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# Meaning of the Model Relationships

1. It is often useful to divide all energy into mechanical and internal pieces.
2. Change in total energy of a system is equal to the energy added to the system. Energy added can be separated into heat, Q, and work, W.
3. When changes in energy are restricted to only the internal energies, conservation of energy reduces to the statement: the change in internal energy is equal to the energy added as heat and work. This statement is referred to as the First Law of Thermodynamics.

Allowing for a transfer of energy into a physical system from outside the system as either heat or work or both, we can express conservation of energy as:

$\Delta E = Q + W$

which actually means

$\Delta E_{\text{physical system}} = Q_{in} + W_{in}$

This equation is always true as long as we interpret ∆E to include all changes of energy associated with the physical system and we mean by Q and W the addition of energy as heat or as work to the system when Q and W are positive. (A negative Q or W means that energy is removed from the system.)

$\Delta E = \Delta E_{\text{mechanical}} + \Delta U = Q + W$

where ∆U = ∆Ebond + ∆Ethermal + ∆Eatomic + ∆Enuclear and is called the change in internal energy. Eatomic includes changes in the atomic states of atoms that are not involved in chemical bonds. In normal chemical reactions and physical phase changes, there are no changes in nuclear energies, so the last term is zero.

First Law of Thermodynamics

If there are no changes in Emechanical, then the expression representing conservation of energy for a particular physical system becomes the familiar relation used by physicists and chemists known as the first law of thermodynamics:

$\Delta U = Q + W$

In chemistry, the first law of thermodynamics at constant pressure is typically written as:

$\Delta U_{rxn} = q - P \Delta V$

The subscript “rxn” signifies that this is a change in energy due to a reaction and the lower case “q” signifies that heat is not a state function, but rather a transfer of energy. The work is expressed explicitly as -P∆V, since only this kind of work is assumed to occur. Rest assured, however, that we are talking about exactly the same thing. You should practice “seeing through or seeing past” the particular symbols to the content represented by an equation. These last two equations for ∆U are perfect examples of this point. Do you look at them and see two different strings of symbols to memorize, or do you immediately recognize them as expressing the same physical content?

It is sometimes useful to write the first law of thermodynamics in differential form:

$dU = dQ + dW$

(dQ is interpreted here to be a small transfer of energy, rather than a true differential)

4) The entropy, S, of a closed system can never decrease in any process (the closed system could be a combined open system and the surroundings with which it interacts). If the process is reversible, the entropy of the closed system remains constant. If the process is irreversible, the entropy of the closed system increases. These relationships are referred to as the Second Law of Thermodynamics.

Definition of Temperature

In thermodynamics, the real definition of temperature is different than what one might expect. Rather than an operational definition (a definition based on experiment), such as defining temperature based on the readings of thermometers, the definition comes from a discussion of entropy and heat. $T \equiv \frac{dU}{dS}$

More commonly written as $\frac{1}{T} \equiv \frac{dS}{dU}$

Because dS/dU is the rate at which entropy changes when heat is added, whereas dU/dS implies a strange causal relationship (that changing the entropy changes the thermal energy, when the thing we can directly affect is heat, not entropy).

You might learn about this definition in greater detail in a later course. For now, think of it as a useful relationship that is worth remembering.

5) When a small amount of heat, dQ, is added to a system the differential change in entropy, dS, varies in direct proportion to dQ. The constant of proportionality is the reciprocal temperature, 1/T . (dS = dQ/T).

6) For processes in which all work is reversible, the heat added to or removed from the system can be expressed in terms of the change in entropy using the previous relationship (dQ = T dS).

The restriction of the reversible process applies only to calculating ∆S (using ∆S = Q/T or dS = dQ/T) as a system moves from one state to another. The actual change in ∆S of the system is the same, regardless of whether the change is reversible or not. That is the meaning of S being a state function. But to calculate ∆S, we must move along a hypothetical path reversibly. This is the beauty or “magic” of thermodynamics: we can calculate values of changes in state functions for very complicated processes by imagining that we move the system between the two states in a manner that makes it easy to do the calculations. But the answers we get for the changes in the state functions using this approach are the same even when the system doesn’t actually change in this way!

Closer Look at Entropy from a thermodynamic viewpoint

The fundamental relationship between entropy and energy is the following. Imagine adding heat to a system reversibly, i.e., no friction occurs and no energy is transferred as heat across a finite temperature difference. Under these conditions there is a change in entropy (entropy is created or destroyed) according to the following relationship:

$$dS = \frac{dQ}{T}$$ (reversible process)

Now S is a state variable, so $$\Delta S = S-{f} - S_{i}$$

For constant T, $$T \Delta S = Q$$

or $$\Delta S = \frac{Q}{T}$$ (reversible process, constant T)

If Q is positive (heat is added to the system) ∆S is positive. On the other hand, if heat is removed, the entropy is reduced. We also see from the above relationship that the SI units of entropy are joules per kelvin.

Calculating ∆S

Since ∆S is a state variable and depends only on the initial and final state, we should be able to directly express it as a function of other state variables. We can use the first law of thermodynamics to do this.

$dQ = dU - dW = dU + PdV$

Substituting this into the our expression for entropy, dS = dQ / T, we get:

$dS = \frac{dU}{T} + P \frac{dV}{T}$

If the temperature is constant, then we can also write

$\Delta S = \frac {\Delta U}{T} + P \frac{\Delta V}{T}$

Experimentally, the most direct route to the determination of values of ∆S is through heat capacity data. In general, C = dQ/dT, so

$dS = \frac{dQ}{T} = C \frac{dT}{T}$

and $\Delta S = \int_{T_{i}}^{T_{f}} \frac{C(T)}{T} dT$

Either Cp or Cv can be used in the above relation. If Cv is used, the two state variables describing the system are T and V (V constant) and if Cp is used, the two state variables describing the system are T and P (P constant).

If C(T) is approximately constant over the temperature range in question, it can be pulled out of the integral and we have the useful result

$\Delta S = C \cdot ln( \frac{T_{f}}{T_{i}})$

7) New energy state functions can be formed from combinations of other state functions. Two common examples are enthalpy, H, and Gibbs energy, G. (H = U + PV; G = H – TS)

8) At constant pressure, the energy added to a system can be expressed as a change in enthalpy (Q = ∆H).

This is a very useful relationship! This is because so much experimental work is done in open containers at one atmosphere of pressure. This is worth looking at more closely to see how it comes about.

Enthalpy

The enthalpy of a physical system is a function of state variables. It is defined as

$H = U + PV$

Note that the three variables that define H are all themselves state variables

We note for future reference that P is an energy density, which, when multiplied by volume, is an energy. Why would a function equal to the internal energy plus the product of the two state variables pressure and volume be useful? As with energy, it is the changes in H that occur during a process that are important. Changes in enthalpy are most useful under certain conditions: specifically, situations in which pressure is held constant. Chemists and biologists conduct many of their experiments (any reaction carried out in a container open to the atmosphere) under constant pressure.

The following derivation shows why enthalpy is a convenient variable when considering changes occurring under constant pressure:

$\Delta H = \Delta U + P \Delta V$

or $dH = dU + P dV$

For ∆U we can substitute the expression for internal energy we obtained from the first law of thermodynamics (at constant pressure): ∆U = Q - P∆V. This gives us

$\Delta H = Q - P \Delta V + P \Delta V$

and ∆H = Q or dH = dQ (constant pressure process and only PV work)

So at constant pressure, the enthalpy change during a reaction is simply equal to the heat entering the system. Likewise, if a phase change is allowed to occur at constant pressure, the heats of melting and vaporization are simply equal to the changes in enthalpy. So now we see why heats of fusion and vaporization are often listed as changes in enthalpies. When you measure the energies transferred as heat during a phase change at constant pressure, you are directly measuring the change in the state function H. A negative ∆H means heat is transferred out, and the reaction is exothermic. A positive ∆H means heat is transferred in and the reaction is endothermic.

∆H < 0 exothermic

∆H > 0 endothermic

It is because dH = dQ (constant pressure and only PV work) for the conditions so common in chemistry and biochemistry, and technology in general, that enthalpy is so useful. Most measurements are carried out at constant pressure, and for many systems, the only work involved comes from compressing or expanding a gas.

9) At constant pressure and temperature, the change in the Gibbs energy expresses a system’s ability to satisfy both the conservation of energy and the tendency of entropy increase, i.e., satisfaction of both the 1st and 2nd law of thermodynamics.
∆G = ∆H – T ∆S. ∆G represents the amount of energy available to do useful work in a reaction or process.

10) The direction in which a reaction or process proceeds is determined by the sign of ∆G for the process:

If ∆G < 0, the reaction is spontaneous

If ∆G > 0, the reaction is not spontaneous

Note: A full discussion of Gibbs Energy is included in the discussion of the Intro Statistical Model of Thermodynamics at the end of this chapter:

11) The values of a small number of State Functions completely define the state of a thermodynamic system. These values are independent of the path that took the system to that state.

12) The amounts of heat and work involved in going from State A to State B will depend on the particular path taken.

13) Values of state functions at state B can be readily calculated, if they are known at state A, because a reversible path can be used for the calculation to get from A to B, even if the actual path the physical system takes, is irreversible.

The end points define the initial and final states. Those states do not depend on the path taken. All real processes have some irreversibility associated with them. However, it is usually possible to move along a path on an appropriate state diagram that is reversible, and thus possible to calculate the changes in the various state variables. So, calculations along reversible paths can allow you to find the values of the state variables. But the point of relationship (12) is that the split between the amount of energy transferred as heat and transferred as work will depend on the actual path taken, as well as the amount of irreversibility in the real process.

## More on Heat Capacities

Under the assumption that there are no bond energy changes and the only energy systems are the thermal and bond systems, the first law of thermodynamics reduces to

$dU = dQ + dW$

$dE_{th} = dQ + dW \text{, since} E_{bond} = 0$

Solving for dQ and differentiating with respect to T we obtain a general expression for the heat capacity in terms of our model:

$C = \frac{dQ}{dT} = \frac{dE_{th}}{dT} - \frac{dW}{dT}$

Previously, we wanted the constant volume heat capacity. In that case, the dW/dT term was zero, so we had for constant volume:

$C_{v} = \frac{dQ}{dT} = \frac{dE_{th}}{dT}$

For constant pressure, the work term is not zero, so we need to evaluate it for the particular substance we are concerned about.

The first step is to write dW as -PdV, since we are concerned with fluids (liquids and gases). The expression for Cp then becomes

$C_{p} = \frac{dQ}{dT} = \frac{dE_{th}}{dT} + P \frac{dV}{dT}$ or

$C_{p} = C_{v} + P \frac{dV}{dT}$

To go any further we need to evaluate dV/dT, which will depend on the substance. We will evaluate this within the Ideal Gas Model shortly.

## Algebraic Representations

Relationships 1-3

$$\Delta E = \Delta E_{\text{mechanical}} + \Delta U$$

$$\Delta E_{\text{physical system}} = Q_{\text{into physical system}} + W_{\text{done on physical system}}$$

or simply $\Delta E = Q + W$

$$\Delta U = \Delta E_{bond} + \Delta E_{thermal} + \Delta E_{atomic} + \Delta E_{nuclear}$$

The 1st Law of Thermodynamics

∆U = Q + W in differential form: dU = dQ + dW and for simple fluids: dU = dQ – PdV
and for reversible fluid systems or along reversible paths: dU = TdS – PdV

Because dU = TdS – PdV involves only functions of state (Q’s and W’s have been replaced by state functions), it is true for all processes, whether they are reversible or not.

Expression for work done on a fluid

$W = - \int_{V_{i}}^{V_{f}} P(V) dV \text{ or in differential form:} dW = -P(V) dV$

Relations 4-6

The Second Law of Thermodynamics:

In a closed system, $$\Delta S \geq 0$$

$dS = \frac{dQ}{T}$ in a reversible process

Extension of the concept of heat capacity:

Heat Capacity

$$C = dQ/dT$$

Heat Capacity at Constant Volume and Constant Pressure (with ∆EBond = 0)

$$C_{v} = dE_{th}/dT C_{p} = C_{v} + P \cdot dV/dT$$

Relations 7-10

Enthalpy

$$H = U + PV$$ , At constant pressure, $$\Delta H = Q$$

Gibbs Energy

$$G = H - TS$$ , At constant pressure and temperature, $$\Delta G = \Delta H – T \Delta S$$

If $$\Delta G < 0$$, the process is spontaneous

If $$\Delta G > 0$$, the process is not spontaneous

If $$\Delta G = 0$$ , the forward and reverse directions are in equilibrium

Equations of Sate:

Ideal Gas Law: $$PV = nRT$$ or $$PV = N k_{B} T$$