)(-182.7 K)
= - 4.6 x 103 J
• and DE2 = nCvDT = -4.6 x 103 J
• and DH2 = nCpDT = -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
DE = DE1 + DE2 = 1.5 x 103 J
DH = DH1 + DH2 = 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 DT:
DT = DPV/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 = nCvDT = (2 mol)(3/2 R)(-60.9 K)
= -1.5 x 103 J = DE1
DH1 = nCPDT = (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.
DT = PDV/nR = (1 atm)(20.0 l)/(2 mol)R
= 121.8 K
Pathway 2 (cont.)
• Then, calculate the rest:
q2 = qp = nCPDT = (2 mol)(5/2 R)(121.8 K)
= 5.1 x 103 J = DH2
DE2 = nCvDT = (2 mol)(3/2 R)(121.8 K)
= 3.1 x 103 J
w1 = -PDV = -20 l.atm = -2.0 x 103 J
Pathway 2 (end)
• 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
DE = DE1 + DE2 = 1.5 x 103 J
DH = DH1 + DH2 = 2.5 x 103 J
Comparison of Path 1 and 2
• Pathway 1
q = 5.5 x 103 J
w = -4.1 x 103 J
DE = 1.5 x 103 J
DH = 2.5 x 103 J
• Pathway 2
q = 3.6 x 103 J
w = -2.0 x 103 J
DE = 1.5 x 103 J
DH = 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
Thermodynamic State Functions
• Thermodynamic State Functions:
Thermodynamic properties that are dependent on
the state of the system only. (Example: DE and
DH)
• 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.