Experiments
Experiment 3
Heat Capacity Ratio For Gases1
Summary - The purpose of this experiment is to determine the
value for the heat capacity ratio for two gases, a monatomic gas and a
diatomic gas. The adiabatic expansion method, a relatively crude
method, will be employed. The results are compared to those predicted
from simple theory. For the diatomic gas comparison is made to both the
classical and quantum mechanical predictions.
Heat Capacity2
The heat capacity, C, of a substance is defined as
C = lim T0 (q/T) = (dq/dT)
Since q is not a state function heat capacity is also in general not a state
function. However, if we restrict ourselves to certain processes we can
make C a state function. In particular, we may show the following
CV = (dq/dT)V = (U/T)V
constant volume heat capacity
Cp = (dq/dT)p = (H/T)p
constant pressure heat capacity
Since the right-hand terms above are state functions both CV and Cp are
also state functions.
General Relationships and Definitions
Consider one mole of an ideal gas. Using a general relationship
that applies to all substances3 we may show the following to be true
Cp,m - CV,m = R ; or Cp,m = CV,m + R
Cp,m and CV,m are the constant pressure and constant volume molar heat
capacities, respectively, that is, the heat capacities per mole of ideal gas.
Finally, we may define , the heat capacity ratio, as
 = (Cp,m/CV,m) = (Cp/CV)
Note that we can (but are not required to) use the molar heat capacities to
define . The heat capacity ratio is a useful quantity in the discussion of
the adiabatic reversible expansion of an ideal gas.4
Total Energy For One Mole of an Ideal Gas
We may use the relationship CV,m = (U/T)V,m to find the
constant volume molar heat capacity for an ideal gas. We start by saying
that the total energy for one mole of an ideal gas is
Utotal = Etrans + Erot + Evib
where total = total energy
rot = rotational energy
trans = translational energy
vib = vibrational energy
In writing this we are assuming a sharp separation between rotational and
vibrational energy and neglecting electronic energy.
We now look at the contributions to the total energy according to
classical mechanics. After this, we consider the effects of quantum
mechanics on our results.
Classical Result For Total Energy
The average translational, rotational, and vibrational energy of
one mole of an ideal gas may be found using the equipartition theorem,5
a general result that can be derived from statistical mechanics. This
theorem predicts the energy per translational, rotational, and vibrational
degree of freedom.
Etrans = 1/2 RT per translational degree of freedom
Erot = 1/2 RT per rotational degree of freedom
Evib = RT per vibrational degree of freedom
For a gas molecule composed of N atoms, there are 3N degrees of
freedom, which may be assigned as follows.
translation = 3
rotation = 2 (linear molecule); 3 (non-linear molecule)
vibration = 3N - 5 (linear molecule); 3N - 6 (non-linear molecule)
CV,m, Cp,m, and 
For a monatomic ideal gas there is only translational energy, so
Utotal = Etrans = 3/2 RT
CV,m = (U/T)V,m = 3/2 R
Cp,m = CV,m + R = 5/2 R
 = Cp,m/CV,m = 5/3 = 1.66...
For a diatomic ideal gas there is translational, rotational, and
vibrational energy, so
Utotal = Etrans + Erot + Evib = 3/2 RT + RT + RT = 7/2 RT
CV,m = (U/T)V,m = 7/2 R
Cp,m = CV,m + R = 9/2 R
 = Cp,m/CV,m = 9/7 = 1.28...
Quantum Mechanics
The above relationships were derived using the virial theorem, a
classical result that assumes that translational, rotational, and vibrational
energy are continuous (can take on any value within a particular range).
However, a consequence of quantum mechanics is that these
energies are quantized, that is, that they can only take on particular
values. At room temperature this only affects the expression for
vibrational energy.
At low temperatures (relative to vib, the
characteristic temperature for a particular vibration6), Evib  0. As
temperature increases the molecule will gain vibrational energy, though
the amount will in general be smaller than predicted by the virial
theorem.
Quantum Effects on 
We now consider the effects of quantum mechanics on our
theoretical value for .
Monatomic ideal gas. Since there is only translational energy,
including quantum mechanics has no effect on the values for CV,m, Cp,m,
and .
Diatomic ideal gas. Since a diatomic molecule has one vibration,
quantum mechanics does have an effect on the values for CV,m, Cp,m, and
.
“low T”
“high T”
CV,m
5/
2
R
7/
2
R
Cp,m
7/
2
R
9/
2
R

7/
5
= 1.40...
9/
7
= 1.28…
In the above table “low T” corresponds to T << vib, and “high T” to T
>> vib. For T ~ vib (“intermediate T”) CV,m, Cp,m, and  will fall
Experimental
You should do a minimum of five trials for both your monatomic
gas (Ar) and your diatomic gas (N2). If you can do more trials you will
get better statistics in your data analysis. You should compare your
experimental results to the theoretical values expected for a monatomic
gas and a diatomic gas. For the diatomic gas you should compare with
both the “classical” and “quantum” theoretical result.
You do not need to answer the questions posed in the discussion
section of the lab text. You should, however, discuss possible random
and systematic errors associated with carrying out the adiabatic
expansion of the gas.
If separate groups carry out the measurements for Ar and N2 you
should share your data with one another, and carry out the data analysis
for both gases.
References
1) Experiments in Physical Chemistry, 8th Edition, 2009. C. W. Garland,
J. W. Nibler, D. P. Shoemaker. p. 106-114.
2) Physical Chemistry, 9th Edition, 2010. P. W. Atkins, J. de Paula. Heat
capacity is discussed on p. 57-63.
3) Atkins, p. 78; 84-85 (Further information 2.2); 122-124.
4) Atkins, p. 63-64; 68.
5) Atkins, p. 9; 47.
6) Atkins, p. 599 for a definition of vib. A table of values for vib is
given in the Appendix (Table 12.2, p. 933).
Experiment 6
Heat of Combustion1
Summary - The purpose of this experiment is to determine the
value for the heat of combustion for naphthalene or another organic
compound. The experimental method that will be used is bomb
calorimetry. Data on the combustion of benzoic acid will be used to
determine the heat capacity for the calorimeter. This will then be used to
analyze the experimental data for naphthalene. The experimental value
for the enthalpy of combustion for naphthalene will be compared to a
literature value.
The Combustion Reaction2
By definition, combustion refers to the reaction of one mole of a
single substance with oxygen (O2(g)) to form combustion products. The
combustion products for a few common elements are as follows
carbon (C) – CO2(g)
nitrogen (N) – N2(g)
hydrogen (H) – H2O()
sulfur (S) – SO2(g)
halogens (X) – X2
When a small sample of a compound is burned in a large excess of
oxygen carbon and hydrogen will be essentially completely converted
into CO2(g) and H2O(). For elements such as nitrogen, sulfur, and the
halogens a complex mixture of products is likely obtained, and so the
“combustion products” listed here are chosen for convenience in doing
thermochemical calculations.
Examples
1) Combustion of n-octane (C8H18()).
C8H18() + 25/2 O2(g)  8 CO2(g) + 9 H2O()
2) Combustion of aniline (C6H5NH2(s)).
C6H5NH2(s) + 31/4 O2(g)  6 CO2(g) + 7/2 H2O() + ½ N2(g)
Enthalpy of Combustion
By definition Hcomb, the enthalpy of combustion (sometimes
also called the heat of combustion), is the enthalpy change when one
mole of a single compound undergoes combustion for standard
conditions. Standard conditions3 are usually taken to be p = 1.00 bar
(approximately the same as 1 atm) and T = 25.0 C, though different
standard temperatures are sometimes used. For example, for the
combustion of n-octane
C8H18() + 25/2 O2(g)  8 CO2(g) + 9 H2O()
Hcomb(C8H18()) = - 5471. kJ/mol
Note that the enthalpy of combustion is negative, meaning the reaction is
exothermic. This is commonly the case for combustion reactions.
Bomb Calorimetry
The most common experimental technique for determining the
enthalpy of combustion for a substance is bomb calorimetry. A sample
of the compound in question is burned under conditions where a large
excess of oxygen is present and the heat that is evolved is measured.
Bomb calorimetry is discussed in detail in the text. Since the combustion
reaction is carried out at constant volume in bomb calorimetry it is
Ucomb, the energy of combustion, that is measured, from which Hcomb
can be found as discussed below.
Calculations
There are two key equations for q, the value for heat for the
combustion reaction
q = mC Ucomb(C) + mW Ucomb(W)
(6.1)
q = - Cbomb T
(6.2)
In the above equations q is the value for heat for the combustion
reaction. C and W refer to the compound (C) and wire (W) (a small
amount of wire, used to initiate combustion, reacts with oxygen and must
be accounted for in the equations), which have energies of combustion
Ucomb(C) and Ucomb(W), respectively (these are usually given in units
of J/g) and masses mC and mW. Cbomb is the heat capacity of the bomb,
and T is the experimental change in temperature observed for the
combustion reaction.
Determination of Cbomb
A convenient method for finding the value for Cbomb, the heat
capacity of the calorimeter, is to burn a known amount of a compound
for which Ucomb is known. Benzoic acid (C6H5COOH(s)) is a
convenient compound to use because it has an accurately known value
for the energy of combustion4, can be obtained at high purity, can be
easily pressed into pellets, and burns completely to form combustion
products.
If we burn a known amount of benzoic acid (and wire)4 and
measure the temperature change associated with combustion, we may
use equation (6.1) above to find q. Equation (6.2) can then be used to
find Cbomb. Assuming that the value for Cbomb is insensitive to the
compound being burned, this value can be used to determine q for the
combustion of a known mass of a different compound, such as
naphthalene, using equation (6.2). Substituting this value for q into
equation (6.1) gives a relationship where the only unknown is Ucomb for
the compound, which can then be found from the equation.
Conversion of Ucomb To Hcomb
The experimental data in this experiment are sufficient to
determine Ucomb, the energy of combustion. However, literature
values are usually given as Hcomb, the enthalpy of combustion. These
two quantities are related by the approximate expression5
Hcomb  Ucomb + ngRT
where Ucomb and Hcomb are given on a per mole basis, and  ng is the
change in the number of moles of gas per mole of reaction (which would,
for example, be equal to – 9/2 for the combustion of n-octane). The
expression approximate because it is derived assuming all gas reactants
and products behave ideally and that the volume of solid and liquid
reactants and products is negligible.
Experimental
You should do a minimum of three trials with benzoic acid to
determine the heat capacity of the calorimeter, and three trials with
naphthalene to determine the enthalpy of combustion. For each trial
there are only three final experimental results that are obtained and used
in the calculations; mC, the mass of compound burned, mW, the mass of
wire burned, and T, the temperature change for combustion. There are
several methods that can be used to determine T, and so you should
clearly indicate the method you have used.
After finding the Ucomb for naphthalene you will need to
convert this to Hcomb by the method indicated above. This will require
the balanced combustion reaction for naphthalene, which should be
given in the lab report. You should compare your value for Hcomb for
napthalene to a literature value (available in Atkins) and discuss what
you find. You should think about, but do not need to answer, the
questions in the discussion section of your lab manual.
References
1) Experiments in Physical Chemistry, 8th Edition, 2009. C. W. Garland,
J. W. Nibler, D. P. Shoemaker. p. 145-158.
2) A good general discussion of thermochemistry, including combustion,
is given in Physical Chemistry, 9th Edition, 2010. P. W. Atkins, J. de
Paula, p. 65-72.
3) Atkins, p. 5; 65.
4) Ucomb(benzoic acid) = - 26.41 kJ/g; Ucomb(wire) = - 6.68 kJ/g, from
Garland.
5) Atkins, p. 57-59.
Experiment 13
Vapor Pressure of a Pure Liquid1
Summary - The purpose of this experiment is to determine the
value for the enthalpy of vaporization for water. The boiling point
method, in which the pressure at which a liquid boils is measured as a
function of temperature, will be used. Two independent trials will be
carried out and compared to each other and to the literature value for the
enthalpy of vaporization.
Vaporization
Vaporization refers to the process
M()  M(g)
where M is a pure chemical substance. The amount of heat that must be
added to convert one mole of a liquid in equilibrium with its vapor into
gas is called Hvap, the enthalpy of vaporization. When vaporization
occurs at a pressure p = 1.00 atm, then the temperature is called the
normal boiling point for the liquid, and the enthalpy change is called
Hvap, the standard enthalpy of vaporization.
Thermodynamics of Vaporization
For general phase transitions occurring among the solid, liquid, or
gas phases of a pure substance the Clapeyron equation2 applies
dp = Spt = Hpt
dT
Vpt
T Vpt
(13.1)
where “pt” stands for “phase transition” (fusion, vaporization,
sublimation, or a solid to solid phase transition). The above equation can
be solved for Hpt and therefore used to find the enthalpy change for a
phase transition.
The Clausius-Clapeyron Equation
For vaporization (or sublimation) where the final phase is the
vapor phase, the Clapeyron equation can be rewritten based on the
following assumptions:
1) The enthalpy of vaporization is approximately independent of
temperature.
2) The vapor phase can be described by the ideal gas law.
3) The volume of the liquid phase is negligible.
Based on these assumptions one obtains the Clausius-Clapeyron
equation3
d ln p = - Hvap
(13.2)
d (1/T)
R
This suggests that the value for the enthalpy of vaporization can be found
by finding the slope of a plot of ln p vs 1/T.
Effect of Nonideal Behavior of Water Vapor
The Clausius-Clapeyron equation assumes that water vapor
behaves as an ideal gas. However, since water is a polar molecule,
nonideal behavior is expected even at low temperature.
If we define the compressibility factor, Z, as
Z = pVm
RT
(13.3)
then the Clausius-Clapeyron equation may be rewritten4 as
d ln p = - Hvap
(13.4)
d (1/T)
RZ
where Z is the compressibility factor for water vapor. At the normal
boiling point, T = 100.0 C, the experimental value for the
compressibility factor for water vapor is Z = 0.986.5
Temperature Dependence of Hvap
The Clausius-Clapeyron equation also assumes that the value for
the enthalpy of vaporization, Hvap, is independent of temperature. In
fact the enthalpy of vaporization is temperature dependent, as is true for
enthalpy changes for other processes.6 This leads to curvature in a plot
of ln p vs 1/T. In the present experiment the amount of curvature
expected is small compared to the experimental error in the
measurements, and so likely will not be noticed.
The enthalpy of vaporization measured in the present experiment
corresponds to the value at the average temperature of your data. It can
be adjusted to the value expected at T = 100.0 C as discussed below.
Accounting For the Dependence of Hvap on T
The general expression for the temperature dependence of the
enthalpy of reaction,6 applied to the process of vaporization, gives
Hvap(T) = H(Tave) + TaveT (Cp,m(g) - Cp,m()) dT
where Hvap(T) is the enthalpy of vaporization at the normal boiling
point temperature T, H(Tave) is the experimental enthalpy of
vaporization, corresponding to the value at the average temperature of
the data, and Cp,m() and Cp,m(g) are the constant pressure molar heat
capacities of the liquid and gas.
Accounting For the Dependence of Hvap on T
If we assume that Cp,m() and Cp,m(g) are independent of
temperature (true if T - Tave is small) then
Hvap(T)  H(Tave) + (Cp,m(g) - Cp,m()) (T - Tave)
You should use the above equation to adjust your value for the enthalpy
of vaporization to the value expected at T = 100.0 C, the normal boiling
point for water. Use Tave = 90.0 C, Cp,m() = 75.29 J/mol.K, and Cp,m(g)
= 33.58 J/mol.K in doing this correction.
Experimental
Two independent trials should be carried out for water. Each trial
should be analyzed independently, and a value for Hvap, along with the
confidence limits for this value, should be found from a plot of ln p vs
1/T, and using Z = 0.986 for the compressibility factor for water vapor.
The experimental values should then be compared to each other, and to a
literature value for the enthalpy of vaporization (available in Atkins).
The pressure data do not need to be corrected for the small change in the
density of mercury with temperature.7
References
1) Experiments in Physical Chemistry, 8th Edition, 2008. C. W. Garland,
J. W. Nibler, D. P. Shoemaker. p. 199-207.
2) Physical Chemistry, 9th Edition, 2010. P. W. Atkins, J. de Paula, p.
146-147.
3) Atkins, p. 148-149.
4) Garland, p. 200.
5) Garland, p. 201.
6) Atkins, p. 73-74; Garland, p 201.
7) These corrections are discussed in Garland. The correction for the
difference in the density of mercury at room temperature (about 20. C)
and 0. C is small (~ 0.4 %), and to a first approximation cancel out.