From WolfWikis

LECTURE 7

Spontaneity and entropy

There are many spontaneous events in nature. See e.g. fig 20.1 and 20.2. If you open the valve in both cases a spontaneous event occurs. In the first case the gas fills the evacuated chamber, in the second the gases will mix. The state functions U and H do not give us a clue what will happen. You might think that only those events are spontaneous that produce heat.

Not so:

1. If you dissolve KNO₃ in water, it does so spontaneously, but the solution gets cold.
2. If you dissolve KOH in water, it does so spontaneously, but the solution gets hot.

Clearly the first law is not enough to describe nature.

A new state function named entropy

Suppose we have a small reversible change dU in the energy of an ideal gas. We know that U only depends on temperature:

dU = C_v.dT

We also know that any reversible work would be volume work.

δw_rev = -PdV

This means that we can write:

δq_rev = dU - δw_rev

δq_rev = C_v.dT + PdV

Let us examine if this represents an exact differential. If that were so we could take cross derivatives and arrive at the same answer:

∂ C_v/∂V should be equal to ∂P/∂T.

∂C_v/∂V= 0 because C_v does not depend on volume (only T, just like U: it is its derivative).

But:

∂P/∂T.= ∂(nRT/V)/∂T = nR/V and this is not zero!!

Clearly, δq_rev is not a state function, but look what happens if we multiply everything with an 'integration factor' 1/T:

δq_rev/T = (C_v/T).dT + (P/T)dV

∂(C_v/T)/∂V = 0 because C_v/T still does not depend on volume

But:

∂(P/T)/∂T.= ∂(nR/V)/∂T = zero!!

Thus, the quantity dS = δq_rev/T is an exact differential, so S is a state function. Its name is the ENTROPY.

Circular integrals

Because it is a state function it integrates to zero over any circular path going back to initial conditions, just like U and H

As shown in section 20-3 we can use this fact to revisit the isotherm + isochore + adiabat circular path

A: isothermal expansion, B:adiabat, C:isochore

On the adiabat B there is no heat so q_rev = 0.

On the isochore C: δw=0, temperature is changing from T₂ back to T₁ this requires heat c_V.ΔT.

Along the isotherm A we have seen that :::q_rev,A = +nRT ln[V₂/V₁]

The two quantities q_rev,A and q_rev,B+C are not the same, which once again underlines that heat is a path function. How about entropy?

First consider the path B+C:

We had seen this integral before, albeit from T₁ to T₂ (See Eq. 19.21):

ΔS_B+C = + nRln V₂/V₁

(Notice the + sign instead of the - sign in 19.21)

Along the isotherm A:

q_rev,A = +nRT ln[V₂/V₁].

T is a constant so that we can just divide and

ΔS_A is also +nRln[V₂/V₁].

Clearly entropy is a state function where q_rev is not.

Spontaneity of an isolated system

Let us consider a zero law process. We have two identical blocks of metal, say aluminum. They are each at thermal equilibrium, but at different temperatures. They are brought into contact with each other but isolated from the rest of the universe.

• Zeroth law: Heat will flow from hot to cold
• First law: There is no change in total energy

so:

dU_A= -dU_B

There is also no work so:

dU_A = δq_A + 0

Because U is a state function this makes q a state function as well, otherwise this equality does not hold. As there is only one term on the right there is only one path (along q). So we could write:

dU_A = dq_A

This implies that we do not need to worry about reversible and irreversible paths as there is only one path.

As

dS = δq_rev/T => TdS = δq_rev = dU in this particular case.

Thus we get:

Clearly as long as the two temperatures are not the same dS is not zero and entropy is not conserved. Instead it is increasing. Over time the temperatures will become the same (if the blocks are identical, the final temperature is the average of T_A and T_B) and the entropy will reach a maximum.

For our two identical blocks of metal, we can in fact derive that the entropy change ΔS = C_v. ln[(T_A² + T_B²)/4T_AT_B]. This is indeed a postive quantity (See page 837).

In general we can say for an isolated system:

dS ≥ 0

Thus iff we are dealing with a spontaneous (isolated!) process dS >0 and entropy is being produced. This gives us a criterium for spontaneity.

Entropy exchange of an open system

In an isolated system dS represents the produced entropy dS_prod and this is a good criterium for spontaneity. Of course the requirement that the system is isolated is very restrictive and makes the criteriun as good as useless... What happens in a system that can exchange heat with the rest of the universe? We do have entropy changes in that case, but part of them may have nothing to do with production, because we also have to consider the heat that is exchanged.

dS = dS_prod + dS_exchange

IF the process is reversible (that is completely non-spontaneous) we are dealing with δq_rev so that dS_exchange =δq_rev /T, but that is also what dS_tot is equal to (by definition). This leaves no room for entropy production.

So we have:

Isolated: dS = dS_prod + 0

Reversible dS = 0 + δq_rev/T

Notice that this demonstrates that for non-isolated systems entropy change is not a good criterium for spontaneity at all...

In the case the heat exchange is irreversible part of the entropy is entropy production by the system:

Irreversible: dS = dS_prod + δq_irr/T

Clearly dS > δq_irr/T in this case.

Generalizing the isolated, irreversible and reversible cases we may say

Second law

The second law of thermodynamics can be formulated in many ways, but in one way or another they are all related to the fact that there is a state function S that at least in isolated systems tends to increase.

For a long time people have looked at the entire universe as an example of an isolated system and concluded that its entropy must be steadily increasing until δS_universe becomes zero. Of late cosmologists like Hawkins have begun to question this assumption. The key problem there is the role of gravity and relativity in creating black holes. We will not consider this problem in any detail.

As we will see below the second law has important consequences for the question of how we can use heat to do useful work.

Vacuum expansion

Let's compare two expansions from V₁ to V₂ for an ideal gas, both are isothermal. The first is an irreversible one, where we pull a peg an let the piston move against vacuum:

The second one is a reversible isothermal expansion from V₁ to V₂ (and P₁ to P₂) that we have examined before.

In the first case the is no change in energy because T does not change. There is also no volume work because -P_extdV = -0dV integrates to zero. The piston has nothing to perform work against until it slams into the right hand wall. At this point V=V₂ and then dV becomes zero.

No energy, no work that means: no heat!.

Clearly the zero heat is irreversible heat q_irr = 0 and this makes it hard to calculate the entropy of this spontaneous process. But then this process ends in the same final state as the reversible expansion from V₁ to V₂ . We know that dU is still zero, but now δw_rev = -δq_rev is nonzero. We calculated its value before:.

q_rev = nRTln[V₂/V₁]

To find ΔS we just can divide out the (constant) temperature:

ΔS= nRln[V₂/V₁]

As S is a state function this equation also holds for the irreversible expansion against vacuum.

For the irreversible expansion into vacuum we see that

ΔS = ΔS_prod + ΔS_exchange = nRln[V₂/V₁] + 0

For the reversible one:

ΔS = ΔS_prod + ΔS_exchange = 0+ nRln[V₂/V₁]

Mixing two gases

Consider two ideal gases at same pressure separated by a thin wall that is punctured. Both gases behave as if the other one is not there and again we get a spontaneous process, mixing in this case.

If the pressure is the same the number of moles of each gas should be proportional to the original volumes V_a- and V_b and the total number of moles to the total volume V_tot.

For gas A we can write: ΔS_A = n_ARlnV_tot/V_A = n_AR ln n_tot/n_A

ΔS_B = R.n_B ln V_tot/V_B = R.n_B ln n_tot/n_B

The entropy change in total per mole of gas is:

ΔS/n_tot =R[n_B ln n_tot/n_B+ n_A ln n_tot/n_A]/n_tot

Using mole fractions y_a = n_A/n_tot and the fact that ln[x/y] = - ln[y/x]

ΔS_mola = -R[y_Alny_A +y_Blny_B]

In the case of more than two gases:

ΔS_mola = -RΣ[y_ilny_i]

This entropy is known as the entropy of mixing. Its existence is the major reason why there is such a thing as diffusion and mixing when gases, but also solutions (even solid ones) are brought into contact with each other.

Carnot cycles

Sadi Carnot was a French engineer at the beginning of the 19th century. He considered a cyclic process involving a cylinder filled with gas. This cycle the Carnot cycle contributed greatly to the development of thermodymanics and the improvement of the steam engine Carnot demonstrated that the cold temperature on the right is as important as the heat source on the left in defining the possible efficiency of a heat engine

As we saw, heat and work are both forms of energy and under the right circumstance one form may be transformed into the other. However, the second law of thermodynamics puts a limitation on this.

To go from work to heat is called dissipation and there is no limitation on this at all. In fact it was through dissipation (by friction) that we discovered that heat and work were both forms of energy.

To go the other way is much harder. We can see that if we consider a circular reversible path in an ideal PV diagram. We start bottom left. The path consists of four steps:

1. isochoric heating: q= +C_vΔT (C_v is constant) from T_c to T_h
2. isothermal expansion q_hot =-w: we get work out: w_hot =-RT_hln [V₂/V₁]
3. isochoric cooling q= -C_vΔT back to T_c
4. isothermal compression q_cold =-w: we must put work in: wcold =+RT_cln [V₂/V₁]

The total fourstep process produces work because w_hot is larger in magnitude than w_cold.

(See here for an animation. Notice anything different?)

The work is the integral under the upper isotherm minus the one under the lower curve, i.e.the surface area in between. Notice that this yellow area vanishes is we chose T_hot= T_cold. Obviously how cold the cold side is of great importance!.

The amount of work is also equal to the difference in the heat picked up at high temperature q_hot and dumped at low temperature q_cold. The isochoric heats cancel. The problem is that q_cold is only zero if the cold temperature is zero (Kelvin!). That means that we can never get all the heat we pick up at high temperatures to come out as work.

Efficiency

We can use the state function S to calculate how efficiently we can use heat to get useful work done. As the path is circular the circular integrals for both U and S are zero. This means that

Of course we spend good money on the fuel to start the cycle by heating things up. So how much work do we get for the heat we put in? If we define the efficiency of our heat engine as

η≡-w/q_hot

we get that

η= 1+ q_cold/q_hot = 1-T_c/T_h.

As you see we can only get full efficiency if T_cold is 0K (i.e. never).

Another implication is that if T_c = T_h no work can be obtained. no matter how much energy is available in the from of heat. Or in other words, if one dissipates work into heat isothermally, none of it can be retrieved.

The above formula is not very forgiving at all. Just imaging that you have a heat source of say 400K (a superheated pool of water, e.g. a geyser) and you are dumping in the river at room temperature 300K. The best efficiency you'll ever get is:

η= = 1-300/400= ¼.

Sadly, you'd be dumping three quarters of your energy as heat in the river.. (And that is best case scenario. There are always more losses, e.g. due to friction).

The arrow saying Q_C in the diagram below should then be three times as fat as the one that says W.

Heat pumps

So getting our work from heat is hard and always less than 100% succesful. The other way around should be easy. After all, we can dissipate work into heat freely even under isothermal conditions!

What happens if we let the heat engine run backwards? Consider reversing all the flows in the above diagram. Obviously we must put in work to make the cycle run in reverse. The heat will now flow from cold to hot, say from your cold garden into your nice and warm appartment. The amount of heat you get in your humble abode will be the sum of all the work (say 100 Joules) you dissipate plus the heat you pumped out of the garden (say 300 Joules). Thus if you are willing to pay for the energy you dissipate (100 Joules), you may well end up with a total of 400 Joules of heat in your appartment!.

Obviously if it is heat you after this is a better deal than just dissipating the work in your appartment (by burning some oil). Then you'd only get 100J for your precious buck.

Dissipating it as electrical heating is even worse because you would

1. first burn (a lot more!) oil to generate heat
2. use this heat to produce electrical work at great expense because a lot of the heats gets dumped at the low temperature side (the river or so)
3. dissipate the work again in your appartment (without using it to pump any heat out of the garden)

I heat my appartment electrically.

To my defense I should add that I do not have a garden to pump heat from and I certainly do not have the (expensive) system of pipes, pumps and exchangers that would be required to make a heat pump work.

I do have a refrigerator. This is a heat pump too. It heats my room by pumping heat from its innards into my room. If I keep the door open, however, all it does is dissipate precious electrical work, because the pumped heat will flow back into its innards and spoil my milk.

All of this makes perfect sense if you own stock in an oil company. I'm sure my pension fund does.