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Ideal Gases and First Law of Thermodynamics
Ideal Gases:
After studying heat transfer processes and phase transitions, we have some idea about the relation
between temperature and kinetic energy. Now, we will study a very special system of molecules in
order to further understand these ideas. What could be the easiest example of a molecular system?
To further understand the concept of temperature, heat and other thermodynamic variables that will
be needed on the way, we need to start from the simplest system of molecules we can think of. Imagine
a box of rigid walls filled with N molecules, but these molecules have a very special characteristic.
These molecules can only interact with the rigid walls of the box. The molecules within this box, do
not interact between each other. Interaction means, force. So, imagine a system of molecules, where
each molecule does not push or pull any other molecule in the system. Can you imagine what would
happen?
Let’s take an example a little bit out of context. Since we have already studied gravitation, we know
that planets, stars, galaxies constitute a system of N celestial bodies that attract between each other
according to Newton’s law of gravitation. So planets are not free to move at their will, since their
paths will be modified by the forces of attraction between them and other celestial bodies. For
example, the Earth does not follow a straight path along our universe, because the Sun, is pulling the
Earth; the same thing happens with the other planets of our solar system and even more, the sun can’t
follow a straight path because there is a black hole within the center of our galaxy pulling it and
pulling any other star within our galaxy. Even more, our galaxy doesn’t follow a straight path because
there are other galaxies pulling it and changing its trajectory, and so on.
If we could observe the “universe” of molecules inside the box, these molecules also attract between
each other, but it is not a gravitational force. It is a force with a completely different nature; however,
what is important to understand, is that like planets, molecules are not free to move around without
being influenced by others. Unfortunately the forces between molecules are much more complicated
than gravitational forces, even for the simplest molecules. So the first natural simplification would be
to study systems of molecules where there are no forces between them. This is, the forces of
attraction or repulsion are zero. This may seem an oversimplification of the problem, however as
we will mention later, there are situations where real molecules can be modeled as noninteracting molecules!!!
Let’s first try to imagine a system non-interacting molecules. Take a look at this animation on the
web:
http://mutuslab.cs.uwindsor.ca/schurko/animations/idealgas/idealGas.htm
Take a large number of molecules in the java applet on this web page, and to see better what happens,
tune the pressure at 1.00 and a low speed for the molecules. What you see, is that molecules hit the
walls of the box, but the molecules do not see each other. Since there is no repulsion or attraction
between molecules, one molecule could pass just through another, like ghosts passing through
objects. This is because there are no forces between them. This is what we know as an ideal gas.
Of course, this is not what happens in reality. But we can model some molecular systems in this way,
because the conditions allow us to treat the system as an ideal gas. An example is pure oxygen at
1atm of pressure and 293 Kelvins. At this pressure, and temperature oxygen atoms are very far from
each other, so that molecular attraction is extremely weak. Even though, it is a real molecule, oxygen
seems to behave very much like an ideal gas at this pressure and temperature.
We have seen that absolute temperature is a direct measure of average kinetic energy. Since there is
no force between molecules of an ideal gas, the average potential energy is zero. Therefore, the
total average energy in a system of non-interacting molecules (ideal gas) is equal to the average
kinetic energy. We discussed before that the average kinetic energy per molecule is equal to:
Ekavg 
3
k BT
2
If this were an ideal gas, this is also equal to the total energy per molecule.
Example 1. If we have 1 mol of ideal gas at 300 Kelvins, what is the total energy of this gas?
Remember that 1 mol has N A  6.022 1023 molecules, where NA is Avogadro’s constant. Therefore
the total energy of the gas is equal to the number of molecules within the gas times the total energy
per molecule. That is:
E
3
N A k BT
2
Note: It turns out that the product R  N A k B  (6.022 1023 molec)(1.38 1023 J / K ) is the
universal gas constant, R  N AkB  8.314 J / mol  K .
The expression for energy given above is only valid for an ideal gas. In fact, it only works for a
monoatomic ideal gas. For diatomic, triatomic, etc; non-interacting molecules, the total energy is:
E
d .o. f
N A k BT
2
Where d.o.f is the number of degrees of freedom. If we have a molecule of one atom (Helium, any
noble gas, etc) the molecule can only experience translations. It can move along the x, y and z axis.
Boltzmann showed that for each translational degree of freedom, there is an average kinetic energy
1
k BT . Therefore, in a three dimensional space there are three translational degrees of
2
3
freedom that correspond to an average energy of Ekavg  k BT .
2
of Ekavg 
For example, if we have a molecule like Oxygen, O2 that is formed by 2 oxygen atoms that looks like
the figure below, the center of mass in this molecule has 3 translational degrees of freedom plus 2
possible rotations along the y and the z axis. For each rotational degree of freedom there is a
corresponding average kinetic energy of Ekavg 
1
k BT . The same term as for translational degrees
2
of freedom. So for an oxygen molecule that has 3 possible translations and 2 possible rotations, there
are a total of 5 degrees of freedom, therefore the average kinetic energy of an oxygen molecule is:
Ekavg 
5
k BT
2
y
C.M
x
z
Can you guess the possible rotations in a molecule of water?
Equation of State:
In the late XIX century Boltzmann found an equation that relates pressure and volume with
the temperature (energy) of a system made of non-interacting molecules (ideal gas). This
relation is also known as the equation of state of an ideal gas. Since the ideal gas is the most
simple of all molecular systems, the equation of state is also very simple and with it, is
possible to find the thermodynamic state of any ideal gas. According to this, volume and
pressure are related with temperature by:
PV  nRT
THIS EQUATION OF STATE IS ONLY VALID FOR SYSTEMS OF MOLECULES THAT
BEHAVE VERY MUCH LIKE THE IDEAL GAS!!! It is very important to stress that in this
expression T must be used in Kelvins and the pressure P, is the absolute pressure, not the gauge
pressure. R is the universal gas constant discussed above and n, is the number of moles.
If this equation represents the relation between thermodynamic variables that belong to a
system that physically can’t exist then, why do we study these molecular systems? These systems
(ideal gases) are studied, because as we said before, there are many materials like oxygen, hydrogen,
helium, methane, etc; that behave very much like an ideal gas. Let’s work out an example:
Example 2. Imagine we have 2 moles of real oxygen at 1atm of pressure and 293.15 Kelvins. What
is the density of oxygen at this pressure and temperature? If we want to know the real answer, then
we have to measure oxygen’s density in a Laboratory. At this pressure and temperature, oxygen’s
density is:
  1.331kg / m3
That is the real value! To figure out how the theory of ideal gases works, let´s calculate the density
of oxygen by using the ideal gas equation of state.
Let’s go back to   M / V . Where M is the mass of the gas. The total mass is just equal to the
number of molecules within the gas times the mass per molecule. Since we have 2 moles of oxygen,
then, there are N  2 N A  2(6.022  1023 ) molecules in the gas of this example. We can take the
mass of one molecule of oxygen from any periodic table. Remember that an oxygen molecule has
two atoms of oxygen, therefore one molecule of oxygen has a mass equal to 2 times the mass of one
atom. The molecular mass is then: m  2(2.656 1026 kg )  5.312 1026 . The total mass of two
moles of oxygen is:
M  mN  (5.312 1026 kg / molecule)(1.2044 1024 molecules)  0.063978kg
Now, we need the equation of state to obtain the volume of this gas. If you observe from the equation
of state, the volume is a function of temperature and pressure, solving for V we have:
V  nRT / P  (2mol )(8.314 J / mol  K )(293.15K ) / (101325Pa)  0.048108m3
Therefore, according to the equation of state of an ideal gas, the density of oxygen in this
case is equal to:   0.063978kg / 0.048108m3  1.329883kg / m3 . It is remarkable how this
number is very close to the real value! This is why we use the ideal gas approximation. Even more,
this not only works for oxygen at 1atm and 293.15 Kelvins, it also works for many values of P and
absolute temperature T. Even more, this equation of state also works very well for many materials
like: Helium, Hydrogen, Noble Gases, Methane, CO2, Ozone, CO, etc. and for several values of
P and T as well!!!
Unfortunately, this equation of state does not work for liquid or solid oxygen, for example. Actually
it only works for gaseous phases, and not in all materials. For example, water vapor, which is in a
gaseous phase, can’t be described with the ideal gas equation! Even at 1atm, water does not behave
like an ideal gas. If you do not think so, just compare the real density of water vapor at 1atm and 100
Celsius, with the value you calculate by using the ideal gas equation. Therefore, there are a lot of
substances that can´t be approximated by an ideal gas equation.
Remark: A solid is a state of matter where atoms or molecules move around an equilibrium
position, so that one can observe a symmetrical structure. Take the example of ice. Water
molecules move a small distance about their equilibrium position, because they feel attractive
and repulsive forces from their neighboring molecules.
Note that if no forces exist between molecules, this can’t happen. Therefore the ideal gas can
never be in a solid state. And a similar situation happens for liquid states. So in an ideal gas
there aren’t phase transitions. Not even vapor-vapor transitions. This is why, it is always
called gas.
There are other types of approximations that try to get closer to reality. For example, since
the work of Stephan Boltzmann, some physicists have obtained equations of state for a gas
of hard spheres. A hard sphere gas, is the next logical step. In this gas, molecules interact
with repulsive forces only. The repulsive force is very simple, it is infinitely large when two
molecules are at a distance equal or smaller than the molecule diameter, but the force is zero
at distances larger than the molecule’s diameter. There is an equation of state for this type of
gas, but since there are no attractive forces, it will always be in the vapor state as well. No
matter how cold it is. There are also mathematical models of gases where molecules repel
between each other with a soft force. These gases are called soft sphere gases, and since there
are not attractive forces, soft sphere gases can’t have phase transitions either. All of these are
models that try to get closer to the real deal, but still, without the attractive force between
molecules, the equations of state are limited.
Up to this day, there are no equations of state that can be obtained mathematically, for
molecules that have repulsive and attractive forces. Therefore, to model phase transitions,
liquids and solids, data has to be obtained from experiments or computer simulations. In
future thermodynamic courses, you will obtain densities, internal energies, enthalpies and
entropies from experimental data recorded on tables or experimental equations of state like
the Wan der Waals and Clausius Clapeyron equations of state. For example, the ice images
shown in the last notes are actually, real images from a computer simulation.
Remember: Even though, ideal gases, are a significant simplification of reality; they
are widely used because, a) they offer a simple mathematical manipulation and b) can
be applied to some real situations with an excellent degree of approximation.
Let us make one more example, in order to learn how the equation of state of an ideal gas
can be used.
Example 3. Physics of Diving. A diver uses a tank of 18 Liters of capacity filled with air.
The diver has to calculate the mass of air that has to be injected into the tank, so that he can
breathe for 45 minutes at a depth of 20m below the ocean surface. If an average person takes
10 Liters of air every intake, and inhales air 10 times per minute, obtain the pressure of the
tank after it has been filled with all the necessary air if the outside temperature is 30 Celsius.
The interesting part of this problem is that air can be treated as an ideal gas, and the answer
we obtain will be very precise, since air behaves very much like an ideal gas. This is because
air is mostly composed of Nitrogen, Oxygen and some noble gases. How do we approach
this problem?
Remember the equation of state: PV  nRT . The pressure of the tank after being filled is;
P  nRT / V
The volume of air inside the tank is equal to the volume of the tank which is 5 Liters. And air will be
injected to reach an equilibrium temperature of 30 Celsius, equal to the outside temperature. The
tricky part of this problem is to obtain the number of moles n, necessary to allow the diver a 20m dive
for 45 minutes.
One thing to note, is that when a diver is submerged, he can only take air from the tank by
using a regulating valve that delivers air to the diver’s lungs at the same pressure than the
outside pressure. Therefore, air delivered by the regulator must be at an absolute pressure of:
Poutside  1atm   gh  101325Pa  (1030kg / m3 )(9.81m / s 2 )(20m)  303411Pa
This is about 3atm of absolute pressure. So, he must inhale 6 Liters of air at this pressure in one intake.
We can use the equation of state, to calculate the number of moles of air on one intake:
n1  PV / RT  (303411Pa)(0.01m3 )  /  (8.314 J / mol  K )(303.15K )   0.722296mol
So, on one intake he breathes 0.72 moles of air. Now is just common sense to get the total number of
moles he will need during his trip underwater. Since, he takes this amount of air 10 times every
minute, he will need: moles / min  10(0.722296mol )  7.22296moles
Therefore, for the whole dive, he will need to breathe:
n  (7.22296moles / min)(45 min)  325.0332moles
So, if we use the equation of state one more time, but now for the tank; we have to inject 325.0332
moles of air inside a tank of 5L. Then:
P  nRT / V  (325.0332moles)(8.314 J / mol  K )(303.15K ) / (0.018m3 )  4.55 107 Pa
This is equal to P  449.165atm . This is the pressure of the tank when the diver starts his trip! If
you are interested, you can check the effects of breathing high pressure air for long dives, if the subject
is not decompressed properly.
This biological effect is extremely painful and is called decompression sickness, basically caused by
the fast liberation of excess nitrogen absorbed by the blood stream, due to the intake of high pressure
air.
Example 4. Physics of Free Diving. Imagine now a free diver that has much more lung
capacity than the scuba diver of the last example. An average free diver can expand his lungs
up to a volume of 15L in one intake. If a free diver takes 15L of air at the beginning of his
dive when the air temperature is 26 Celsius, calculate the volume of his lungs when the diver
reaches the world record of 101m of depth if the outside temperature at this depth is 4 Celsius.
We use the ideal gas equation of state to obtain the moles of air he takes into his lungs in just
one intake. Then:
n  PV / RT  (101325Pa)(0.015m3 )  / (8.314 J / mol  K )(299.15K )   0.611moles
This is the amount of air he takes at the beginning of his dive. Now, when he gets to 101m of depth,
the outside pressure is:
Poutside  1atm   gh  101325Pa  (1030kg / m3 )(9.81m / s 2 )(101m)  1.1219 106 Pa
Which is about 11 atm of pressure! Now, we can use the equation of state again to obtain the
volume of his lungs at this depth:
V  nRT / V  (0.611moles)(8.314 J / mol  K )(277.15K ) / (1.1219 106 Pa)  0.001255m3
His lungs are compressed to a volume of 1.255L, which is 11.95 times smaller at that depth!
Example 5. Does water vapor behaves like an ideal gas? Even though water vapor is in its gaseous
state, it may not behave like an ideal gas. So let us calculate the density of water vapor at 120 Celsius
and 1 atm of absolute pressure by using the ideal gas equation of state. From this equation we can
obtain the molar density, which is the number of moles n, divided by the volume:
n / V  P / RT  101325Pa / (8.314J / mol  K )(393.15K )  30.999moles / m3
Where 30.999 moles of water molecules are equivalent to a total number of:
N  nN A  (30.999moles / m3 )(6.022 1023 molecules / mol )  1.866762 1025 molecules / m3
Given that one molecule of water has one oxygen atom with an atomic mass number of 16
and two hydrogen atoms with an atomic mass number of 1, then one water molecule has a
molecular mass of 18 units of atomic mass. Then, the mass of a water molecule is:
m  18(1.66053892 1027 kg )  2.98897 1026 kg / molecule
Therefore, water vapor density at this temperature and pressure according to the ideal gas equation of
state is:
  mn / V  (2.98897 1026 kg / molecule)(1.866762 1025 molecules / m3 )  0.55797kg / m3
Now let us obtain the real value. This has to be obtained from steam tables (what you will do
in future thermodynamic courses). Let’s compare with superheated steam. At 1 atm and 120
Celsius, the density of steam is: 0.557724kg/m3. This value is very close to the value obtained
by assuming ideal gas behavior. So superheated steam at 120 Celsius behaves very much like
an ideal gas. But let us now compare with saturated steam. From the tables we have that
saturated steam at 1 atm of absolute pressure and 120 Celsius, has a density of: 1.121 kg/m3,
which is approximately twice the value obtained by assuming ideal gas behavior!!!
Therefore, the ideal gas equation of state can’t be applied for saturated steam. This happens
because molecular interactions between water molecules can’t be neglected. That is why you
will be forced to use steam tables.
First Law of Thermodynamics.
The first law of thermodynamics is a generalization of conservation of energy. When we
studied energy conservation, we only considered one type of energy, which is mechanical
energy. The first law of thermodynamics, takes into account two types of energy: Mechanical
energy and heat. We have seen that heat is another manifestation of kinetic energy, but
conservation of energy in heat transfer mechanisms is not considered by Newton´s principle
of energy conservation.
It is easier to understand this law, when applied to gases. Mainly because, we can apply the
ideal gas equation of state as a first start. But keep in mind that, this law can be applied to
liquids, solids, binary mixtures, etc. Let’s start by considering a gas (it doesn’t have to be an
ideal gas) inside a cylinder. The top of this cylinder is a movable piston, very much like the
piston from a car engine. The figure below shows a schematic picture of a piston.
Δx
T2
Q
T1
T1
F1=P1A
F2=P2A
If we put heat (thermal energy) Q, and allow the gas inside the piston to expand, some of this
energy will be used to change the gas temperature, (average kinetic energy of the molecules)
and the gas will use the rest of the energy to expand, which is mechanical work. In the figure,
before we put energy, the gas in a thermodynamic state, called state 1. In this state, the gas
has an average energy U1, which is exactly the same as the average kinetic energy of:
d .o. f
U1 
Nk BT1 . This type of energy is also known as internal energy.
2
For a gas that doesn’t behave like an ideal gas (for example, the saturated steam of example 5), there
is another contribution to the internal energy coming from the interaction forces between molecules.
With this thermodynamic variable we can say in which state is the gas. After we add all the energy
Q, the gas ends up in another thermodynamic state with internal energy U 2 
d .o. f
Nk BT2 . (Note
2
that these are not states of matter, since the gas does not change to liquid or solid, by state here it is
only meant a thermodynamic state.) So after we added all the heat, the gas has increased or decreased
its internal energy by an amount equal to:
U  U 2  U1
Example 6. Assume that the gas inside the piston shown in the figure above is Helium (which behaves
like an ideal gas for many values of temperature and pressure). Assume that have 3 moles of Helium
inside the cylinder. If the temperature in the initial state is 300 Kelvins and the temperature in the
final state is 600 Kelvins, what is the change in internal energy?
Since Helium is a monoatomic gas, its internal energy is: U 
3
3
Nk BT  nN A k BT . So in its
2
2
initial state the internal energy is:
U1 
3
3
Nk BT1  (3moles )(6.022 1023 molecules / mol )(1.38 10 23 J / K )(300 K )  11218.986 J
2
2
Its energy in the final state is:
U2 
3
3
Nk BT2  (3moles)(6.022 1023 molecules / mol )(1.38 1023 J / K )(600 K )  22437.972 J
2
2
The change in internal energy is: U  U 2  U1  22437.972 J  11218.986J  11218.986J .
However, as we can see from the figure above, the gas not only increased its internal energy, it also
increased its volume. To increase volume or decrease volume, the gas has to do some work. Work is
a form of mechanical energy given by W  F x . The force exerted by the gas displaces the piston
an amount Δx. This process requires some energy which is equal to the work done by the force. This
energy comes from the heat Q, injected to the gas. Therefore, if energy is conserved, the provided
heat Q, must be equal to the work done by the gas and the increase or decrease in Internal Energy.
This is the first law of thermodynamics:
Q  U  W
We can find an equivalent expression for mechanical work that can be applied to a gas. We first start
from the definition of mechanical work by; dW  Fdx . If we observe, the force exerted by the gas
on the piston is equal to F  PA . To move the piston a very small distance dx, the necessary
mechanical work is: dW  PAdx . Finally, notice that the product Adx is equal to a small change in
volume dV  Adx . Therefore the amount of work needed to move the piston a small distance dx is:
dW  PdV . We imagine the piston being displaced a small distance because, as the gas expands,
the pressure changes, so we need to consider small displacements because in this case the force is not
constant. In general, by adding a small amount of heat, it will be used for small changes in internal
energy and do a small amount or work:
dQ  dU  dW  dU  pdV
This is the most general form of the first law of thermodynamics, because it can be applied to any gas
(ideal or not) and consider cases where the force may not be constant. Even more, it can also be
applied to liquids, solids, etc; as written in the above expression.
Now we need to consider 4 possible cases of expansion or compression:
1) Consider that we do not inject energy to the gas, therefore dQ=0J
2) Consider that we add energy to the gas, but there is no change in internal energy
(temperature remains constant) dU=0J
3) Consider that we add energy so the pressure is constant P=const
4) And finally, we can add energy and glue the piston to the cylinder. In this case the
gas does not change in volume. Since dW=PdV; in this case at constant volume,
dV=0m3. Therefore dW=0J.
1. Fast Expansion
Let’s consider the case where no heat flows in or out of the cylinder. Can you imagine a
situation where a gas expands without giving any energy to it or taking energy from it?
Imagine a gas inside a diver’s tank. Suddenly we break its valve so the gas starts to leak very
fast out of the tank. Remember the pressure inside this tank?. It was about 400 atm. So,
suddenly the gas expands by leaking very fast, so that its pressure in the final state is equal
to 1atm. Another example of an extremely fast expansion is an explosion. Like the explosion
of the gas mixture within the cylinder on a car engine. As soon as the spark ignites the
mixture, the gas expands extremely fast. The fast expansion does not allow the superheated
gas to exchange heat with the surroundings, therefore there is no flow of heat.
The gas expands its volume so it must do work! According to energy conservation, the
needed energy to expand and push the piston violently will come from its own internal
energy. The gas will take the necessary energy from the molecules kinetic energy to achieve
the expansion. In other words:
0J  dU  pdV
pdV  dU
This is called an adiabatic process. The above equation from thermodynamics, can be applied to
any substance, for example the exploding mixture we just described, the fast expanding air from the
diver’s tank, etc.
Adiabatic Process for Ideal Gas. Unfortunately without thermodynamic tables of the gas, we can’t
obtain more information (you will do this in future thermodynamic courses). For now, we will apply
this equation for the ideal gas. Because in this case we do not need tables. We have an equation of
state. For an ideal gas we know that dU 
R  N AkB ). Therefore:
dof
dof
nN A k B dT 
nRdT (keep in mind that
2
2
pdV  
dof
nRdT
2
For the left hand side we can use the ideal gas equation of state: p  nRT / V . Therefore:
nRT
dof
dV  
nRdT
V
2
dV
dof dT

V
2 T
We can integrate both sides:
dV
V

dof
2

dT
C
T
Where C, is a constant of integration. The integral is simple, remember it is the natural logarithm:
ln(V )  
dof
ln(T )  C
2
This can be simplified further to:
V  eCT  dof /2  C1T  dof /2
Where C1  eC is still a constant. This constant has a very important physical meaning. For this type
of process, the special combination
V
T
 dof /2
 C1 remains constant. So for example, if a gas is
expanded very fast (adiabatically) from state 1, to state 2, the special combination of volume and
temperature to the power of -dof/2, remains constant along the whole process. Therefore the equality
must be satisfied at state 1 and state 2. Even more, this equality must be satisfied at any state between
1 and 2, during a fast expansion. Then:
V
a
 dof /2
a
T

V
b
 dof /2
b
T
Va Ta dof /2  Ta 

 
Vb Tb dof /2  Tb 
 dof /2
We can use this last expression to obtain different versions to the adiabatic process, depending of the
thermodynamic variables of interest. For example, if we are given the volumes in states a and b, and
the temperature in state a, we can just solve for Tb in this equation to obtain the temperature in state
b. But what if we are given the pressure in state a, instead of the temperature? In this case we can use
the equation of state for an ideal gas in order to obtain an equivalent expression for an adiabatic
process, that relates pressure and volume. So let’s use the equation of state and solve for T:
Ta  PV
a a / nR and Tb  PV
b b / nR . If we substitute these into the expression for the adiabatic
process:
Va Ta dof /2  PaVa 



Vb Tb dof /2  PV
b b 
 dof /2
V 
. Then we arrange terms,  a 
 Vb 
dof /2 1
P 
 b 
 Pa 
dof /2
dof /2
V  P  dof /21
.
; a  b 
Vb  Pa 
So we get an expression for an adiabatic process that relates pressure and volume as:
dof /2
Va  Pb  dof /21
 
Vb  Pa 
We can also use the equation of state to relate pressure and temperature if we are solving a problem
where only we are given the pressures in states a and b, and the temperature in state a. We solve for
the volumes at each state, from the equation of state and substitute these as before to obtain:
Pb  Tb 
 
Pa  Ta 
( dof /2) 1
As an exercise, try to obtain this last expression that relates pressure and temperature at two different
states that belong to an adiabatic process.
Example 7. A tank is filled with oxygen gas at 400 atm of pressure and 293 Kelvins, like the tank
from the diver’s example. Suddenly the valve of the tank is broken so all the oxygen inside, leaks out
of the tanks in a very small amount of time. What is the temperature of the oxygen as soon as it leaks
out of the tank?
Since the oxygen is a diatomic molecule, it has 5 degrees of freedom. Therefore the exponent for the
condition in an adiabatic process is -5/2.
2. Constant Temperature: Extremely slow expansion or compression!
Now imagine that we have a gas in an initial state a. We start to heat the gas, but in this case, we do
it very slowly, so slow, that we are giving enough time to the gas, to exchange heat with its
surroundings so that its temperature will not change. If we do not do this slowly enough, we will not
give time to the gas to exchange energy with its surroundings and its temperature will change! Think
about it, if the temperature doesn’t change, according to the first law of thermodynamics, all the added
energy will be transformed into work, it is the most efficient process in the universe!!! The slower
is the process, more efficiency we get out of it. Imagine we apply this to an engine, then if we move
the piston slow enough more efficient it will be. Then, if it the gas expansion is infinitesimally close
to zero, its efficiency will be 100%. This is impossible, but we can approach to this situation, by
slowing down the expansion as much as we can. According to the first law of thermodynamics, the
added energy Q for this process is then: dQ  dW  pdV . If we use know the equation of state for
an ideal gas, we have: dW 
nRT
dV . Since the temperature during this process doesn’t change, T
V
is constant, we can integrate to get the work during the entire process. The gas will expand from an
initial state with volume Va to a final state with volume Vb. Then:
Vb
W  nRTa 
Va
V 
dV
 nRTa ln  b 
V
 Va 
This is the total work done by the gas during a process at constant temperature, (also called isothermal
process), and according to the first law of thermodynamics, since internal energy does not change in
this process, the work is equal to the total heat added Q. Q  W .
Example 8. Suppose that the oxygen from Example 7 expands also, but now the expansion
is very slow, so that its temperature remains constant. You can imagine a very small hole
where the gas leaks out, but the amount of gas leaking is very small. Finally after all the gas
has leaked out of the tank, its pressure is equal to the atmospheric pressure. a) If the gas tank
has a volume of 18L, what is the volume of the gas in its final state? and b) What is the total
work done by the gas during its expansion?
a) To obtain the volume of the gas in its final state, we can use the equation of state for an
ideal gas. First, we calculate the number of moles of oxygen within the tank:
n
PV
(400atm)(101325Pa / atm)(0.018m3 )
a a

 299.4825moles
RTa
(8.314 J / mol  K )(293K )
Now for the volume in its final state:
Vb 
nRTb (299.4825moles)(8.314 J / mol  K )(293K )

 7.2m3
Pb
(101325Pa )
b) Then the total work done by the gas is:
V 
 7.2 
W  nRTa ln  b   299.4825moles(8.314 J / mol  K )(293K ) ln 
  4371012.86 J
 0.018 
 Va 
This work is the energy that the oxygen needed to expand from a volume of 0.018m3 to a final volume
of 7.2m3. If the surroundings are at higher temperature than the oxygen, then this energy comes from
the surroundings. As the oxygen leaks out of the tank, the gas absorbs energy from its surroundings
to keep its temperature constant; otherwise the gas would get colder. Keep in mind, that this expansion
must be very slow, so the oxygen has enough time to absorb heat from its surroundings and not get
colder. This process is opposite to the adiabatic expansion. In this process, heat flows into the gas to
keep the temperature constant, and is very slow. In the adiabatic process the gas is thermally isolated
and expands very fast.
3. Constant Pressure:
In this process energy flows, in such a way that the gas is expanded or compressed without changing
the pressure. According to the first law of thermodynamics, in this process, energy added to the gas
is used to change its internal energy and to do some mechanical work.
dQ  dU  pdV
If we substitute this into the first law of thermodynamics we get:
dQ  dU  pdV 
dof
nRdT  pdV
2
Now, using the equation of state pdV  nRdT we have:
dQ  dU  pdV 
dQ 
dof
nRdT  nRdT
2
dof  2
nRdT  nC p dT
2
Where we have defined an specific heat at constant pressure of an ideal gas: C p 
dof  2
R . After
2
we describe the process at constant volume, we will discuss the physical meaning of this
thermodynamic variable. For now, it will be enough to understand that this variable represents the
amount of energy needed to change the temperature of 1 mole of an ideal gas. The similar concept
for the specific heats we used when studying energy transfer.
Since the pressure remains constant in this process, we can obtain the total work very easily from:
Vb
W  Pa  dV  Pa (Vb  Va )
Va
Sterling engines in jets, use this process as a part of their thermodynamic cycles. Since jets have
external combustion engines (not internal combustion), the expansion of the air-gas mixture happens
outside of the propeller, and its pressure is the same as the atmospheric pressure at the altitude of
flight.
Example 9. Suppose the air-gas mixture on a jet engine is at the atmospheric pressure of 0.4atm. This
pressure is found at altitudes of 10000m above sea level. After ignition, the gas expands and its
reaches about 1400K. a) What is the amount of mechanical work done by 1 mole of mixture, if the
temperature of the gas after before ignition is equal to the surroundings temperature of -500C? and b)
If the engine provides 100 moles of mixture per second, what is the power of this engine in horse
power units?
a) Assuming that the molecular weight of the air-gas is approximately 28g/mol (this is a big
assumption! Because we are just assuming that the air-gas mixture is composed of nitrogen and
oxygen only) we can use the equation of state to obtain the volumes of the mixture in its initial and
final state. Then:
Va 
nRTa (1mole)(8.314 J / mol  K )(223K )

 0.04574m3
Pa
(0.4atm)(101325 Pa / atm)
The volume of the mixture in its final state is:
Vb 
nRTb (1mole)(8.314 J / mol  K )(1400 K )

 0.287m3
Pb
(0.4atm)(101325Pa / atm)
The mechanical work done by 1 mole of mixture during the expansion is:
W  Pa (Vb  Va )  0.4atm(101325Pa / atm)(0.287 m3  0.04574m3 )  9778.2678 J
We have assumed that the air-gas mixture behaves like an ideal gas, and this is not true, because the
gas (fuel) is in its liquid state.
b) For 1 second, the engine burns 100 moles of air-gas (fuel) mixture, therefore the total work done
in 1 second is:
W  Pa (Vb  Va )  (100moles / s)(9778.2678J / mole)  977826.78J / s
Which is 977826.78 Watts of power of 1311.28HP, during this process.
4. Process at Constant Volume:
When the gas is not allowed to expand or compress, the volume doesn’t change, and all the energy
added to the gas is used to change its internal energy:
dQ  dU
dof
nRdT  nCV dT , where we have defined a
2
dof
constant volume specific heat for an ideal gas CV 
R.
2
If the gas behaves as an ideal gas, then: dQ 
What is the difference between the specific heat at constant volume and the specific heat at constant
pressure? It is very intuitive to see this difference. The first law of thermodynamics is very helpful if
we want to understand the difference between these two. If we expand a gas at constant pressure, the
gas needs energy to increase its temperature and also to do mechanical work. Therefore if we want to
increase the temperature by 1 kelvin of 1 mole of an ideal gas at constant pressure, we need energy
to increase its temperature, and some more so the gas can expand. On the other hand, if we do not
allow the gas to expand, we only need energy to increase its temperature. Therefore the specific heat
of the gas at constant pressure must be greater than its specific heat at constant volume. In fact:
CP 
dof  2
R  Cv  R . To increase the temperature by 1 kelvin, of 1 mole of an ideal gas at
2
constant pressure, we need an extra of 8.314Joules, than we would if the gas is not allowed to expand.
Those 8.314Joules extra are needed by the gas to expand.
Ideally car engines and diesel engines use constant volume (isocoric) processes in their
thermodynamic cycles.
Thermodynamic Cycles and the First Law of Thermodynamics.
A thermodynamic cycle is a set of different thermodynamic processes that take the gas through a set
of different states, and the cycle ends in the same state it started from. The figure below shows an
example of an entire cycle. The thermodynamic states are labeled as a, b and c. In this example, the
gas starts at state a. Pressure is plotted on the vertical axis and volume on the horizontal axis. The
process between state a and state b, is a vertical straight line, indicating that the volume of the gas is
constant through the process (it is an isocoric process). The curve connecting state b and c, represents
an expansion, since the volume is increased. In the figure is shown that during this process the flow
of heat dQ=0J, therefore this process is a fast expansion (adiabatic expansion). Finally the gas is taken
from state c to its initial state a through an horizontal straight line; since the volume is decreased, the
gas is compressed during this process. An horizontal line in a pV diagram like the one shown on this
figure, means that the gas pressure is constant through the entire process (isobaric process).
So let us review the cycle. The process from state a, to state b is a constant volume process (isocoric),
the process from state b to state c is a fast expansion (adiabatic expansion) and finally, the process
from state c to the initial state a, is a process where the gas is compressed at constant pressure (isobaric
compression).
Pressure
T1
b
Pb
T1
ConstVolume
T1
Pa
T1
a
VaT1
c
Vc
T1
Const Pressure
T1
T1
Volume (L)
T1
Procedure to solve thermodynamic cycles.
1. For any cycle that we can think of, the first thing to do, is to obtain the thermodynamic
variables at each state, which are: pressure, volume and temperature. To obtain these
variables, we will use the first law of thermodynamics and the equation of state. (It is
important to note, that these procedure applies to any substance, ideal gas or not) In the case
of an ideal gas, we will use the equation of state for an ideal gas.
b
T1
2. If we are asked to obtain the total work during the cycle, we may use the definition of work
as dW=PdV in each process, or it may be more practical to use the first law of
thermodynamics. For an adiabatic and isothermal processes, for example, it is much faster to
calculate the work done by the gas, by using the first law of thermodynamics. For an isobaric
process, it is straightforward to obtain the work by using dW=PdV.
3. We must apply the first law of thermodynamics to obtain the heat added to the gas during the
cycle, QH; and the remaining heat during the cycle QC.
4. If we want to know the efficiency of the cycle, it can be determined from the heat added and
the total work.
The heat added during a thermodynamic cycle can be determined by applying the first law of
thermodynamics in each cycle. Only positive values of Q contribute to the heat added QH.
Example 10. a) Calculate the amount of energy added to 1.5moles of Helium gas if it
experiences an isobaric expansion to twice its original volume of 20L at 5atm, b) if the gas
experiences an adiabatic expansion, c) If the gas experiences an isothermal expansion and d)
if the gas is heated at constant volume to twice its initial temperature.
a) For an isobaric expansion, according to the first law of thermodynamic, the energy
added is:
Q  nC p T
Since Hydrogen behaves very much like an ideal gas, we can use the equation of state for an ideal
gas to obtain the initial and final temperatures of the process;
Ta 
PV
(5atm)(101325Pa / atm)(0.020m3 )
a a

 812.485K and
nR
(1.5moles)(8.314 J / mol  K )
Tb 
PV
(5atm)(101325Pa / atm)(0.040m3 )
b b

 1624.9699 K
nR
(1.5moles)(8.314 J / mol  K )
Then the heat added during this process is:
Q  nC p T  (1.5moles )(5 / 2)(8.314 J / mol  K )(1624.97 K  812.485K )  25331.25109 J
Note: if the value turns out to be negative, it means that the energy was taken from the gas, not added
to it.
You may remember, that the specific heat at constant pressure Cp for an ideal gas is:
dof  2
5
Cp 
R . In this case, since Helium gas is a monoatomic molecule, then: C p  R . If the
2
2
energy added is positive, it means that we are giving this energy to the gas, and the gas will use this
energy to increase its temperature (internal energy) and to do work by expanding itself.
b) According to the definition of adiabatic process, the added heat is 0J, (remember that the gas
expands very rapidly and it does not have enough time to interchange energy with its
surroundings), then Q=0J
c) For an isothermal expansion, by using the first law of thermodynamics, we have:
dQ  dW
In this type of process, the temperature of the gas doesn’t change, therefore dU=0J. According to the
first law of thermodynamics, in this case, all the energy added to the gas will be used as work.
Therefore, if we calculate the work done during this process, it is exactly equal to the energy added
to the gas.
V 
 0.040 
W  nRTa ln  b   (1.5moles)(8.314 J / mol  K )(812.485K ) ln 
  7023.314J ,
 0.020 
 Va 
therefore, in this case, the heat added is Q=7023.314J.
d) If the gas is heated at constant volume, all the energy given to the gas will be used to
increase its internal energy, since no work is done during this process. Then,
according to the first law of thermodynamics; Q  nCv T
Q  nCv T  (1.5moles)(3 / 2)(8.314 J / mol  K )(1624.97 K  812.485K )  15198.751J
For an entire cycle, if we use the first law of thermodynamics, the total heat Q flow during
the cycle is: Q  W . There is no change in internal energy during an entire cycle, because the gas
starts and ends in the same thermodynamic sate. So, for an entire cycle, the total heat is equal to the
amount of work done by the gas in the cycle.
The total heat is equal to the added heat during the cycle, which we will denote by QH; minus the heat
that remained during the cycle, QC. This heat, QC; represents the amount of energy that could not be
used as work. If QC is zero, then we would have a cycle with a 100% of efficiency in converting all
added heat into work. (By the way, there are two other laws in thermodynamics that forbid the
existence of cycles with a 100% efficiency).
Can you guess now, how to calculate the efficiency of a thermodynamic cycle? Efficiency is defined
as the percent of the added heat that is used as mechanical work. Then:  
W
, where η is the
QH
thermodynamic efficiency. We can obtain the efficiency by subtracting the remaining energy (the
energy that is not used to do work) from the added energy, QH  QC . This quantity represents the
amount of energy that was used to do work in the cycle. Therefore, this quantity is equal to the total
work. If we calculate the remaining heat, instead of the total work, we can obtain the efficiency as
well:

QH  QC
QH
Example 10. 2.4 moles of an ideal monoatomic gas at 330K are heated at constant volume to
a second state. Then the gas experiences an adiabatic expansion from state 2, to state 3 with
a temperature of 510K. Then, the gas is compressed at a constant pressure of 1atm from state
3 to its original state. a) Find the pressure, volume and temperature at each state, b) Draw a
PV diagram for the entire cycle, c) Find the total work done during the cycle and d) Calculate
the heat QH and efficiency of the cycle.
a) First we have to use the equation of state for an ideal gas, to find the temperature, pressure
and volume of the gas in each state. In the first state the gas has a temperature of 330K and
pressure of 1atm. With the equation of state we can obtain the volume of the gas:
V1  nRT1 / P1  2.4(8.314 J / mol  K )(330 K ) / (101325 Pa)  0.064986m3
Now we need the temperature for state 2. Since the volume in state 2 is equal to the volume of state
1, then:
V2  0.064986m3
We do not know the pressure and temperature of the gas in state 2. The equation of state is not enough,
so we need to apply the first law of thermodynamics for an adiabatic expansion, since the process
form state 2 to state 3 is adiabatic. We can use the equations for an adiabatic process that relate state
2 with state 3:
V2  T2 
 
V3  T3 
3/2
Then let’s solve for T2
V 
T2   2 
 V3 
2/3
 0.064986m3 
T3  
3 
 0.100433m 
2/3
510 K  681.723K
Now the pressure in state 3 can be found by using the equation of state:
P2  nRT2 / V2  2.4(8.314 J / mol  K )(681.723K ) / (0.064986m3 )  209319.3619 Pa
Which is 2.066 atm approximately.
b) The resulting PV diagram is shown in the Figure below.
Pressure (atm)
T1
2
P2
T1
ConstVolume
T1
1atm
T1
1
V1T1
Const Pressure
T1
3
VT
3
1
T1
Volume (L)
T1
c) To find the total work during the cycle, we can use the first law of thermodynamics in each
process. Since the process from state 1 to state 2, is at constant volume, the gas doesn’t do
any work during this process, W12  0 J . For the adiabatic process from state 2 to state 3,
according to the first law of thermodynamics; 0J  dU  dW , since no heat flows in or out of the
gas during this process, the total work is equal to the total change in internal energy,
3
3
W23  U 23  (U 3  U 2 )  ( nRT3  nRT2 ) ,
which
is:
2
2
3
3
W23   nR(T3  T2 )   (2.4moles)(8.314 J / mol  K )(510 K  681.723K )  5139.74 J
2
2
Finally the gas is compressed at constant pressure, in this case:
W31  P1 (V1  V3 )  101325Pa(0.064986m3  0.100433m3 )  3591.6673 J
The total wok done during this cycle is then:
W12  W23  W31  0 J  5139.74 J  3591.6673J  1548.0727 J
d) To obtain the total heat added to the cycle, we use again the first law of thermodynamics in each
process. For the process at constant volume we have: dQ  dU , all the heat added is used to increase
the internal energy of the gas, since the gas doesn’t do any work in this process:
Q1 2  U1 2  (U 2  U1 ) 
3
3
nR(T2  T1 )  (2.4moles)(8.314 J / mol  K )(681.723K  330 K )  10527.21J
2
2
From state 2 to state 3, the process is adiabatic, therefore, no heat is added or subtracted from
the gas during this process. Q23  0 J . Finally, the gas experiences an isobaric compression
(constant pressure), where energy is extracted from the gas:
Q31  U 31  W31 
3
nR(T1  T3 )  P1 (V1  V3 )
2
3
(2.4moles)(8.314 J / mol  K )(330 K  510 K )  101325Pa (0.064986m3  0.100433m3 )
2
 5387.472 J  3591.6673J
 8979.1393J

Therefore, the added energy to the cycle is QH  Q12  10527.21J and the energy subtracted
from the cycle (the remaining heat) is: QC  Q31  8979.1393J . If we use the first law of
thermodynamics for the entire cycle, just to check our answers, we can see that the net heat added to
the cycle must be:
Q  U  W . Where the change in internal energy for the entire cycle is 0J, because the initial and
final state during a cycle is the same state, which in this case is state 1. Then the total heat during the
cycle Q=QH-QC is equal to the net work done by the cycle:
QH  QC  10527.21J  8979.1393J  1548.07 J
Which is exactly the same as the total work calculated in part c). The heat added during the
cycle is QH=10527.21J. The efficiency of the cycle is by definition, the percent of the total added
heat or energy given to the cycle that can be used as mechanical work. In this case:

W
1548.07 J

 0.14705
QH 10527.21J
So the cycle has an efficiency of 14.705%. This means that of the total heat or energy added to the
engine, only 14.7% will be used as work to perform some task. For example, only 14.7% of the energy
input could be used by a car engine, to actually move the car.