1 Gases
2 Microscopic Energies
3 First Law
4 2nd & 3rd Law
5 Phase transitions
6 Mixtures
7 Phase Diagrams
8 Equilibrium
9 Molecular Interactions
Second Law of thermodynamics
The second law of thermodynamics can be stated in several ways. One early postulated version by Kelvin was quite interesting.
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This makes quite clear that we cannot simply collect energy from random motion (heat) and spontaneously organize it into work. While many engines do in fact use a heat source (combustion of fuel), it must also have a heat sink (the environment) to operate.
We can, however, convert work (organized translational energy) into heat (random translational energy). it's done via friction, for example. When we apply the brakes on our car or bike, we convert the energy of motion in to heat in the brake pads and disks. In fact, the "power", of brake systems is defined as the rate at which they can dissipate this extra heat into the environment.
A more general statement of the second law is
A process is spontaneous only if the total entropy increases. (The total entropy means of the whole universe.)a more useful definition might be
The entropy of an isolated system increases during a spontaneous process (ΔS > 0)We've already seen that an infinitesimal change in the entropy of a system can be expressed (equation 4.6) as and thus, for a measurable change in entropy we integrate between initial and final states.
So, we need to find the reversible path between initial and final states to calculate the entropy change using heat. We didn't really need to specify the subscript "rev" here for reversible. The fact that we are doing the integral implies reversible. However, just to be complete, we include it here.
Example: What is the change in entropy for an isothermal expansion of an ideal gas?
To do this, re recall that entropy is a state function, which mean that we need to use the reversible path to evaluate the energy transferred as heat (equation 4.10), which simplifies after integration to
| ΔSsys = qrev/T |
In an isothermal reversible expansion, we know that ΔU = 0 = qrev + wrev ., i.e., qrev = – wrev , so we can substitute in the expression for reversible isothermal work (eq. 3.11) to get
| and, we can also express the change in entropy for the isothermal change in pressure of an ideal gas as (piVi = pfVf). | |
Now, we can also calculate the change in the entropy of the surroundings. if we consider the surroundings to be a sink of constant temperature and volume (systems are infinitesimally small by comparison) then we can calculate the change in entropy of the surroundings by using the energy transferred as heat,
| ΔSsur = – q/T |
The negative sign is because any heat lost by the system is gained by the surroundings and the subscript "rev" is missing because the surroundings knows nothing about the system process. It only knows that amount of heat transferred in or out.
We know that entropy is a state function from the equations we derived from statistical thermodynamics. Another useful relationship we can consider is the efficiency of a process. The efficiency ε is defined as
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This equation tells us that if a process can convert all the input heat energy into work, then it is 100% efficient. But the second law already tells us this is impossible. So how do we calculate the efficiency of a process, in theory, at least. Efficiency less than 100% means some of the energy we put into the process will be lost as heat. We need a place to put this heat, a cold sink. Thus, we can break down an engine that takes in heat energy and performs work into three parts, a heat source, a cold sink and the work output.
In this configuration, we can calculate the efficiency of the engine using the equation
Obviously, the greater the difference between the temperature of the heat source and the cold sink, the greater the efficiency. The only way we can have a process operating at 100% efficiency though would be to have a cold sink at absolute zero, which is impossible.
In the real world, the cold sink is often the environment and the heat source involves the combustion of fuel.
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If we wish to reverse the process and remove heat energy from a cool source and put it into a hot sink, we will obviously need to do some work. The same concepts work here. The flow of heat from the cold source to a hot sink is not spontaneous since the decrease in entropy of the cold source is greater than the increase in entropy of the hot sink and so the net change in entropy will be negative, contrary to the second law.
where qc is the amount of heat transferred. Since this is always a negative quantity, it cannot happen without some other source of energy.
We will need to add work to the mix to make this process happen.
the minimum amount of work needed would just counter balance the differential such that the previous sum equals zero. In this case, we would need to distinguish the value qc (heat transferred out of the cold source) from the value qh (heat transferred into the hot sink). However, they are related via |qh| = |qc| + |w|. obviously, the smaller the amount of work we need to do per unit of energy transferred, the more efficient the refrigerator is. We define the coefficient of performance as
now
as long as the transfer is done reversibly, we can also express this coefficient using temperatures alone.
There would be additional energy requirements in a real refrigerator since in addition to transferring energy to cool the "cool source" we would also need to apply work to transfer enough energy to overcome the heat "leaking" into the cool source from the environment since there is no perfect insulator.
Now let's consider the total entropy of the universe
dStot = dSsys + dSsurr. This can be expanded using equations 4.11 and 4.13 to be
dStot will only be zero for system processes that occur reversibly.
This relationship can be written as
In this form the equation is called the Clausius inequality. It is very important in our discussion of spontaneity of chemical reactions. Now, if our system is totally isolated from the surroundings such that dq = 0 then
which tells us that the change in entropy for a spontaneous process cannot be negative. This is in essence, the second law of thermodynamics.
Measurement of Entropy
We now have the means to measure the change in entropy of any substance from it's pure form at absolute zero up to what ever state we find it by adding up all the changes. So, a substance with melting point Tf and boiling point Tb will have as its entropy above the boiling point
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The main problem with this method will be the determination of heat capacity near T = 0. For this we use the Debye approximation that the heat capacity is proportional to T3 near absolute zero. Thus, we find the value of a that fits the parameterized equation Cp = aT 3 to the experimental data near absolute zero to extrapolate all the way to zero.
Now, according to the statistical equations we were deriving earlier, it should be possible to determine the entropy of a system exactly. This is covered by the third law of thermodynamics.
The Third Law of Thermodynamics
The entropy of a pure crystal at absolute zero is zero.
This make sense from the statistical point of view. If the crystal is pure, then every atom position is identical and identified and at absolute zero, there is no energy to populate anything but the groundstate energy, thus there is only one microstate possible and hence entropy is zero.
Thus, S(0) from equation 4.18 is zero.
Example: What is the molar entropy of nitrogen at 25ºC?
Here are the results of each step in the journey from zero K to room temperature for nitrogen
| Process | Sºm / J K-1 mol-1 for the process | |
| Debye extrapolation |
1.92 |
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| Integration from 10K to 35.61K |
25.25 |
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| phase transition (solid –> solid) |
6.43 |
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| Integration from 35.61K to 63.14K |
23.38 |
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| Fusion |
11.42 |
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| integration from 63.14K to 77.32K |
11.41 |
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| Vaporization |
72.13 |
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| Integration from 77.32 to 298.15 |
39.20 |
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| total |
191.14 |
but N2 is not an ideal gas |
| correction |
0.92 |
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| grand total |
192.06 |
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Helmholtz and Gibbs Energies
Recall equation 4.16, the Clausius inequality. We can rewrite this expression for a process at constant volume as
| or |
Similarly, we can write this expression for a process at constant pressure as
| or |
Thus, by equation 4.19, we can write that at constant internal energy or constant entropy, we have
| dSU,V > 0 and dUS,V < 0, respectively |
This relates changes in system state functions to the spontaneity of the process. The first one tells us that in an isolated constant volume system, the entropy of the system must increase for a spontaneous process. The second equation is less obvious and seems to say that spontaneous processes at constant entropy and volume tend to loose energy. Actually, what's happening is that the lost internal energy is going to the surroundings as heat and increasing it's entropy.
By equation 4.20, we can write that at constant enthalpy or constant entropy, we have
| dSH,p > 0 and dHS,p < 0, respectively |
These two equations again tell us about the connection between changes in state functions of the system and the spontaneity of the process. The first one says that for constant enthalpy and pressure, the entropy of the system must increase. The second one tells us that if the entropy of the system does not change then the enthalpy must decrease so that the surroundings will increase it's entropy.
These two concepts lead us to the Helmholtz energy A, defined as
| A = U – T S |
The Gibbs energy
| G = H – T S |
Where all the symbols refer to the system.
Infinitesimal changes at constant temperatures in these energies are expressed as
| dA = dU – T dS and dG = dH – T dS |
we can now see that for a spontaneous process, we must have the following conditions met:
| dAT,V < 0 and dGT,P < 0 |
Clearly, if there is a way for the system at constant volume and temperature (isochoric and icothermal) to lower AT,V, it will. dAT,V = 0 is the conditions for equilibrium, when no further macroscopic changes occur.
Similarly, for isothermal and isobaric conditions a spontaneous process will lower it's Gibbs energy GT,P . When dGT,P = 0 then the system will be at equilibrium.
