Lecture 2: Enthalpy
• Reading: Zumdahl 9.2, 9.3
• Outline
– Definition of Enthalpy (H)
– Definition of Heat Capacity (Cv and Cp)
– Calculating E and H using Cv and Cp
What is Enthalpy?
Thermodynamic Definition of Enthalpy (H)
H = E + PV
E = energy of the system
P = pressure of the system
V = volume of the system
Consider a process carried out at constant pressure:
If work is of the form w= -PV, then:
E = qp + w = qp - PV
Rearranging,

E + PV = qp
where qp is heat transferred at constant pressure.
Recall that
H = E + PV
Then
H = E + (PV)
Since
(PV) = PV + VP = PV (P constant)
Substituting,
But we just showed
So finally,
H = E + PV
E + PV = qp
H = qp
In words, the change in enthalpy (H) is equal to the heat
transferred at constant pressure (qp)
What about changes in enthalpy?
Consider the following expression for enthalpy change in a
chemical process:
H = Hprods - Hreacts
If H >0, then qp >0, and the reaction is endothermic.
If H <0, then qp <0, and the reaction is exothermic.
Picturing Enthalpy Changes

Enthalpy
(Similar to previous discussion for
energy changes)
Hinitial
q out
Hfinal
Enthalpy
H < 0
Hfinal
Hinitial
H > 0
q in
If heat comes out of system, the
enthalpy decreases (ex. cooling
water).
If heat goes into the system, the
enthalpy increases (ex. heating
water)
Heat capacity at constant volume
• Recall from Chapter 5 (section 5.6):
(KE)ave = (3/2) RT (for an ideal monatomic gas)
• Temperature is a measure of molecular speed.
• In thermodynamic terms, an increase in system
temperature corresponds to an increase in average kinetic
energy of a gas ( i.e., T is proportional to KE)
Heat capacity at constant volume
• (KE)ave = 3/2 RT (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 at constant volume. Cv is
referred to as the constant volume heat capacity.
Q: Heat capacity at constant P?
What about at constant pressure? In this case,
(PV)-type work can also occur:
PV = nRT
PV = nRT = RT (for 1 mole)
= R (for T = 1 K)
Cp = “heat into translation” + “work”
= Cv + R = 5/2R (for an ideal monatomic gas)
Cv for Monatomic Gases
z
1/2 mv 2 = 3/2 RT
x
y
Q: What are the energetic
degrees of freedom for a
monatomic gas?
Ans: Just translations(x,y,z),
which contribute (3/2)R to
Cv.
Cv for Polyatomics
Iz
Iy
N2 C v = 5/2 R (approx.)
Iy
Iz
Ix
3
NO2 C v = 7/2 R (approx.)
Q: What are the
energetic degrees of
freedom for a
polyatomic gas?
Ans: translations, and
rotations, and
vibrations… all of
which may contribute
to Cv (depends on T).
Variation in Cp and Cv
Ar, He, Ne
H2
CO2
Units: J/mol.
Cv
Cp
12.47
20.8
20.54
28.86
28.95
K
37.27
• Monatomics:
Cv = 3/2 R
Cp = 5/2 R
• Polyatomics:
Cv > 3/2 R
Cp > 5/2 R
But, Cp = Cv + R
(always!)
Heating an ideal monatomic gas at
constant volume
Recall
So
Recall
So,
Eave = 3/2 nRT (av trans. energy)
E = 3/2 nRT = (3/2R)nT
Cv = 3/2 R
E = nCvT
Q: Why Cv? We envision heating our system at constant
volume. As such, all heat goes towards increasing E
(no work is done, since no volume change is possible)
What if we heat the ideal gas at
constant pressure?
In this case, we have a volume change and (PV)
work can occur:
recall
Cp = Cv + R
q p = n Cp T
q p = n (Cv + R) T
recall
E = nCvT
so
q p = E + nRT = E + PV
q p = H = n Cp T
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 = nCvT
H = nCpT
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 liters and pressure of 2.0 atm is
heated until a volume of 10 liters is reached with
pressure unchanged?
Pinit = 2 atm
Pfinal = 2 atm
Vinit = 5 l
Vfinal = 10 l
Tinit = ? K
Tfinal = ? K
Using the ideal gas law PV = nRT
PV = nRT, so we can determine T.
V = (10 L - 5 L) = 5 L
Substituting values,
(2 atm (5 L 
(
(1 mol) .0821 L atm mol K
 121.8 K  T

So, now we can calculate ∆E, ∆H, w and q:
(

E  nCv T  (1 mol) 12.5 J mol.K (121.8 K   1522.5 J
(

H  nC p T  (1 mol) 20.8 J mol.K (121.8 K =2533.4 J
(

w  Pext V   (2 atm (5 L  101.3 J L.atm = -1013.0 J
q  E  w  1522.5 J- (-1013.0 J   2535.5 J