Chapter

The Laws of Thermodynamics

Blinn College - Physics 1401 - Terry Honan

. - Work and the First Law

Thermodynamic Work

As we saw in Chapter F, the work done in one dimension is W=∑ F Δ x.  Consider a gas at a pressure P expanding in a piston. If the piston has a cross-sectional area of A and it expands by the infinitesimal amount dx, then the infinitesimal work done by the gas is F Δ x=P A Δ x=P Δ V.  Here we have used F=P A and Δ V=A Δ x.

This result is more general; if any system at a pressure P expands by Δ V, the small amount of work is P Δ V.

The work done by a thermodynamic system is

W=∑ P Δ V.

When a system expands it does work on the environment. Our sign convention is that W is the work done by the system, where positive work decreases the energy of the system. The work done on the system is -W. The historical reason for this convention is that the focus was on getting work from engines, where the engine was the system.

For the simple case of expansion with a constant pressure we get

W=P Δ V.

Generally, when the volume increases the work is positive; it is negative when volume decreases.

PV-Diagrams and Thermodynamic States

Suppose there is some thermodynamic system, for instance a fixed quantity of some substance.  Knowledge of the pressure P and volume V uniquely specifies the thermodynamic state.  For example, one could find the temperature uniquely from P and V.  The equation that relates P, V and T is called the equation of state.  For complex thermodynamic systems this is not a simple function that one can write down, but for an ideal gas it is just the ideal gas law.

A PV-diagram is a graph of P vs. V (P is the y-axis and V is the x-axis.)  Work has a simple interpretation in terms of PV-diagrams; it is the area under the curve.

W= ∑ P Δ V=±Area Under

Since P is always positive the sign of P Δ V is the same as on the sign of Δ V.

In thermodynamics we often deal with cycles. A cycle is a closed path in a PV-diagram. For a cycle there is a positive and a negative contribution.  If the path is clockwise the positive contribution dominates and the result is positive and it just then enclosed area. When the path is counterclockwise the result is negative.

(Subscript[W, cycle])=±Area Enclosed

Example . - The PV-Diagram

Solution

1(QuantityUnit)=1.013×(10^5)(QuantityUnit)   ,and,  1(QuantityUnit)=(10^(-3))(((QuantityUnit))^3),   ⟹   1(QuantityUnit) (QuantityUnit)=101.3(QuantityUnit)

(Subscript[W, A,B])=+(1/2) 4(QuantityUnit)  (2(QuantityUnit)+8(QuantityUnit)) =+20(QuantityUnit) (QuantityUnit)=+2026(QuantityUnit)

(Subscript[W, A,C,B])=+ 4(QuantityUnit)×8(QuantityUnit) =+32(QuantityUnit) (QuantityUnit)=+3242(QuantityUnit),   and  ,(Subscript[W, A,D,B])=+ 4(QuantityUnit)×2(QuantityUnit) =+8(QuantityUnit) (QuantityUnit)=+810(QuantityUnit),

Solution

(Subscript[W, B,A])=-(Subscript[W, A,B])=-2026(QuantityUnit)

Solution

(Subscript[W, A,D,B,C,A])=- 4(QuantityUnit)×6(QuantityUnit) =-24(QuantityUnit) (QuantityUnit)=-2431(QuantityUnit),

(Subscript[W, A,B,D,A])=+(1/2) 4(QuantityUnit)×6(QuantityUnit) =+12(QuantityUnit) (QuantityUnit)=+1215(QuantityUnit),

Quasi-static Processes

This discussion of clear paths in a PV-diagram implies that we are undergoing a quasi-static process. A process is quasi-static if it is performed sufficiently slowly, so that at each intermediate state the system stays in thermal equilibrium. As an example of a process that is not quasi-static, consider the free expansion of a gas. Start with a container containing two regions separated by a partition. On one side of the partition is a pressurized gas and on the other side is a vacuum. When the partition is removed, the gas undergoes a free expansion. The initial and final states are in equilibrium but the intermediate states are not. When a system is not in equilibrium, pressure and temperature are not well-defined quantities.

It is possible, theoretically, to do this expansion with a quasi-static process by slowly dragging a piston from initial to final positions.

In this example with an expanding gas, approximating a quasi-static process need not be too slow. To be in thermal equilibrium the system must be at a uniform temperature; there can be no temperature gradients. Suppose heat is added to a solid block. To be in equilibrium every point in the block must be at the same temperature. An approximately quasi-static process here would be very slow; each time after a small amount of heat is added, one must wait for the system to equilibrate.

Adiabatic and Isothermal Processes

A process is said to be adiabatic when it is sufficiently well-insulated so that no heat enters or leaves the system. So for an adiabatic process Q=0. If a process is done quickly, like the free-expansion of a gas discussed above, then there is not enough time for heat to flow and it is adiabatic. The word isothermal implies the temperature is constant, so an isothermal process is at constant temperature.

Work for Isothermal Expansion of an Ideal Gas

For an ideal gas we can use the ideal gas law, the equation of state, to write the pressure as a function of volume.  When the gas undergoes an isothermal expansion its temperature stays constant.

P(V)=((n R T)/V)  (T is constant)

The work is the area under this curve. Using integral calculus one can show that this becomes:

W=n R T  ln ((Subscript[V, f])/(Subscript[V, i])).

Note that the natural log function guarantees that result of the work formula has the correct sign, since the natural log of a ratio less than one is negative.

Example . - Isothermal Expansion

Solution

m=0.050(QuantityUnit) , T=20(QuantityUnit)=293(QuantityUnit)   and   ((Subscript[P, f])/(Subscript[P, i]))=3

(Subscript[m, (Subscript[O, 2])])=2 (Subscript[m, O])=2×16(QuantityUnit)=32(QuantityUnit)  ⟹  (Subscript[m, mol])=32(QuantityUnit)/(QuantityUnit)=0.032(QuantityUnit)/(QuantityUnit)

m=n (Subscript[m, mol])  ⟹  n=(m/(Subscript[m, mol]))=1.56(QuantityUnit)

P V=n R T  (Underscript[⟹, n,T const.])  (Subscript[P, i])(Subscript[V, i])=(Subscript[P, f])(Subscript[V, f])  ⟹  ((Subscript[V, f])/(Subscript[V, i]))=((Subscript[P, i])/(Subscript[P, f]))=(1/3)

W=n R T  ln ((Subscript[V, f])/(Subscript[V, i]))  and  R=8.314 (((QuantityUnit))/((QuantityUnit)·(QuantityUnit)))  ⟹  W=-4180(QuantityUnit)

State Functions

A state function is a function of pressure and volume.  It has a unique value at each point in a PV-diagram.  Temperature is a state function, since it follows from the equation of state.

f=f(V,P) is a state function.

When a system changes from ((Subscript[V, i]),(Subscript[P, i])) to ((Subscript[V, f]),(Subscript[P, f])) a function f has a change of Δ f given by

Δ f=f((Subscript[V, f]),(Subscript[P, f]))-f((Subscript[V, i]),(Subscript[P, i]))

A cycle is some process that forms a closed loop in a PV-diagram; it follows that for a state function in a cycle: Δ f=0.

Internal Energy and the First Law

The energy of a thermodynamic system can be increased by adding thermal energy in the form of heat or by adding mechanical energy by doing work on the system.  We will write the internal energy of a thermodynamic system as U.  The first law of thermodynamics is

Δ U=Q-W.

Δ U is the change in the internal energy of the system.  Q is the heat added to the system.  W is the work done by the system.  It follows that -W is the work done on the system.  Internal energy is a state function.  Since it is a state function then Δ U is independent of the path taken.  Since the work is path dependent it follows from the first law that the heat is also path dependent, and not a state function.

Example . - The PV-Diagram (Continued)

Solution

W=(Subscript[W, A,B,D,A])=+12(QuantityUnit) (QuantityUnit)=+1215(QuantityUnit),

0=Δ U=Q-W   ⟹  Q=W=+1215(QuantityUnit),

Solution

Q=-0.30(QuantityUnit)×((4186(QuantityUnit))/((QuantityUnit)))=-1256(QuantityUnit)

W=(Subscript[W, B,A])=-2026(QuantityUnit)

Δ U=Q-W=+770(QuantityUnit)

. - Entropy

Irreversibility and Reversible Processes

Thermodynamics puts an arrow on time. Take a video of dropping an egg on the floor and run it in reverse. Clearly, the reversed video would show itself as unphysical. All real-world physical processes are irreversible but we can consider idealized reversible processes.

A quasi-static process is reversible. As we saw last chapter a quasi-static process is one where we have an equilibrium state for each intermediate state in a process; they are given by clear paths in PV-diagrams. By following backward along the path in a PV-diagram we can restore the system and its surroundings to their original state. A thermodynamic system undergoing an irreversible process can always be returned to its original state but in both the process and return, a net quantity of heat will flow to the environment; this cannot be undone. Whenever heat flows spontaneously across a temperature gradient we have an irreversible process. The free expansion of a gas, as discussed earlier is not quasi-static and thus not reversible. This chapter will discuss entropy and the second law of thermodynamics. For a system and its environment, a reversible process has zero change in the total entropy and an irreversible process has a net increase in entropy.

A reversible (quasi-static) process is an idealization. When the state of a system is changed by a small amount it moves slightly away from equilibrium, but if done sufficiently slowly the deviations from equilibrium can be negligible and the process can be arbitrarily close to being reversible.

Definition of Entropy

Entropy S is a measure of disorder.  One is not used to thinking of disorder as a precise notion that can be quantified, but it turns out that we can give it a precise definition.  Doing so is beyond the scope of this course.

If we add heat to a system then we increase its disorder.  For example, water at 0°C is a more disordered state than ice at 0°C; to go from ice to water requires only heat.  We define entropy as the state function; for a small Q entropy is defined by

Δ S=(Q/T).

When heat is added to a system its entropy increases.  When it is removed (Q<0) it decreases.

Constant Temperature and Latent Heats

If the temperature T is a constant then summing all the small Δ S above gives the same expression

Δ S=(Q/T)

But now the Q is the total amount of heat added to the system. When some substance changes phase, its temperature stays constant. In that case the heat can be written in terms of the latent heat and mass.

Q=±m L

Temperature Change and Specific Heats

We can find entropy changes in cases of changing temperatures using the specific heat formula Q=m c Δ T. This can be done with calculus by summing over small changes in temperature Δ T.

Δ S=∑((m c Δ T)/T)

The result from calculus becomes.

Δ S=m c ln ((Subscript[T, f])/(Subscript[T, i]))

Example . - Making Ice Cubes

Solution

m=0.45(QuantityUnit) ,  22(QuantityUnit)=295(QuantityUnit) ,  0(QuantityUnit)=273(QuantityUnit) ,  -18(QuantityUnit)=255(QuantityUnit) ,
(Subscript[c, W])=4186(((QuantityUnit))/((QuantityUnit) (QuantityUnit))) , (Subscript[c, I])=2100(((QuantityUnit))/((QuantityUnit) (QuantityUnit))),  and,  (Subscript[L, f])=3.34×(10^5)(((QuantityUnit))/((QuantityUnit)))

Water ,at ,22(QuantityUnit)   (Overscript[⟶, 1])   Water, at ,0(QuantityUnit)  (Overscript[⟶, 2])   Ice, at, 0(QuantityUnit)   (Overscript[⟶, 3])   Ice, at -18(QuantityUnit)

Δ (Subscript[S, 1])=m c ln ((Subscript[T, f])/(Subscript[T, i]))=m (Subscript[c, W])ln((273(QuantityUnit))/(295(QuantityUnit)))=-146.99(((QuantityUnit))/((QuantityUnit)))

Δ (Subscript[S, 2])=((±m L)/T)=((-m (Subscript[L, f]))/(273(QuantityUnit)))=-550.55(((QuantityUnit))/((QuantityUnit)))

Δ (Subscript[S, 3])=m c ln ((Subscript[T, f])/(Subscript[T, i]))=m (Subscript[c, I])ln((255(QuantityUnit))/(273(QuantityUnit)))=-64.46(((QuantityUnit))/((QuantityUnit)))

Δ (Subscript[S, ice tray])=Δ (Subscript[S, 1])+Δ (Subscript[S, 2])+Δ (Subscript[S, 3])=-761.0(((QuantityUnit))/((QuantityUnit)))

. - The Second Law of Thermodynamics

Statement of the Second Law

A thermally isolated system is one with no heat entering or leaving it.  This could consist of smaller systems exchanging heat between them but no heat is allowed to leave or enter the larger system.  The total change in entropy for a system is written as Δ (Subscript[S, tot]).

For a thermally isolated system the total change in entropy cannot be negative.

Δ (Subscript[S, tot])≥0

Heat Reservoirs

In thermodynamics we define a reservoir to be something sufficiently large to maintain a constant temperature given whatever amount of heat exchange as is relevant.  If the amount of heat transferred is sufficiently small then a reservoir could be as small as a glass of water.  If the amount of heat is large then we must consider reservoirs that are very large.

Example . - Making Ice Cubes (Continued)

Solution

0=(Subscript[Q, tot])=(Subscript[Q, freezer])+(Subscript[Q, ice tray])  ⟹  (Subscript[Q, freezer])=-(Subscript[Q, ice])

m=0.45(QuantityUnit) ,  22(QuantityUnit)=295(QuantityUnit) ,  0(QuantityUnit)=273(QuantityUnit) ,  -18(QuantityUnit)=255(QuantityUnit) ,
(Subscript[c, W])=4186(((QuantityUnit))/((QuantityUnit) (QuantityUnit))) , (Subscript[c, I])=2100(((QuantityUnit))/((QuantityUnit) (QuantityUnit))),  and,  (Subscript[L, f])=3.34×(10^5)(((QuantityUnit))/((QuantityUnit)))

(Subscript[Q, 1])=m (Subscript[c, W])Δ (Subscript[T, W])=-14441(QuantityUnit)

(Subscript[Q, 2])=±m L=-m (Subscript[L, f])=-150300(QuantityUnit)

(Subscript[Q, 3])=m (Subscript[c, I])Δ (Subscript[T, I])=-170010(QuantityUnit)

(Subscript[Q, ice])=(Subscript[Q, 1])+(Subscript[Q, 2])+(Subscript[Q, 3])=-208751 (QuantityUnit)

(Subscript[Q, freezer])=-(Subscript[Q, ice])   and   Δ (Subscript[S, freezer])=(Q/T)=((Subscript[Q, freezer])/(255(QuantityUnit)))=-((Subscript[Q, ice])/(255(QuantityUnit)))=818.6(((QuantityUnit))/((QuantityUnit)))

Δ (Subscript[S, total])=Δ (Subscript[S, freezer])+Δ (Subscript[S, ice tray])=57.6(((QuantityUnit))/((QuantityUnit)))

The Second Law and the Direction of Heat Flow

In any spontaneous process, heat flows from hot to cold.  This is an immediate consequence of the second law.  Consider an amount of heat Q flowing from a reservoir at temperature (Subscript[T, 1]) to a second reservoir at temperature (Subscript[T, 2]).  Since heat is leaving (Subscript[T, 1]) its entropy decreases, while the entropy of (Subscript[T, 2]) increases.  The total change in entropy is the sum of these two; by the second law it must be nonnegative.

0≤Δ (Subscript[S, tot])=Δ (Subscript[S, 1])+Δ (Subscript[S, 2])=((-Q)/(Subscript[T, 1]))+(Q/(Subscript[T, 2]))   ⟹  (1/(Subscript[T, 1]))≤(1/(Subscript[T, 2]))

Cross-multiplying and remembering that absolute temperatures are always positive gives

(Subscript[T, 2])≤(Subscript[T, 1]).

Thus, heat flows from a higher to a lower temperature.

Refrigerators, Air Conditioners and Heat Pumps

It is possible however, to have a situation where heat is taken from a cold reservoir and vented to a hotter reservoir.  The key phrase in the above discussion is spontaneous process.  A refrigerator removes heat from a cold interior and vents it to the warm coils behind.  The crucial point is that more heat is vented to the back than is removed from the interior.  The additional energy is provided by some motor (a compressor) doing work; it pumps the coolant around some cycle.

Consider a cold reservoir at temperature (Subscript[T, C]) and a hot reservoir at (Subscript[T, H]).  (Subscript[Q, C]) is removed from (Subscript[T, C]) and (Subscript[Q, H]) then is vented to (Subscript[T, H]).  The conservation of energy implies that

(Subscript[Q, H])=(Subscript[Q, C])+W.

This says that the additional energy comes from the work done by the motor.  The change in the total entropy is

Δ (Subscript[S, tot])=Δ (Subscript[S, C])+Δ (Subscript[S, H])=((-(Subscript[Q, C]))/(Subscript[T, C]))+((Subscript[Q, H])/(Subscript[T, H])).

It should be clear that if W is large enough that the (Subscript[Q, H]) may be sufficiently large to cancel any negative Δ (Subscript[S, C]).  Heat is moved from cold to hot.

An air conditioner is simply a refrigerator operating between the cold interior of a house and the warmer exterior.  A heat pump is an air conditioner running in reverse on a cold winter day.  The outdoors is air conditioned!  Heat (Subscript[Q, C]) is removed from the cold exterior and more heat (Subscript[Q, H]) is vented inside.  The amount of electrical energy that is used is determined by the work W.  The heat vented inside is larger than the amount of electrical energy that is used.  The (Subscript[Q, C]) is essentially a bonus.

. - Heat Engines

Heat Engines and Efficiency

A heat engine is essentially an air conditioner running in reverse.  Some fuel is burned creating an input heat of (Subscript[Q, H]).  The point of the engine is to convert this heat to work W.  We will define the efficiency of a heat engine to be the energy output of the engine, the work, divided by the energy input, (Subscript[Q, H]).

e=(W/(Subscript[Q, H]))

A perfect heat engine would be 100% efficient.  Unfortunately this is, because of the second law, impossible.  There is a decrease in entropy without any increase to cancel it.

With a heat engine we have a hot reservoir at temperature (Subscript[T, H]) producing the input heat (Subscript[Q, H]).  Some of that heat is converted to work W, but necessarily some of it must be vented to the environment.  Take the environment to be a cold reservoir at (Subscript[T, C]) receiving the vented heat (Subscript[Q, C]).

Conservation of energy gives

(Subscript[Q, H])=(Subscript[Q, C])+W.

With this we can rewrite the efficiency

e=1-((Subscript[Q, C])/(Subscript[Q, H])).

The Second Law and Efficiency

The second law has the following form

Δ (Subscript[S, tot])=Δ (Subscript[S, C])+Δ (Subscript[S, H])=((Subscript[Q, C])/(Subscript[T, C]))+((-(Subscript[Q, H]))/(Subscript[T, H]))≥0.

This gives a fundamental upper limit on the efficiency of a heat engine.

((Subscript[Q, C])/(Subscript[T, C]))+((-(Subscript[Q, H]))/(Subscript[T, H]))≥0  ⟹  ((Subscript[Q, C])/(Subscript[Q, H]))≥((Subscript[T, C])/(Subscript[T, H]))  ⟹  1-((Subscript[Q, C])/(Subscript[Q, H]))≤1-((Subscript[T, C])/(Subscript[T, H]))

It follows that

e≤(Subscript[e, max])  ,where,  (Subscript[e, max])=1-((Subscript[T, C])/(Subscript[T, H])).

A Carnot Engine

A Carnot engine is a theoretical heat engine of maximum efficiency.  Its efficiency (Subscript[e, c]) is then given by

(Subscript[e, c])=(Subscript[e, max])=1-((Subscript[T, C])/(Subscript[T, H])).

Example . - Heat Engine

Solution

(Subscript[Q, H])=3800(QuantityUnit) ,  (Subscript[Q, C])=2400(QuantityUnit) ,  (Subscript[T, H])=350(QuantityUnit) =623(QuantityUnit)   and   (Subscript[T, C])=40(QuantityUnit) =313(QuantityUnit)

e=1-((Subscript[Q, C])/(Subscript[Q, H]))=0.368=36.8 %

Solution

Δ (Subscript[S, tot])=Δ (Subscript[S, H])+Δ (Subscript[S, C])=-((Subscript[Q, H])/(Subscript[T, H]))+((Subscript[Q, C])/(Subscript[T, C]))=-6.0996(((QuantityUnit))/((QuantityUnit)))+7.6673(((QuantityUnit))/((QuantityUnit)))=+1.568(((QuantityUnit))/((QuantityUnit)))

Solution

(Subscript[e, max])=1-((Subscript[T, C])/(Subscript[T, H]))=0.498=49.8 %