Notes on the course of
Physical Chemistry
Chem 230
For the second term
of the scholar year 1427/28
Text Book:
3d Edition of “Physical Chemistry” book
written by P.W. ATKINS
Dr. Ammar Tighezza
Office: 2A19 B5
Phone: 4675954
-1-
‫القواعد المنظمة لسير الدراسة والتقويم‬
‫في المقرر ‪ 032‬كيم‬
‫‪ .1‬هناك إختباران فصليان بـ ‪ 02‬درجة لكل واحد منهما (األول يوم اإلثنين ‪ 4101/3/41‬و‬
‫الثاني يوم اإلثنين ‪ )4101/1/02‬و اختبارنهائي من ‪ 02‬درجة‪.‬‬
‫‪ .2‬هناك ثالثة إختبارات فجائية بمجموع عشرة درجات (‪ ،)3 + 3 + 4‬لذا يجب اصطحاب آلة‬
‫حاسبة في كل المحاضرات‪.‬‬
‫‪ .3‬من غاب أي اختبار بدون عذر شرعي مقبول فله صفر في ذلك اإلختبار الذي غاب عنه‪.‬‬
‫‪ .4‬من كان له عذر شرعي مقبول فتحتسب له نفس درجة اإلختبار الموالي (أو نسبة منها)‪.‬‬
‫‪ .5‬ال تعطى أي درجات على الحضور في المحاضرات ولكن تطبق لوائح الجامعة المعروفة‬
‫بالنسبة للحرمان من دخول اإلختبار النهائي‪.‬‬
‫‪ .6‬يستفيد الطالب الذي لم تزد غياباته عن ‪ 5‬غيابات (بعذر أو بدون عذر) من مساعدة بمقدار‬
‫درجتين للحصول على األمتياز األعلى أو ‪ 3‬درجات للنجاح‪.‬‬
‫‪ .7‬ال يستفيد الطالب الذي زادت غياباته عن ‪ 5‬غيابات (بعذر أو بدون عذر) من أي مساعدة و‬
‫ال يستقبل في الساعات المكتبية‪.‬‬
‫‪ .8‬لن ينفع أي تسول في نهاية الفصل من أجل زيادة الدرجات‪ ،‬بل سيلغي اإلستـفادات المذكورة‬
‫في النقطة السادسة‪.‬‬
‫‪ .9‬ال يتميز الطالب الحامل للمادة عن الطالب المستجد و تطبق القواعد المذكورة على الجميع‪.‬‬
‫‪-2-‬‬
Contents of Chem_230
1. Molecular kinetic theory of gases
 the kinetic-molecular gas model
 the pressure of a gas
 kinetic energies and temperature
 numerical values of molecular energies and molecular speeds
 distribution of molecular velocities
 the main free path, collision diameter, and collision number
 numerical values of collision properties
 van der Waals’ equation
2. First law of thermodynamics
 measurement and relation of thermal and mechanical enrgy
 the first law of thermodynamics
 determination of ∆E: reversible and Irreversible processes
 Work for expansions and contractions of the system
 the enthalpy function
 some properties of State functions
 Dependence of E and H of an ideal gas on P and T
 Adiabatic expansions of ideal gas
3. Thermochemistry
 measurement of heats of reaction
 Internal-energy and enthalpy changes in chemical reaction
 Relation between ∆E and ∆H
 Indirect determination of heats of reaction
 Standard heats of Formation
 Temperature dependence of heats of reaction
 Bond enthalpies and Bond Energies
4. Entropy and the second and third laws of thermodynamics
 General Statements of 2nd Law of thermodynamics
 Entropy and another Statement of 2nd Law of thermodynamics
 The Carnot Cycle
 The efficiency of the transformation of heat into work
 The 3rd law of thermodynamics
5. Free Energy and chemical equilibria
 A convenient measure of the driving force of a reaction: the Free Energy
 Standard Free Energies
 Dependence of Free Energy on P and T
 Quantitative relation of ∆G and the equilibrium constant
6. Adsorption and heterogeneous catalysis
 Classifications of adsorptions of gases on solids
 Heat of adsorption
 The Langmuir adsorption isotherm
 Determination of surface areas
-3-
Reviewing Ideal Gases Laws
PV = Constant
when
T and n are
Constant
P/T = Constant
when
V and n are
Constant
V/T = Constant
when
P and n are
Constant
PV/T = Constant
when
n is Constant
PV=nRT
PM=dRT
d=m/V ,
n=m/M ,
-4-
n = N / NA
Exercises
1) A sample of pure gas at 27°C and 380 torr- occupied a volume of 492 mL. What
is the number of molecules of this sample?
2) The supersonic airplane produces 6.6xl04 kg of C02 every hour of flight. The
volume (in liters) of this amount at 4.1x105 atm and 627°C is:
3) A 0.30 g sample of a certain gas occupies a volume of 82 mL at 304 kPa and
27°C. What is its molar mass and molecular mass ?
4)
Given a 2.8 g sample of N2 in a rigid container at STP.
a: What is the volume of this sample ?
b: What will be the pressure when heated to 57 oC ?
5) At 500 oC and 700 Torr, the density of sulfur vapor is 3.7 g L-1. What is the molar
mass of sulfur under these conditions ?
6)
A balloon that contains 5.41 L of helium at 25 °C has a pressure of 101.50 kPa. If the
balloon is heated to 35 °C and the pressure is changed to 102.80 kPa, what would be the final
volume in L?
7)
The mass percentage composition of dry air is: N2, 75.5; O2, 23.2; Ar, 1.3.
a: The mole fraction of O2 is:
A) 0.79
B) 0.33
C) 0.21
D) 0.54
b: The total pressure of air is 3 atm, then what is the partial pressure of O2 ?
8) A mixture of cyclopropane and oxygen can be used as an anesthetic. If the partial
pressures of cyclopropane and oxygen are 170 torr and 570 torr, respectively, in a gas
cylinder, then what is the mole ratio n(C3H8)/n(O2) and the mass ratio m(C3H8)/m(O2)?
9) A 50.0 L container is initially divided into two equal halves by a partition. One
half contains CO gas at a temperature of 25 oC and a pressure of 4.7 atm. The other half
contains O2 gas at the same temperature and a pressure of 3.5 atm. The partition is
impenetrable to gas molecules of both kinds. The partition is then removed and the
gases are allowed to mix.
a. Assuming no reactions occur and that the gases can be treated as ideal gases,
what is the total pressure in the container after the gases have mixed.
b. The container is then heated to 500 C and the two gases react to form CO2 gas
according to the equation:
2 CO + O2  2 CO2
What is the final pressure in the container after all reactions have ceased?
Additional exercises: (Page 34)
A1.1
(a) 2570 torr (b) 3.38 atm
A1.2
401 K
A1.3
4.27 kPa
A1.4
(a) 3.14 L
(c) 2.83x104 Pa
(b) 3.47 kPa
-5-
Kinetic Molecular Theory of Gases
Kinetic molecular theory, KMT for short, is very different from
thermodynamics although it deals with some of the same variables, such as
pressure, temperature, volume, and density.
Thermodynamics does not care whether molecules exist or not.
Essentially all the results of thermodynamics can be worked out without any
assumptions concerning the particle nature of matter. KMT, however, is a
molecular theory (or a molecular model) of matter. (Molecular theories of
matter are sometimes called "microscopic" models as opposed to the
"macroscopic" model of bulk material. KMT is a microscopic model of a
gas.)
1. The Kinetic-Molecular Gas Model:
Assume that the gas consists of:
 N molecules (N is large, on the order of Avogadro's number, NA =
6.023 × 1023 molecules/mol)
 The molecules have mass, m.
 The molecules are contained in a volume, V.
 The gas is at some temperature, T (which is the Kelvin temperature).
 The dimensions of the molecules are small compared to the average
distance between molecules and compared to the size of the container.
 There is no potential energy of interaction between the molecules.
 The molecular motion is random, or chaotic, so that the gas is
isotropic. "Isotropic" means that the properties of the gas are the same
in every direction.
2. The Pressure of a Gas
Pressure is defined as the force per unit area:
In the kinetic-molecular theory of gases, pressure is the force exerted
against the wall of a container by the continual collision of molecules
against it. To calculate the pressure we need to determine the force exerted
by gas molecules colliding with wall A.
-6-
The force exerted by a molecule of mass m colliding with wall A can
be calculated from
.
The last quantity in the above equation can be determined if we know the
change in velocity per collision with wall A and the time between collisions
with wall A.
A collision with wall A will reverse only the x-component of the
velocity. If we assign the average initial x-component of the velocity before
collision as -ux and the final x-component of the velocity after collision to
ux then the change in velocity with each collision with wall A is
∆u = ux – (-ux) = 2 ux
The time between collisions with wall A will again depend upon the
x-component of the velocity and the distance travelled by the gas molecule
(along x) between collisions. In our box a gas molecule, after colliding with
wall A, would have to travel along x to the opposite wall, a distance of a,
and back again to wall A, for a total distance travelled along x of 2a. Thus
the time between collisions with wall A would be
and the force exerted by one gas molecule of mass m colliding with wall A
becomes
-7-
Rearranging we find:
pV = mux2
per gas molecule.
For N molecules we have:
N
P
i
(mux2 )i m N 2
m
  (ux )i  N ux2
V
V i
V
Recognizing that the velocity is related to its components by the
Pythagorean Theorem and that, on average, each of the components are
equal ( ux  u y  uz ) we find:
u 2  ux  u y  uz  3ux
2
Then we obtain:
Rearranging we find:
P
2
2
2
m u2
N
V
3
PV 
This equation can also be written as:
1
N mu 2
3
PV 
-8-
1
n M u2
3
(1)
(1bis)
3. Kinetic Energies and Temperature:
The average kinetic energy of one molecule is:
ke 
1
mu 2
2
Equation (1) can then be changed to:
PV 
PV 
2 1
N ( mu 2 )
3 2
2
2
N ke  n( N A ke)
3
3
PV 
2
n KE
3
(2)
where KE is the kinetic energy of Avogadro’s number of molecules.
Comparing this last equation (2) with:
we can find that:
KE 
PV = nRT
3
RT
2
(3)
and:
ke 
with: k =
KE
3 R
3

T  kT
NA
2 NA
2
(4)
R
called Boltzmann’s constant.
NA
4. Molecular Speeds:
The kinetic energy of Avogadro’s number of molecules (that’s one
mole of gas) can be written:
KE  N A (
1
1
mu 2 )  M u 2
2
2
-9-
(5)
where M = m NA , is the molar mass.
From equations (3) and (5) we can obtain:
u2 
3RT

M
3kT
m
(6)
The term u 2 is known as the root-mean square speed (rms)
Equation 6 is the first of three quantities that we will define and use to
describe the velocities of molecules in a gas. We will see that all three of
the velocity measures will have the same order of magnitude but will
differ slightly in their exact value.
Example:
As an example, let's calculate the rms velocity of a nitrogen molecule
at 25oC. Sometimes it is easier to work with a mole of molecules
instead of individual molecules. It is easy to see that.
rms 
u2 
3RT

M
3x8.314 x 298
 515 m s 1
0.028
5. Distribution of Molecular Speeds:
-
Average speed ( u ):
u
-
8RT
8kT

M
m
(7)
Most probable speed (α):

2 RT
2kT

M
m
(8)
The three speeds (root-mean square, average speed and the most
probable speed) are in the ratios:
rms : u : α = 1.00 : 0.92 : 0.82
- 10 -
(9)
Example 1:
The root-mean square speed, rms, of N2 at 298 K is 515 m s-1.
What are the average speed and the most probable speed of this gas at
the same temperature?
Answer: by using equation (9) we can obtain
u
0.92

 u  0.92 rms  0.92 x 515  473.8 m s 1
rms
1

rms

0.82
 u  0.82 rms  0.82 x 515  422.3 m s 1
1
Example 2: Calculate the ratio
u
for the following 7 gas molecules:
rms
Molecule
1
2
3
4
5
6
7
Speed / m s-1
370
400
380
420
300
440
480
Answer:
u = (370 + 400 + 380 + 420 + 300 + 440 + 480) / 7 = 398 m s-1
rms = (3702 + 4002 + 3802 + 4202 + 3002 + 4402 + 4802) / 7 = 402 m s-1
u / rms = 398 / 402 = 0.99
- 11 -
6. The free mean path and collision number:
Let us consider a particular molecule A with diameter d and moving
with a speed u . This molecule will collide in 1 s all molecules that have
their centers in the cylinder of the figure below:
The volume of the cylinder whose radius is equal to diameter of the
molecule d is:
V  d2u
The number of molecules in the cylinder is:
N   d 2 u N*
where N* is the number of molecules per cubic meter:
N* 
N PN A

V
RT
The free mean path, L, is known as the distance traveled between collisions;
that's the length of the cylinder, u , divided by the number of collisions
occurring in the cylinder:
L
u
1

*
 d u N  d 2 N*
2
Taking in account that collisions goes from glancing collisions (a) to headon collisions (b), the average speed is 2 u for collisions in right angles (c)
as we can see it on the figure below:
The free mean path becomes:
- 12 -
u
1

(10)
 d 2 2 u N*
2 d 2 N *
The number of collisions that a molecule A makes per second is denoted Z1:
L
Z1 
u
 2 d 2 u N*
L
(11)
The molecule A, in relation to the other molecules, travels with an effective
speed equal 2 u .
The number of collisions occurring in a unit volume per unit time is denoted
by Z11:
1
1
Z11  Z1 N * 
 d 2 u ( N * )2
(12)
2
2
The factor 1/2 ensures that each collision will not be counted twice.
7. Numerical values of collision properties:
Let us consider again the N2 (d = 3.74x10-10 m) gas at 25 0C and 1 atm. First
we have to calculate N* (number of molecules per cubic meter), starting
from:
PV = nRT
and using the relations:
N* 
N
V
and n 
N
NA
then we get:
N* 
P N A 101325 x6.022 x1023

 2.46 x1025 m 3
RT
8.314 x 298
We can calculate now the free mean path, L:
L
1
1

 6.50 x108 m
10 2
25
2
*
1.414 x3.14 x(3.74 x10 ) x 2.461x10
2 d N
and the number of collisions of molecule A per second ( u was already
calculated):
Z1  2  d 2 u N *  1.414 x3.14 x(3.74 x1010 ) 2 x515 x 2.461x1025  7.31x109 collisions s 1
and the number of collisions per m3 per second:
Z11 
1
 d 2 u ( N * )2  8.99 x1034 collisions m3 s 1
2
- 13 -
8. Real gases: Van Der Waals' equation.
The Dutch chemist van der Waals has attributed the failure of the PV = nRT
relation to duplicate the behavior of real gases to the neglect of:
a) The volume occupied by the gas molecules
b) the attractive force among the molecules
and has proposed these corrections:
 Correction of the volume
The presence of molecules of nonvanishing size means that a certain
volume, called the excluded volume, is not available for molecules to move
in. The excluded volume for one mole is represented by b, then the more
appropriate equation would be:
P( V – nb) = nRT
b is characteristic of each gas and must be determined empirically.
The relation between b and the size of the molecule can be seen on the
figure below:
Molecular volume
Excluded volume
(per pair of molecules)
1
excl
V
 4  d 3 
14
3
   d   4    
23

 3  2  
and b is equal to:
bN V
1
A excl
 4  d 3 
 4N A     
 3  2  
(13)
Example:
What is the value of b for a molecule of N2 gas (d = 2.88x10-10 m) ?
Answer:
3

 4  d 3 
 2.88 x1010  
23 4
   3.01x105 m3mol 1
b  4 N A       4 x6.023x10  3.14
2
 3

 
 3  2  
- 14 -
 Correction of pressure:
The attraction acts with the confining pressure to hold the molecules
together. the gas is confined, therefore, not only by the external pressure ,
but also by the intermolecular attractions which contribute with a term
proportional to (n / V)2. The proportionality factor is denoted by a . Finally
the equation becomes:
2

n 
 P  a   V  nb   nRT
 V  

(13)
Example:
What is the real pressure of 3 mol of CO2 in a 10 L container at STP ?
Van der Waals constants are : a = 2.01 atm L2 mol-2, b = 0.0319 L mol-1
Answer:
2
2
nRT
3x0.082 x 273
n
3
P
 a  
 2.01   6.6 atm
V  nb   V  10  3x0.0319
 10 
- 15 -
Exercises:
energy at 25 0C of 1 mol of water vapor ?
10) At STP (1 atm, 0oC), what is the rms speed of CH4 ?
(Ans: 652 m s-1)
11) To what temperature (in K) must be heated O2 to get the same average speed of
H2 at 27 oC ?
(Ans : 4800 )
12) At 27 oC and 600 Torr,
a: What is the number of molecules of O2 per m3 (N*)?
(Ans : 1.93x1025 m-3)
b: Calculate the mean free path (L) of O2 (d = 3.84 Å).
(Ans: 7.9x10-8 m)
- 16 -
1st Law of Thermodynamics
The 1st Law of Thermodynamics simply states that energy can be
neither created nor destroyed (conservation of energy). Thus power
generation processes and energy sources actually involve conversion of
energy from one form to another, rather than creation of energy from
nothing. For example:
Automobile Engine
Chemical  Kinetic
Heater/Furnace
Chemical  Heat
Hydroelectric
Gravitational  Electrical
Solar
Optical  Electrical
Nuclear
Nuclear  Heat, Kinetic, Optical
Battery
Chemical  Electrical
Food
Chemical  Heat, Kinetic
Photosynthesis
Optical  Chemical
As you can see conversion between chemical energy and other forms of
energy are extremely important, whether you are veterinarian or a
mechanical engineer. That is what we will focus on for the remainder of this
chapter.
System and Surroundings
The 1st Law of Thermodynamics tells us that energy is neither created
nor destroyed, thus the energy of the universe is a constant. However,
energy can certainly be transferred from one part of the universe to
another. To work out thermodynamic problems we will need to isolate a
certain portion of the universe (the system) from the remainder of the
universe (the surroundings).
- 17 -
Sign Convention
When working numerical problems we will quickly become confused if we
don’t adopt a universal convention for when we use a positive sign or a
negative sign.
Sign Convention for heat, q


Heat is transferred into the system  q > 0
Heat is transferred out of the system  q < 0
Sign Convention for work, w


Work is done upon the system by the surroundings  w > 0
Work is done by the system on the surroundings  w < 0
The 1st Law of Thermodynamics will be much more useful if we can express it
as an equation.
U = q + w



(1)
U  The change internal energy of the system,
q  The heat transferred into/out of the system,
w  The work done by/on the system.
This reformulation of the 1st Law tells us that once we define a system
(remember we can define the system in any way that is convenient) the
energy of the system will remain constant unless there is heat added or
taken away from the system, or some work takes place.
P-V Work
Most chemical reactions either give off or absorb heat, but not all
chemical reactions do a significant amount of work. By far the most common
types of work associated with chemical reactions are:


Electrical work (i.e. batteries, fuel cells, etc.)
Mechanical work done by an expanding or contracting gas
At this point in the course we will not concern ourselves with electrical work.
Therefore, we only have to worry about work when a gaseous product or
reactant is involved.
- 18 -
PV work is represented by the following differential equation:
(2)
where:



W = work done on the system
P = external pressure
V = volume
Therefore, we have:
For an expansion against a constant external pressure Pex we have:
W = - Pex (Vf – Vi) = - Pex ∆V
(3)
Example:
Calculate ∆E for the conversion, at 100 oC and 1 atm pressure, of 5
mole of water to steam. The heat of vaporization of water is 40670 J mol-1,
the density of liquid water can be taken as 1 g mL-1, and water vapor can be
treated as an ideal gas.
Q = 40670 x 5 = 203350 J
- 19 -
W = - Pext (VG – VL)
Pext = 1 atm = 101325 Pa
VL = 5 x 18 = 90 mL = 9x10-5 m3
VG = 5 x 22.4 x 373/273 = 153 L = 0.153 m3
W = - 101325 x (0.153 - 9x10-5) = - 15505 J
∆U = 203350 – 15505 = 187845 J
Isothermal reversible expansion of an ideal gas:
Suppose that we have a certain amount of gas in a cylinder fitted with a piston and
that the piston is fitted with a horizontal tray at its upper end. In that upper tray we now
place a set of small weights. We allow the piston to position itself such that Pin = Pex. The
system is at equilibrium. If we now remove one of the very small weights from the upper
tray, this has the effect of reducing Pex very slightly. The piston will respond by moving
up a little, the value of pin will drop a little and equilibrium will be re-established.
If we remove another one of the very small weights, this has the effect of again
reducing Pex very slightly. The piston will respond by moving up a little more, the value
of Pin will drop a little more and equilibrium will be re-established.
This process can be repeated again and again, removing one small weight in each
step and allowing the piston to move upwards very slowly and gradually, at each and
every step the system is virtually in equilibrium with the surroundings.
(Pin = Pex = nRT/V)
- 20 -
Then, for an isothermal reversible expansion of an ideal gas, we
have:
W = - n R T Ln (Vf / Vi)
(4)
Internal Energy
The internal energy (U) encompasses many different things, including:



The kinetic energy associated with the motions of the atoms,
The potential energy stored in the chemical bonds of the molecules,
The gravitational energy of the system.
It is nearly impossible to sum all of these contributions up to determine the
absolute energy of the system. That is why we only worry about U, the
change in the energy of the system.
Our convention for U is to subtract the initial energy of the system from
the final energy of the system.
∆U = U(final) – U(initial) = q + w
In a chemical reaction the energy of the reactants is U(initial) and the heat
of the products is U(final).
For a transformation at constant volume (∆V = 0), no PV work, the change in
Internal Energy is:
∆U = qv
- 21 -
The Enthalpy function:
For a gas, a useful additional state variable is the enthalpy which is defined
to be the sum of the internal energy U plus the product of the pressure p
and volume V. Using the symbol H for the enthalpy:
H=U+pV
(5)
For a system with heat transfer Q and work W, the change in internal
energy U from state 1 to state 2 is equal to the difference in the heat
transfer into the system and the work done by the system:
U2 - U1 = Q + W
For the special case of a constant pressure process, the work done by the
gas is given as the constant pressure p times the change in volume V:
W = - p [V2 - V1]
Substituting into the first equation, we have:
U2 - U1 = Qp - p [V2 - V1]
Let's group the conditions at state 2 and the conditions at state 1 together:
(U2 + p V2) - (U1 + p V1) = Qp
The (U + p V) can be replaced by the enthalpy H.
∆H = H2 - H1 = Qp = n Cp (T2 - T1)
(6)
State Functions
A state function in thermodynamics is a property of the present state
of a system, and has a value that is independent of how that state was
prepared.
Ex.: P, V, T, U, H …
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A path function in thermodynamics is a quantity that is associated
with processes in which systems undergo a change of state. The value of a
path function corresponding to a particular change in state depends not only
on the initial and final states of the system undergoing the process, but also
on the details of the pathway involved in the change.
Ex.: Q and W
Some Properties of state functions
If Z is a state function and Z = f(x, y) then we have:
 Z 
 Z 
 dy
dZ  
 dx  
 x  y
 y  x
(7)
  Z 
  Z 


  
y  x  y x  y  x
(8)
Exp: for the internal energy, E=f(V,T), and the enthalpy, H=f(P,T), we have:
 U 
 U 
dU  
 dV  
 dT
 V T
 T V
(9)
 H 
 H 
dH  
 dP  
 dT
 P T
 T  P
(10)
  E 
  E 

 
 
T  V T V  T V
and
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  H 
  H 

 


T  P T P  T  P
The dependence of Internal Energy and Enthalpy on Pressure
JOULE’S EXPERIMENT
Expressed in mathematical terms:
 U 

 0
 V T
(11)
NB:
JOULE’S experiment was a crude one. His thermometer, although the
best available at the time, was insufficiently sensitive to detect the small
change in temperature that does accompany the expansion of air under the
conditions of his experiment.
The result of JOULE’S experiment applies only to Perfect Gases.
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From Eq. (9) and Eq. (10) we can deduce that:
 U 
 U 
 U 
dU  
 dV  
 dT  
 dT
 V T
 T V
 T V
and
or
dU  U 


dT  T V
dH  H   H 

 

dT  T  P  T V
Adiabatic Expansion of an Ideal Gas
An adiabatic process is one in which no heat is exchanged with the
environment (Q = 0). Adiabatic processes happen either very rapidly, so
that there is no time for heat exchange, or in a well insulated system.
For an ideal gas, the first law of thermodynamics (dU = δq + δw) for an
adiabatic process (δq = 0) gives:
dU = nCvdT = - PdV
The ideal gas law ( PV = nRT) gives
nCvdT = - nRTdV/V
dT
R dV

T
Cv V
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using (R = Cp – Cv), and  
Cp
CV
we have
dT
dV
 (1   )
T
V
By integration we find
T
Ln f
 Ti

V
  (1   ) Ln f

 Vi



which is equivalent to
Vf
 
Ti  Vi
Tf



(1 )
or
V
 i
Ti  V f
Tf




(  1)
or more generally
TV ( 1)  cons tan t
(12)
We can deduce also that
PV   cons tan t
(
TP
1

)
(13)
 cons tan t
(14)
The work is given by:
w = ∆U = nCv(T2 – T1)
PV  C
 PV
w  nCV  2 2  1 1   V ( P2V2  P1V1 )
nR  R
 nR
w
P2V2  P1V1
 1
(15)
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About γ :
The ratio of specific heats γ can be related via kinetic theory to the
number of degrees of freedom available to the molecules of the gas via

d 2
d
(15)
A monatomic gas (noble gases He, Ne, Ar, e.g.) has three dimensions in
which it can travel and hence only three degrees of freedom. Hence, we
expect
γ = 5/3 = 1.67
(γHe = 1.66, γNe = 1.64, γAr = 1.67).
A gas consisting of diatomic molecules (H2 , N2 , O2 , e.g.) can rotate. A
``dumbbell'' shape has one axis of symmetry, so it can only rotate about
two axes. The bond between the atoms of a diatomic molecule behaves
roughly like a (quantum mechanical) spring. Vibrations generally require
more energy than rotations. In a temperature regime in which only
rotations are excited in the gas, we expect
γ = 7/5 = 1.4
(γH2 = 1.41, γN2 = γO2 = 1.40).
At higher temperatures, vibrations add a degree of freedom, giving
γ = 8/6 = 1.33.
All of these predictions are approximate, because real gases are only
approximately ideal, and it is found empirically that γ varies weakly with
temperature.
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Exercises
1) In a certain process, an ideal gas (it’s our system) has absorbed 900 J. In this
process, the volume of the gas was decreased, by the action of a constant external
pressure of 5 atm, from 20 L to 10 L.
What is the internal energy change (ΔU) of the system in this process?
(Ans: 5965 J )
2) A 3.2-g sample of O2 is allowed to expand isothermally in a vessel maintained at
27 oC from 246 cm3 to 1230 cm3.
Calculate the work done when the system expands:
a:
against a constant pressure of 1.0 atm
(Ans: -99.7 J)
b:
reversibly to the same final volume (1230 cm3)
(Ans: -401 J)
3) A sample of 2 mol of an ideal gas occupies a fixed volume of 15.0 L at 300 K.
When it is supplied with 2.35 kJ of energy as heat its temperature increases to
341K. What are the values of w, ΔU, ΔH for this process?
4) The molar heat capacity at constant pressure (in : J K-1 mol-1) of an ideal gas is
given by: Cp = 27 + 7x10-3 T
How many kJ of heat are needed to increase the temperature of one mole of this
gas, at constant pressure, from 300 K to 500 K?
(Ans: 5.96 kJ)
5) Three moles of an ideal gas are allowed to expand reversibly from an initial
pressure of 10 atm to a final pressure of 2 atm, the temperature being kept
constant at 27 oC.
What are the values of q , w , and ∆H for the gas in this expansion?
6) One mole of an ideal gas is heated at constant volume in a container of 10 liters
from 0 to 200 oC. The molar heat capacity at constant volume (in : J K-1 mol-1) of
this gas is: Cv = 27 + 7x10-3 T – 8x10-7 T2
Calculate q , w , and ∆E for this process.
7) Hydrogen gas (Cp = 29 J K-1 mol-1) is expanded reversibly and adiabatically from
a volume of 2 liters , at a pressure of 6 atm and a temperature of 27 oC, until the
volume is 6 liters.
a - Calculate the volume and temperature of the gas after the expansion.
b - Calculate q , w , ∆E, and ∆H for the gas.
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