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.

Historically, people were interested in understanding this question so that they could understand the efficiency with which heat is converted into work. This was a very important question at the dawn of the industrial revolution since it was easy to conceive of an engine powered by steam, but it turned out to be quite difficult to build one that was efficient enough to get anything done! The problem it turns out is related to the question of the spontaneity of chemical reactions. In an engine, say for example the internal combustion engine in a car, there is a cycle in which fuel is burned to heat gas inside the piston. The expansion of the piston leads to cooling and work. Compression readies the piston for the next cycle. A state function should have zero net change for the cycle. It is only the state that matters to such a function, not the path required to get there. Heat is a path function. As we all know in an internal combustion engine (or a steam engine), there is a net release of heat. Therefore, we all understand that dq ¹ 0 for the cycle.

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.

Imagine we begin with some hot gas (following ignition of some fossil fuels). The piston in the engine expands. Let us divide the expansion conceptually into two phases. Phase I is an isothermal expansion (dU = 0) and phase II is an adiabatic expansion (dq = 0). For the two expansion phases we can write

Phase	Transition	path	Result
I.	1®2	isothermal	dw = -dq
II.	2®3	adiabatic	dU = dw

We have seen that for an isothermal expansion at the hot temperature

dw = -PdV = -nRTln(V₂/V₁)

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 III. isothermal compression at the temperature of the surroundings and IV. adiabatic compression.

Phase	Transition	Path	result
III.	3®4	Isothermal	dw = -dq
IV.	4®1	Adiabatic	dU = dw

For the isothermal compression

dw = -PdV = -nRTln(V₃/V₄)

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.

This is a key fact since is means that T₁/T₄ = T_hot/T_{cold =} T₂/T₃. Further, this means that (V₄/V₁) ^2/3 = (V₃/V₂) ^2/3 and therefore V₄/V₁ = V₃/V₂. We rearrange the indices to obtain V₄/V₃ = V₁/V₂. Note that the piston returns to its original condition at the end of the cycle. A state function would have the same value at the beginning and end of the cycle and D(state function) = 0 for the cycle. A path function depends on the path taken and D(path function) ¹ 0 for the cycle. To illustrate this we calculate the total work dw and heat dq for the cycle.

w = dw_I + dw_II + dw_III + dw_IV

=-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(V₄/V₃) [since dw_II = - dw_IV]

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]

Since T_hot ¹ T_cold dw ¹ 0

q = dq_I + dq_III [since dw_II = 0 and dw_IV = 0]

= - dw_I - dw_III [since dU = 0 for isothermal steps]

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

dq_I/T_hot + dq_III/T_cold = nRln(V₂/V₁) - nRln(V₂/V₁) = 0.

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

dS = dq_rev/T

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.

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

h = |w_total|/q_hot

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

h = | T_cold – T_hot |/T_hot = (T_hot – T_cold)/T_hot = 1 - T_cold/T_hot

Therefore we only need to know the temperature of the engine during the expansion, T_hot and the temperature of the surroundings, T_cold to calculate the thermodynamic efficiency. For example, if the ambient temperature is 300 K and the temperature in the piston of your engine is 500 K we calculate h = 1 – 300 K/500 K = 0.4 or 40%. Suppose a materials scientist discovers a metal that can withstand a 600 K operating temperature. What efficiency could be achieved using this metal to build your engine? The formalism discussed can be used to model other aspects of your car engine.

Thermodynamic temperature scale

Recall that dq_I/T_hot + dq_III/T_cold = 0. We can express dq_I as q_hot and dq_III as q_hot cold. The above equation implies that one of the heats must be negative. It must q_cold since that is a compression step and the work is positive on step III. We can write

q_hot/T_hot = - q_cold/T_cold

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

|q_hot|/T_hot = |q_cold|/T_cold

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

DS = q/(500 K) + q/(300 K) = q[1/500 + 1/300]

which depends entirely on the sign of q. Since q = C_vDT

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

DS = C_v[-100/500 + 100/300] = C_v(1/3 – 1/5) = 2C_v/15 > 0.

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.

The example we have discussed relates to the point made by McQuarrie and Simon on page 825 (Section 20-3) where they state that : because dS = dq_rev/T, energy delivered as heat at a lower temperature contributes more to an entropy (disorder) increase than at a higher temperature.”

The example of the two bricks in thermal contact is a version of the description in Section 20-4 of the entropy. We have shown that for a spontaneous process dS > 0. Clearly, if the two blocks are at equilibrium (say at 400 K) dq = 0 and dS = 0. Based on this example we have

dS > 0 (spontaneous process)

dS = 0 (equilibrium)

The entropy change is also zero for an isolated (adiabatic) reversible process. Consider the expansion discussed above. Phase I is isothermal and phase II is adiabatic. For the adiabatic phase dq_rev = 0 and therefore dS = 0. Note that during this process the volume increases, but the temperature decreases. Apparently, the increase in disorder due to the increasing volume is exactly offset by the decreasing disorder due to the decreasing temperature. This allows us to state that

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.

dS = dq_rev/T = nC_vdT/T at constant volume

To obtain DS we need to integrate both sides

The left hand side is DS = S₂ – S₁ and the right hand side is nC_vln(T₂/T₁). Exactly the same reasoning applies at constant pressure, so that DS = nC_pln(T₂/T₁).

For a constant temperature expansion we can state

dS = dq_rev/T = - dw_rev/T. The logic behind this statement is that the internal energy change is zero for a constant temperature process and so dq_rev = - dw_rev. To calculate the reversible work we simply plug in dw_rev = -PdV. According to the ideal gas law P = nRT/V so dS = nRdV/V.

The result of this equation is that DS = nRln(V₂/V₁) at constant temperature.

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.

a. constant pressure expansion, w = -P_extDV = - P_ext(V_f – V_i)

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

The P-V plot for an expansion has the form

The work for the isothermal expansion is the negative of the area under the blue curve and the work for the constant pressure expansion is the negative of the area under the violet line. The two works are clearly not equal (as we saw earlier).

Regardless of the path chosen, the entropy change for the system will be the same since it is a state function. The entropy along path a. must be calculated in two steps (a constant volume decompression and a constant pressure expansion) in order to get from the initial to the final state.

dS = dq_p/T + dq_v/T = C_pdT/T + C_vdT/T

constant volume step: P_i, V_i, T_i ® P_f, V_i, T_e

constant pressure step: P_f, V_i, T_e ® P_f, V_f, T_f

Note that T_f = T_i since they lie along an isotherm.

DS = C_pln(T_e/T_i) + C_vln(T_i/ T_e) = (C_p – C_v)ln(T_e/T_i)

= nRln(T_e/ T_i).

The ratio T_e/ T_i = V_f/ V_i

We conclude that for the system, DS = nRln(V_f/ V_i) which is exactly what we calculate for the reversible path! The statement that we always should calculate the entropy along a reversible path is really for ease of calculation. We must get the same answer no matter which path we follow.

So far we have been discussing how to calculate entropy for the system. The total entropy change for a process must include the entropy change in the system and the surroundings. We just saw that DS_system does not depend on path. How about DS_surr, the entropy change in the surroundings?

Returning to the expansion. For the reversible path dU = 0, and dq_rev = -dw_rev. The heat flow into the surroundings is q_rev and therefore the heat flow from the surroundings into the system is –q_rev. The entropy change in the surroundings is DS_surr = -q_rev/T. In an isothermal expansion, heat actually flows from the surroundings into the system to maintain constant temperature. The point, however, is that for the reversible path the entropy change in the surroundings is exactly equal and opposite to that in the system.

DS_surr = -DS_system

DS_total = DS_surr + DS_system = 0

For the irreversible process (the constant pressure expansion), we find that for the surroundings the heat flow is –q_irr. For the total path from the initial to the final state we know that dU = 0. The total work done along the irreversible path is w = -P_extDV. Therefore, the heat transferred from the system to the surroundings is equal to q_irr = P_extDV = P_ext(V_f–V_i). To calculate the entropy change in the surroundings we use the fact that

q(surroundings) = - q(system)

DS_surr = -q_irr/T = -P_ext(V_f–V_i)/T is always less than DS_system for an expansion since the reversible work is the maximum work possible (see the P-V plot above). McQuarrie and Simon give the limiting case of zero pressure in section 20-6. If P_ext = 0 then q_irr = 0 and clearly DS_surr = 0.

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

DS_surr = -q_irr/T = -(21.4 L-atm)/(300 K) = - 0.071 L-atm/K

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

DS_surr = nRln(V_f/V_i). In the present example V_f/V_i = 10.

DS_surr = (1 mole)(0.082 L-atm/mole-K)ln(10) = 0.188 L-atm/K

DS_total = DS_surr + DS_system = 0.188 – 0.071 = 0.117 L-atm/K.

To conclude we have shown that

DS_total = 0 for a reversible process (equilibrium)

DS_total > 0 for an irreversible process (spontaneous)

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

We can calculate the entropy change in the system at

Constant temperature	DS = nRln(V2/V1)
Constant volume	DS = nCvln(T2/T1)
Constant pressure	DS = nCpln(T2/T1)

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

S = klnW

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

We use Stirling’s approximation lnN! » NlnN – N.

And using the fact that N = Sn_j we have

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

And since Sp_j = 1, we have

For a collection of N particles or molecules, or

on a per molecule basis.

In terms of the ensemble partition function Q for the canonical ensemble we note that since p_j = e^-bej/Q the above expression becomes

S = U/T + klnQ