CHAPTER 2 LECTURE NOTES
The First Law of Thermodynamics:
The simplest statement of the First Law is as follows:
∆U = q + w.
Here ∆U is the “internal energy” of the system, q is the heat and w is the work.
CONVENTIONS (The System Centered Point of View):
Heat (q):
In numerical calculations, when the numerical value of q turns out to be positive,
heat is absorbed by the system, if the value turns out to be negative, heat is given
out by the system.
If we supply q as “input” to a problem, the sign of q has to be determined based on
whether the heat is absorbed by the system or is lost from the system.
Example: It takes 4.18 J of heat to heat 1.0 g of water through 1°C. If no work is
done, according to the first Law,
∆U = 4.18 J if we heat 1.0 g of water through 1°C.
∆U = –4.18 J if we cool 1.0 g of water through 1°C.
Work (w):
In numerical calculations, when the numerical value of w turns out to be positive,
work is done on the system, if the value turns out to be negative, work is done by
the system.
Example: See Fig. 2.2. It takes 5.0 J of energy to push a 5.0 kg piston through 1.0
m. If no heat is involved, according to the first Law,
w = 5.0 J if the piston is pushed in (by us) through 1.0 m.
w = –5.0 J if the piston is pushed out (by the gas) through 1.0 m.
Therefore, the process determines whether heat is absorbed or given out, and
whether work is done on the system or by the system.
Examples 2.1, 2.2, Problem 2.7.
2
State Functions and Path-Dependent Functions:
A thermodynamic property that depends only on state of the system and not
on how it got there is called a state function. The internal energy, U, is a state
function. Therefore, a change in the internal energy, ∆U, can be calculated simply
by knowing the initial and final values of U:
∆U = Uf –Ui.
On the other hand, q and w are path-dependent functions. We generally denote
state functions with upper case variables and path-dependent functions with
lower-case variables.
Reversible and Irreversible processes:
A “reversible” process is one in which the system and surroundings are
continuously in equilibrium. This implies that the process takes place
infinitesimally slowly so that there is plenty of time at each stage for equilibration
with the surroundings. “Reversible” processes are NOT “cyclic” processes, i.e.,
the system is not required to return to the initial state at the end of the process.
Fig. 2.4(a)
w rev = − ° V PdV
V2
1
Fig. 2.4(b)
w irr = −P 2 (V 2 − V 1 )
3
First Law: dU = dq + dw.
Constant Volume (Isochoric) Processes:
At constant volume, dV = 0; therefore, dU = dqV, where the subscript implies that
volume is being held constant.
The heat required to heat an object (system) through 1°C is called the heat
capacity C of the object (system). Therefore,
dU = dqV = CVdT, or C V =
dq V æ ¹U ö
=
.
dT è ¹T ø V
Constant Pressure (Isobaric) Processes:
From the first law, dqP = dU – dw = dU + PdV. Therefore,
q P = ° U dU + ° V PdV = (U 2 − U 1 ) + P(V 2 − V 1 )
U2
V2
1
1
This yields qP = (U2 + PV2) – (U1 + PV1) and suggests the definition of a new state
function, enthalpy, defined as
H = U + PV.
It follows that dqP = dH and, therefore,
dH = dqP = CPdT or C P =
dq P æ ¹H ö
=
.
dT è ¹T ø P
The molar heat capacities are denoted as CV,m and CP,m, respectively.
For an ideal gas,
dHm = dUm + d(PVm) = dUm + d(RT).
¹H
¹U
Therefore, æ m ö = æ m ö + R, or CP,m = CV,m + R.
è ¹T ø P è ¹T ø V
Here we have exploited the fact that H and U are both state-functions. A more
rigorous derivation is given on p. 83 (see Eq. 2.119).
Problems: 2.3, 2.4, 2.5, 2.6, 2.8, 2.9, 2.10
Assignment: Review basic thermochemistry and Hesse’s Law from Gen. Chem.
Example 2.3, 2.4, Problems 2.14, 2.16, 2.18, 2.21, 2.23 – 2.26, 2.30, 2.31.
4
Measurement of Enthalpy Changes: Calorimetry
Basic principle:
heat lost = –heat gained.
(a) Constant pressure calorimetry: coffee-cup calorimeter:
This simple device directly gives the enthalpy change since the process studied
takes place at constant pressure.
Example: heat of neutralization (Gen. Chem.): the neutralization reaction
generates heat which is lost to the surrounding water, thermometer, stirrer etc., and
results in an increase in their temperature. If the heat capacity of the calorimeter
(water + thermometer + stirrer, etc.), C (J °C–1), is known, then
heat lost = –C × (tf – ti).
Therefore, the heat generated by the neutralization can be calculated and the molar
enthalpy of neutralization determined.
(b) Constant Volume Calorimetry: the bomb calorimeter:
This device yields the ∆U for the process from which ∆H has to be calculated, as
∆H = ∆U + ∆(PV) = ∆U + ∆(nRT) = ∆U + ∆nRT,
where ∆n = no. of moles of gaseous products – no. of moles of gaseous reactants
for the reaction of one mole of the substance of interest.
For instance,
C2H6O(l) + 3O2(g) → 2CO2(g) + 3 H2O(l);
∆n = 2 – 3 = –1.
C2H6(g) + (7/2)O2(g) → 2CO2(g) + 3 H2O(l); ∆n = 2 – 4.5 = –2.5.
Problem 2.55
Standard State:
By general agreement, the standard state of a substance is defined as the most
stable form of that substance at 298.15 K and 1 bar pressure. Therefore, oxygen is
a gas in its standard state and water is a liquid.
Standard state of a solution is defined as 1 mol kg–1 or a 1 m solution.
Calorimetry: Example 2.5, Problems 2.5, 2.10, 2.15, 2.17, 2.32, 2.35
5
Temperature dependence of Enthalpies
Since dH = CPdT, the temperature dependence of enthalpy is related to the
temperature dependence of CP, which is commonly expressed as
CP(T) = d + eT + fT–2,
where the constants d, e and f are experimentally determined for a large number of
gases (See Table 2.1, p. 69). Then,
T2
,H = ° T æè d + eT + f2 öø dT
T
1
= d(T 2 − T 1 ) + e æè T 22 − T 21 öø − f æ T1 − T1 ö
2
è 2
1 ø
Problems 2.19, 2.22, 2.46.
Temperature Dependence of Enthalpies of Reaction
Enthalpy changes during reaction are normally calculated under standard
conditions using tables of standard enthalpies of formation. Consider the reaction
nAA + nBB → nCC + nDD
The standard enthalpy of reaction is calculated as:
, r H = (n C , f H C + n D , f H D ) − (n A , f H A + n B , f H B ).
If the temperature dependence of the CP,m’s of each reactant and product are also
given in terms of the parameters d, e and f, we can define
,d = (n C d C + n D d D ) − (n A d A + n B d B ),
and similarly, ∆e and ∆f. Then, to evaluate the enthalpy of reaction at a certain
temperature θ, we can simply evaluate
, r H D = , r H + ° 298.15 K (,d + ,eT + ,fT −2 )dT
D
More generally, if we know the enthalpy of reaction at a temperature T1, we can
calculate the enthalpy at another temperature T2. See Eq. (2.51) and (2.52).
Example 2.6.
6
Reversible Processes Involving Ideal Gases (Section 2.6)
(a) Reversible isobaric expansion/compression:
w rev = − ° V PdV = −P(V 2 − V 1 ).
V2
1
q rev = ° T C P dT = C P (T 2 − T 1 ), if CP is constant.
T2
Fig. 2.6 (a)
1
,U = q rev + w rev , (first law!) and
,H = q rev = q P .
(b) Reversible isochoric expansion/compression:
w rev = 0!
q rev = ° T C V dT = C V (T 2 − T 1 ) if CV is constant.
T2
1
,U = q rev = q V .
,H = ° T C P dT = C P (T 2 − T 1 ), by definition!
T2
Fig. 2.6 (b)
1
(c) Reversible isothermal expansion/compression:
V2
V2
w rev = − ° V PdV = −nRT ° V dV
1
1 V
V
P
= −nRT ln æ V 2 ö = nRT ln æ P 2 ö
è 1ø
è 1ø
Fig. 2.6 (c)
,U = ,H = 0!
since T2 = T1.
q rev = −w rev , from first law.
7
(d) Reversible adiabatic expansion/compression:
In adiabatic process, by definition, q = 0. Therefore, in an adiabatic expansion, the
internal energy decreases (i.e., the temperature decreases) and in an adiabatic
compression, internal energy increases (i.e., the temperature increases)..
,U = C V (T 2 − T 1 ),
q rev = 0, and therefore,
Fig. 2.6 (d)
w rev = ,U = C V (T 2 − T 1 )
,H = C P (T 2 − T 1 ).
An isotherm is characterized by the equation PV = constant, or P1V1 = P2V2. We
now try to derive an equation that characterizes an adiabat.
From First Law, we have
for ideal gas.
dU = −PdV, or C V,m dT = − RT
V dV,
Rearranging, we get
dV which, when integrated, yields
C V,m dT
T = −R V ,
T
V
C V,m ln æ T 2 ö = R ln æ V 1 ö
è 1ø
è 2ø
This immediately gives the result
(2.86)
T 2 æ V 1 ö R/C V,m
, or using the fact that CP,m – CV,m = R,
T1 = è V2 ø
T 2 æ V 1 ö (C P,m −C V,m )/C V,m æ V 1 ö ?−1
= V
, where γ = CP,m/CV,m.
T1 = è V2 ø
è 2ø
We now eliminate temperature by using ideal gas law to get
P 2 V 2 æ V 1 ö ?−1
P2 æ V1 ö ?
=
,
or
P1 V1 è V2 ø
P1 = è V2 ø ,
which is equivalent to P1V1γ = P2V2γ or PVγ = constant.
Reading Assignment:
Bond Enthalpies (p. 73)
(2.92)
(2.90)
8
Reversible Isothermal Expansion of a Real Gas
We will focus on the van der Waals equation.
2
P = nRT + n 2a .
V − nb V
Therefore,
2
V2
V2
w rev = − ° V PdV = ° V éë nRT + n 2a ùû dV,
V
1
1 V − nb
which, upon integration, yields
V − nb ö
− n 2 a æ V1 − V1 ö
w rev = −nRT ln æ 2
è 2
è V 1 − nb ø
1 ø
(2.123)
From Chapter 3, we use the relationship
2
dU = n 2a dV,
V
(2.124)
2
V2
,U = ° V nV 2a dV = −n 2 a æ V1 − V1 ö
è 2
1
1 ø
(2.126)
and thus get
Now, using the first law,
V − nb ö
.
q rev = ,U − w rev = nRT ln æ 2
è V 1 − nb ø
Also, since initial and final pressures and volumes are known (or can be found),
,H = ,U + (P 2 V 2 − P 1 V 1 )
In order to calculate ∆U, ∆H etc. for isochoric and isobaric processes, we need to
be given the values of CV,m and CP,m for the gases.
It is not easy to derive the equation for an adiabat for a real gas
Example 2.10, Problems 2.37 – 2.45, 2.49 – 2.51, 2.54, 2.56, 2.60, 2.63, 2.64