6
Absorption of Heat
According to the First Law,
∆E = q + w = q - P∆V, assuming P-V work is the only kind that can occur.
Therefore, ∆E = qV. The subscript means that the process occurs at constant volume.
So, if we can control our system and maintain constant volume, we know that if we inject heat
into a constant volume system then we know the change in internal energy.
qV
For an ideal gas, an increase
in E results in a temperature
increase.
∆T
So, what about constant pressure?
∆E = q - P∆V; q = ∆E + P∆V. Let’s define H = E + PV, the
enthalpy.
P
∆H = ∆E + ∆(PV) = ∆E + P∆V + V∆P
V
So, ∆H =q + w + P∆V + V∆P = qP at
constant pressure.
Thus, ∆E = qV
∆H = qP
Internal energy is the appropriate energy variable to use at constant
volume
Enthalpy is the appropriate energy variable to use at constant pressure
Note that at constant pressure, some of the heat goes into expanding (or compressing) the
volume, so for a fixed amount of heat injection, the temperature increase should be smaller for
the constant pressure case.
qP
∆T + work
7
This provides a nice lead in for the concept of heat capacity.
Definitions of Heat Capacities
Suppose we inject a given amount of heat q into a system consisting of one mole of a
substance, and we observe a temperature rise.
We say q = C∆T for one mole. C has units of Joules per mole per degree. This is the
molar heat capacity.
For n moles, we write q = nC∆T
C is the molar heat capacity
Sometimes we can express heat capacity per gram. We call these quantities specific
heats. “Specific” always means per gram. We denote these quantities as c .
For m grams, we write q = m c ∆T.
c has units of Joules per gram per degree.
So, lets go back to heat absorption processes occurring at constant volume or pressure.
At constant volume, Cv = qV/∆T = ∆E /∆T
At constant pressure, CP = qP/∆T = ∆H /∆T
Cp vs Cv
Let’s take two identical systems of gases, inject identical amounts of heat. Let one
system be held at constant volume, the other at constant pressure. Measure ∆T. What
happens?
At constant volume, all heat goes into increasing temperature. At constant pressure,
some of the heat goes into increasing the volume of the gas, the rest into increasing the
temperature. So, we expect ∆T (constant V) > ∆T (constant pressure). Since ∆T =
q/C, that means that Cp > Cv. At constant pressure, the gas must do work on the
surroundings, or the surroundings do negative work on the system.
For one mole of a gas, CP = CV + R
For solids, CP ≈ CV
Let’s do some simple examples:
Take one mole of helium gas, inject 100 J of energy, measure ∆T
8
At constant volume, CV = 12.5 Joules per Kelvin per mole.
∆T =Q/C = 100 J / [(12.5 J K-1 mol-1) ( 1 mol)] = 8.0 degrees. Note that the degree size
is the same for Celsius and Kelvin, so if the gas is initially at 0 °C, T goes to 8.0 °C.
The gas is also at 273.15 K, so its new temperature is 281.15 K.
At constant pressure, CP = CV + R = 12.5 + 8.3 J K-1 mol-1 = 20.8 J K-1 mol-1
Then ∆T =Q/C = 100 J / [(20.8 J K-1 mol-1) ( 1 mol)] = 4.8 degrees. The difference
arises from the fact that the container expands, so some of the energy is converted into
work. This amount of work “steals” 3.2 degrees of the temperature rise.
What does heat capacity mean?:
Heat capacity tells us about the atomic and molecular motions of the system.
In a gas, atoms can absorb heat in their kinetic energy. Remember from kinetic theory
that for a monatomic gas,
KE = 3 kT This is the result for three dimensions.
2
Molecules can also rotate:
and vibrate:
Rotations generally add ½RT of energy per mole per rotation. Linear molecules
contribute two rotations, nonlinear molecules 3 rotations.
Vibrations in molecules in the gaseous state generally do not make a full contribution of
RT to the internal energy. Vibrations in solids are the only modes of internal energy
available to absorb heat. A solid generally has a heat capacity of 3R J K-1 mol-1 ≈ 25 J K1
mol-1
Calorimetry
9
We can use the ideas of the First Law to understand how two objects at different
temperatures reach equilibrium. A styrofoam coffee cup makes a pretty good calorimeter. In
this case, q = 0, since no heat is exchanged with the surroundings.
Example: Place 100 g of H2O( l ) at 25 °C in an insulated cup. Drop 30.0 g of Fe at 150
°C in the cup. Calculate the final state of the system.
q(Fe) + q(H2O) = 0
q(Fe) = moles Fe × heat capacity of Fe × (Tfinal – Tinitial)
q(H2O) = moles H2O × heat capacity of H2O × (Tfinal – Tinitial)
Work with molar heat capacities
C(Fe) = 25 J K-1 mol-1
C(H2O) = 75.3 J K-1 mol-1
Moles Fe = 30.0 g / 55.8 g mol-1 = 0.54 mol
Moles H2O = 100 g / 18 g mol-1 = 5.55 moles
Tinitial (H2O) = 25 °C
Tinitial (Fe) = 150 °C
(0.54)(25)(Tfinal – 150) + (5.55)(75.3)(Tfinal – 25) = 0
417.9 Tfinal – 10448 +13.5 Tfinal – 2025 = 0
431.4 Tfinal = 12473
Tfinal = 28.9 °C
In a laboratory calorimeter,
combustion is often carried out.
The ∆T is measured, and the heat
evolved when M grams of
substance undergoes combustion is
calculated from the heat capacity of
the calorimeter and its contents.
Note that the units are J per degree.
No moles involved in this
calculation.
Example: Suppose 1.00 grams of
a substance with a formula weight
of 100 g /mole undergoes
combustion in a calorimeter. The
10
heat capacity of the calorimeter and its contents is 19,000 J per degree. A ∆T of 2.20 °C
is observed. What is the ∆E for combustion?
qV = 19,000 J/deg × 2.20 °C = 41,800 J, so this is a change in internal energy.
This heat is per 1g/100 g/mol = 0.01 moles of compound
So ∆E = -qV/moles = -4180 kJ/mole.
Why is the sign for the reaction negative? Positive q is absorbed by the surroundings.
Therefore negative q is absorbed by the system.
Properties of Enthalpy
The enthalpy is an especially useful quantity for calorimetry, since most chemical
reactions occur at constant pressure, 1 atm.
The units of ∆H are J mol-1. So, for the reaction
∆H = -283.0 kJ. 283.0 kJ of energy are liberated
CO(g) + ½O2(g) → CO2(g)
when one mole of CO2 is formed. So, 2 × 283.0 = 566.0 kJ is liberated when 2 moles of
CO2 are formed. H is an extensive thermodynamic variable.
H = E + PV. E, P, V are thermodynamic functions of state (they are not interactions
between the system and the surroundings) so H is also a state function.
∆H is independent of path. This is a very important concept.
CO(g)
∆H1
½O2(g) ∆H2
½O2(g)
C(s) + ½O2(g)
CO2(g)
∆H3
We can make CO2 directly via the lower pathway, or indirectly, via the upper, stepwise
pathway.
∆H1 = -110.5 kJ mol-1
∆H2 = -283.0 kJ mol-1
∆H3 = -393.5 kJ mol-1