Entropy

One might be tempted based on the results of thermochemistry to predict that all exothermic reactions would be spontaneous. The corollary this would be the statement that no endothermic reactions are spontaneous. However, this is not the case. There are numerous examples of endothermic reactions that are spontaneous. Of course, heat must be taken up from the surroundings in order for the process to occur. Nonetheless, the enthalpy of the reaction does not determine whether the reaction will occur, only how much heat will be required or generated by the reaction. The observation that gases expand to fill a vacuum and that different substances spontaneously mix when introduced into the same vessel are further examples that require quantitative explanation. As you might guess by now, we are going to define a new state function that will explain all of these observations and define the direction of spontaneous processes. This state function is the entropy.

If we take the molecular point of view it seems that the spontaneous direction is associated with an increase in the disorder or randomness of the system. For example, if we take the expansion of a gas. Suppose a gas in vessel A is allowed to expand into vessel B. The translational partition function is proportional to the volume. Our interpretation of the partition function is that it gives us a measure of the number of energy levels accessible to the system. If we double the volume, we double the number of accessible levels. This implies that our ability to define the energy level of any given particle decreases by a factor of two. There is in effect more randomness in the system. The same type of reasoning applies to mixing.

Likewise, we can suppose that heating an ideal gas will result in an increase in the disorder. More states will be accessible in the microscopic picture. But, as we discussed above, heat alone is not the answer to the puzzle of spontaneity. Heat is not a state function and it is path dependent.

Entropy is defined by a thermodynamic cycle

Let us follow a thermodynamic cycle to understand how heat and work are path functions and then define a state function for the cycle. A scheme for the cycle is shown below

The states are shown in violet numerals. The processes in the expansion are shown in Roman numerals.

For an adiabatic expansion

C_vdT = -(nRT/V)dV

And as we saw previously for an adiabat

(T₃/T₂) = (V₂/V₃) ^2/3

During the compression we can similarly divide the process conceptually into two phases

and for the adiabatic compression that returns the piston to its original state we have:

(T₁/T₄) = (V₄/V₁) ^2/3

Since the initial expansion process followed an isotherm we conclude that T₂ = T₁ = T_hot and likewise for the isothermal compression compression T₃ = T₄ = T_cold.

=-nRT_hotln(V₂/V₁)–C_v(T_cold–T_hot)–nRT_coldln(V₄/V₃)–C_v(T_hot – T_cold)

w = -nRT_hotln(V₂/V₁) – nRT_coldln(V₁/V₂) [since V₄/V₃ = V₁/V₂]

w = -nRT_hotln(V₂/V₁) + nRT_coldln(V₂/V₁) [property of logarithms]

but have already calculated this above for the work

q = nRT_hotln(V₂/V₁) + nRT_coldln(V₄/V₃)

q = nRT_hotln(V₂/V₁) + nRT_coldln(V₁/V₂) [since V₄/V₃ = V₁/V₂]

q = nRT_hotln(V₂/V₁) - nRT_coldln(V₂/V₁) [property of logarithms]

Neither the work nor the heat is a state function for this cycle. Notice that the form that we have used to express the work and heat shows that if we divide by the temperature upon each step we have a function that is exactly zero for the cycle

This suggests the definition of a new state function. It is called the entropy, S.

We specify that q is reversible since in the example we gave all of the steps were reversible. In fact, we shall see that the entropy should always be calculated along a reversible path.

The Second Law

Second law: It is impossible to construct a device that operates in cycles that converts heat into work without producing some other change in the surroundings.

Consider an idealized cycle. In this cycle there are the following four steps

1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression

This is known as a Carnot cycle.

Figure 1. A thermodynamic cycle

Clearly, the volume ratios at points 1, 2, 3, and 4 are established by

so that

The heat q for the adiabatic steps is zero.

Therefore,

For the isothermal steps. The red curve represents the isothermal expansion (hot gas) and the blue curve represents the isothermal compression (cold gas). Since these processes are isothermal we know that

The total work of isothermal expansion and compression is given by

Note that V₂ > V₁ and so w_hot < 0 represents work done by the system.

Note that V₄ < V₃ and so w_cold > 0 represents work done on the system.

Since

And furthermore V₄/V₃ = V₁/V₂ we have

which means that

The latter equation is very important since it defines a thermodynamic temperature scale. The ratio of the temperatures is equal to the ratio of the heats transferred

Note further that we define the thermodynamic efficiency as

Using this definition of efficiency we can prove that the cycle shown in Figure 1 applies not only to an ideal gas but any working substance. The logic is as follows. We show two engines operating such that one is a heat pump and the other is a cooler (i.e. it is a refrigerator which is the reverse of a heat pump). We can show that no matter what the working fluid of engines 1 and 2 (ideal gas, real gas etc.) the thermodynamic efficiency of two ideal engines must be the same. Our negative proof is as follows:

This implies w₁ > w₂ and since

Thermodynamic Efficiency

The cycle just described could be the cycle for a piston in a steam engine or even in an internal combustion engine. The hot gas that expands following combustion of a small quantity of fossil fuel drives the cycle. If you think about the fact that the piston is connected to the crankshaft you will realize that the external pressure on the piston is changing as a function of time and is helping to realize an ideal reversible expansion. If we ignore friction and assume that the expansion is perfectly reversible we can apply the above reasoning to your car. The formalism above for the entropy can be used to tell us the thermodynamic efficiency of the engine. We define the efficiency as the work extracted divided by the total heat input. This is essentially what you do when you think in terms of miles per gallon of gas. However, there you are considering the gas before it is combusted (i.e. not directly in terms of its heat content) and the work after it has been converted into the rotation of the wheels by the crankshaft and the gears. So it should make sense that thermodynamics defines

efficiency = work done/heat used

I use the absolute value sign because I like to think of efficiency as a positive number and the work obtained from the system will be negative. McQuarrie and Simon use a minus sign in the definition to achieve the same end.

We saw above the for a reversible cycle the total work is

w_total = nR(T_cold – T_hot)ln(V₂/V₁)

The heat used is that heat used for the expansion at T_hot (i.e. the heat expelled from the cylinder during the isothermal step or step I. We call this the heat since in an engine fuel is burned and it causes the expansion. There is also a heat associated with the compression in step III. This heat comes from the surroundings at the temperature T_cold (i.e. T_cold must be the temperature of the surroundings). So for step I we have

q_hot = nRT_hotln(V₂/V₁)

Putting these two expressions together we find

Thermodynamic temperature scale

Since q_cold is negative we can combine the minus sign with q_cold and write the expression as

and finally

|q_hot|/|q_cold| = T_hot/T_cold

The ratio of the heats is equal to the ratio of temperatures for two steps in a thermodynamic cycle. This defines a temperature scale and allows one to measure temperature as well (i.e. this scheme represents a thermometer). Both this expression and the thermodynamic efficiency further imply that there is an absolute zero of temperature.

The entropy change is a measure of a spontaneous process

To examine the function that we have just defined, let us imagine that we place to identical metal bricks in contact with one another. If one of the bricks is at equilibrium at 300 K and the other at 500 K, what will the new equilibrium temperature be?

Intuitively, you would say 400 K and you would imagine that heat flows spontaneously from the warmer brick to the colder brick. The entropy function makes these ideas quantitative.

Our system consists of two bricks in contact. Let us calculate the entropy for the system.

We define the entropy as

and it can be either positive or negative.

q = C_v(T_f – T_i) = C_v(400 – 500) = -100 C_v for the brick at 500 K

q = C_v(T_f – T_i) = C_v(400 – 300) = 100 C_v for the brick at 300 K

where C_v must be greater than zero. We substitute these values into the above expression

The entropy is greater than zero for a spontaneous process. In this case, the process is heat flow, but the interpretation is general. Note that this statement is true for any substance since the heat capacity is always positive.

dS > 0 (spontaneous process)

dS = 0 (equilibrium)

dS > 0 (spontaneous process in an isolated system)

dS = 0 (reversible process in an isolated system)

These two equations are a statement of the second law of thermodynamics. Since the universe can be considered as an isolated system we can state that the entropy of the universe tends toward a maximum. That means that because there is a spontaneous direction for changes in the universe dS > 0. Clausius defined the first and second laws of the thermodynamics with the following words.

Die Energie des Universums ist konstant; die Entropie strebt ein maximum zu.

The energy of the universe is constant; the entropy tends toward a maximum.

Calculation of the entropy change

To really understand these statements we need to consider how to calculate the entropy change in a system that is not isolated. Let us consider the entropy change as a function of the temperature first.

It is important to reiterate that the calculation of the entropy always follows a reversible path. You might ask, well what happens if I do not calculate the entropy along a reversible path? Since S is a state function, the answer should be the same. To put this to the test, let us return to the example of expansion of gas in a cylinder. We saw that the process can occur along different paths.

b. reversible isothermal expansion the work w = -nRTln(V_f/V_i).

The P-V plot for an expansion has the form

The ratio T_e/ T_i = V_f/ V_i

q(surroundings) = - q(system)

In our case (see P-V plot above) with P_ext = 10 atm, V_i = 0.246 L and V_f = 2.46 L, we have

w = -10 atm (2.46 L – 0.246 L) = -21.4 L-atm

To calculate the entropy we need the temperature. We have not discussed how many moles are in the system either. We must specify one of these variables to complete the problem. Let us assume that the temperature is 300 K. The number of moles is given by

n = PV/RT = (10 atm)(2.46 L)/(0.082 L-atm/mole-K)(300 K)

n = 1.0 in this case. Well at least that was easy!

So the entropy change in the surroundings is

To calculate the entropy change in the system we choose the reversible path along the isotherm.

To conclude we have shown that

The total entropy change depends on the entropy of the system and the surroundings.

We can calculate the entropy change in the system at

We can use the combined first and second law expression to derive a large number of important thermodynamic expressions.

dU = TdS - PdV

For example,

This means that

We can integrate the above expression to determine the entropy change at constant volume. The entropy change at constant pressure can be derived by an exactly analogous procedure starting with

dH = TdS + VdP

Finally, we have

Statistical Entropy

So far we have taken a macroscopic view of the entropy. However, just like energy, pressure, and heat capacity we can calculate the entropy from a microscopic view of matter. In the microscopic view we regard the "disorder" of the system in terms of the population of energy levels. In fact, this view is fundamental and is actually the starting point for the derivation of partition function we used earlier.

To understand this let us consider a quantity W that describes the number of ways we can distribute particles among available energy levels. Suppose we have N particles and M available energy levels. The number of ways of distributing the particles among these levels is given by a binomial distribution. We are asking how many ways we can put n₁ particles in level 1 and n₂ particles in level etc. up to level M if we have N total particles where N = n₁ + n₂ + … n_M. We call the number of ways W.

Where the factorial is N! = N(N-1)(N-2)…(2)1. The validity of this equation is perhaps best seen using the example of dice.

Note that there is significant but subtle difference between the presentation here and in Simon and McQuarrie section 20-5. Simon and McQuarrie discuss how many systems have the same energies while we discuss how many particles can occupy the same energy levels. The difference is the following

Systems with the same energy (microcanonical partition function)

Particles in the same energy level (molecular partition function)

The discussion in 20-5 is actually discussed in terms of the micro-canonical ensemble (constant N,V,E) without explicitly defining it. The partition function Q that we introduced earlier is in the canonical ensemble (constant N,V,T) and this is the one that is relevant for both Chemistry 431 and 433. The logic of the definition of the entropy is the same in both descriptions.

The entropy increases as the number of ways of arranging the system increases. In other words as W increases the energy increases. The formulation of Boltzmann states

This is motivated by the fact that the Ws for different systems are multiplicative while the entropy must be additive. To further interpret this definition we substitute for W

The probability that a given level is occupied is p_j = n_j/N. Substituting this into the expression where n_j appears we find

For a collection of N particles or molecules, or

on a per molecule basis.

S = U/T + klnQ