To the Members of Mr. Jefferson’s University:

By the driven dispositions that brought us to this school, we hold ourselves back from “sticky” situations much too often, preferring to not ask about potentially dangerous relationships, eating habits, or abuse of any kind, lest we be cast as meddlesome. While it is much easier to lower our blinds and ignore what might not be “our business,” the well-being of every individual within this community is, in fact, our business. We write today in the hopes of instilling a better understanding of the available resources, particularly CAPS and 295-TALK. We hope that students will come to see these lines as places they feel really comfortable calling just to ask any questions about student well-being, not just a number to report suspicions, and we encourage you to take advantage of them as a means to nourish the UVA community.

We wish you the best of luck as you commence your examination preparations.

Lecture 25

The Laws of Thermodynamics

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Reversible Thermal Processes

We also assume they are reversible (frictionless pistons, etc.):

For a process to be reversible, it must be possible to return both the system and its surroundings to the same states

they were in before the process began.

We will discuss 4 idealized processes with Ideal Gases:•Constant Pressure•Constant Volume•Constant Temperature•Q= 0 (adiabatic)

We will assume that all processes we discuss are “quasi-static” – they are slow enough that the system is always “in equilibrium” (fluid volumes have the same temperature throughout, etc.)

Constant pressureWork done by an expanding gas, constant pressure:

Work is area under the PV graph

Isobaric process

so changing volume implies changing temperature

Examples: piston against atmosphere, or vertical piston with constant weight on top

Work is area under the PV graph

imagining any general process as approximated by a number of constant pressure processes:

Constant VolumeIf the volume stays constant, nothing moves and no work is done.

so changing pressure implies changing temperature

Change in internal energy is related only to the net heat input

Isovolumetric process

Constant Temperature

If the temperature is constant, the pressure varies inversely with the volume.

Isothermal processes

Constant TemperatureA system connected to a large heat reservoir is usually thought to be held at constant temperature. Volume can change, pressure can change, but the temperature remains that of the reservoir.

if W < 0 (work done on the system)than Q<0 (heat flows out of the system)

if W > 0 (work done by the system)than Q>0 (heat flows out into the system)

T = constant W = Q

Work in an Constant-Temperature Process

The work done is the area under the curve:

For you calculus junkies:

Adiabatic ProcessAn adiabatic process is one in which no heat flows into or out of the system.

One way to ensure that a process is adiabatic is to insulate the system.

The adiabatic P-V curve is similar to the

isothermal one, but is steeper.

Q = 0

Another way to ensure that a process is effectively adiabatic is to have the volume change occur very quickly.

In this case, heat has no time to flow in or out of the system.

Rapid Adiabatic Process

Thermal Processes

The different types of ideal thermal processes

Specific Heat for an Ideal Gas at Constant Volume

Specific heats for ideal gases must be quoted either at constant pressure or at constant volume. For a constant-volume process,

for an ideal gas (from the kinetic theory)First Law of Thermodynamics

Specific Heat for an Ideal Gas at Constant Pressure

At constant pressure, (some work is done)

Some of the heat energy goes into the mechanical work, so more heat input is required to produce the same ΔT

for an ideal gas (from the kinetic theory)First Law of Thermodynamics

Specific Heats for an Ideal Gas

Although this calculation was done for an ideal, monatomic gas, the difference Cp - Cv works well for real gases.

Both CV and CP can be calculated for a monatomic ideal gas using the first law of thermodynamics.

Specific Heats and Adiabats In Ideal Gas

The P-V curve for an adiabat is given by

for monotonic gases

Work of a Thermal Cycle

In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:

a) positive

b) zero

c) negative

V

P

In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:

a) positive

b) zero

c) negative

The gas expands at a higher

pressure

and compresses at a lower

pressure.

In general, clockwise = positive

work;

counterclockwise = negative work.

V

P

Work of a Thermal CycleWork of a Thermal Cycle

How much work is done by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

a) 4 P1V1

b) 7 P1V1

c) 8 8 P1V1

d) 21 P1V1

e) 29 29 P1V1

a) 4 P1V1

b) 7 P1V1

c) 8 8 P1V1

d) 21 P1V1

e) 29 29 P1V1

How much work is done by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

Area under the curve:(4 V1)(P1) + 1/2 (4V1)(2P1) = 8 V1P1

How much internal energy is gained by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

How much internal energy is gained by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

Ideal monatomic gas: U = 3/2 nRT

Ideal gas law: PV = nRT

P2V2 = 15 P1V1

Δ(PV) = 14 P1V1

U = 3/2 PV

ΔU = 21 P1V1

How much heat is gained by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

How much heat is gained by the gas in this process, in terms of the initial pressure and volume?

One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1.

P1

V1 V2 =5V1

P2 = 3P1

a) 7 P1V1

b) 8 P1V1

c) 15 15 P1V1

d) 21 P1V1

e) 29 P1V1

First Law of Thermodynamics

W = 8 P1V1

a) zero zero

b) -153 J -153 J

c) -41 J

d) -26 J

e) 41 J

Reading QuizReading Quiz Internal Energy Internal EnergyAn ideal gas is taken through the four processes shown. The changes in internal energy for three of these processes is as follows:

The change in internal energy for the process from C to D is:

Reading QuizReading Quiz Internal Energy Internal Energy

a) zero zero

b) -153 J -153 J

c) -41 J

d) -26 J

e) 41 J

An ideal gas is taken through the four processes shown. The changes in internal energy for three of these processes is as follows:

The change in internal energy for the process from C to D is:

PV = nRT

so in a PV cycle, ΔT = 0

ΔT = 0 means that ΔU = 0

ΔUCD = -41 J

a) at constant pressure at constant pressure

b) if the pressure increases in if the pressure increases in proportion to the volumeproportion to the volume

c) if the pressure decreases in proportion to the volume

d) at constant temperature

e) adiabatically

Internal EnergyInternal Energy

An ideal gas undergoes a reversible expansion to 2 times its original volume. In which of these processes does the gas have the largest loss of internal energy?

a) at constant pressure at constant pressure

b) if the pressure increases in if the pressure increases in proportion to the volumeproportion to the volume

c) if the pressure decreases in proportion to the volume

d) at constant temperature

e) adiabatically

Internal EnergyInternal Energy

An ideal gas undergoes a reversible expansion to 2 times its original volume. In which of these processes does the gas have the largest loss of internal energy?

Since U = 3/2 nRT, and PV=nRT, the largest loss in internal energy corresponds to the largest drop in temperature, and so the largest drop in the product PV.a) PV doubles. Ufinal = 2Uinitial

b) (PV)final = 4 (PV)initial Ufinal = 4Uinitial

c) PV is constant, so U is constantd) U is constante) Adiabatic, so ΔU = -W. This is the only process which reduces U!

The Zeroth Law of ThermodynamicsIf object A is in thermal equilibrium with object B, and object C is also in thermal equilibrium with object B, then objects A and C will be in thermal equilibrium if brought into thermal contact.

That is, temperature is the only factor that determines whether two objects in thermal contact are in thermal equilibrium or not.

Object B can then be a thermometer, providing a scale to compare objects:

Temperature

The First Law of Thermodynamics

The change in a system’s internal energy is related to the heat Q and the work W by conservation of

energy:

It is vital to keep track of the signs of Q and W.

Combining these gives the first law of thermodynamics.

Reversible (frictionless pistons, etc.) and quasi-static processes

For a process to be reversible, it must be possible to return both the system and its surroundings to the same states

they were in before the process began.

V

P

Quasi-static = slow enough that system is always effectively in equilibrium

area under the curve

W =W

The Second Law of Thermodynamics

We observe that heat always flows spontaneously from a warmer object to a cooler one, although the opposite would not violate the conservation of energy.

When objects of different temperatures are brought into thermal contact, the

spontaneous flow of heat that results is always from the high temperature object to the low temperature object. Spontaneous heat flow never proceeds in the reverse

direction.

The Second Law of Thermodynamics:

Heat EnginesA heat engine is a device that converts heat into work. A classic example is the steam engine. Fuel heats the water; the vapor expands and does work against the piston; the vapor condenses back into water again and the cycle repeats.

All heat engines have: a working substance a high-temperature reservoir a low-temperature reservoir a cyclical engine

Efficiency of a Heat Engine

An amount of heat Qh is supplied from the hot reservoir to the engine during each cycle. Of that heat, some appears as work, and the rest, Qc, is given off as waste heat to the cold reservoir.

The efficiency is the fraction of the heat supplied to the engine that appears as work.

Assumption: ΔU = 0 for each cycle, else the engine would get hotter (or colder) with every cycle

Efficiency of a Heat Engine

The efficiency can also be written:

In order for the engine to run, there must be a temperature difference; otherwise heat will not be transferred.

This is an idealization; no real engine can be perfectly reversible.

The maximum-efficiency heat engine is described in Carnot’s theorem:

If an engine operating between two constant-temperature reservoirs is to have maximum efficiency, it must be an engine in which all processes are reversible. In addition, all reversible engines operating between the same two temperatures, Tc and Th, have the same efficiency.

Show the efficiency of the Carnot Cycle:

Maximum Work from a Heat Engine Cycle

The maximum work a heat engine can do is then:

If the two reservoirs are at the same temperature, the efficiency is zero.

The smaller the ratio of the cold temperature to the hot temperature, the closer the efficiency will be to 1.

Perfect efficiency for Tc =?

Heat Engine Heat Engine

The heat engine below is: a) a reversible (Carnot) heat engine

b) an irreversible heat engine

c) a hoax

d) none of the above

Heat Engine Heat Engine

The heat engine below is: a) a reversible (Carnot) heat engine

b) an irreversible heat engine

c) a hoax

d) none of the above

Carnot e = 1 − TC/TH = 1 − 270/600 = 0.55.

But by definition e = 1 − QL/QH

= 1 − 4000/8000 = 0.5, smaller

than Carnot e, thus irreversible.

4-stroke Automobile Engine

While heat will flow spontaneously only from a higher temperature to a lower one, it can be made to flow the other way if work is done on the system. Refrigerators, air conditioners, and heat pumps all use work to transfer heat from a cold object to a hot object.

Refrigerators, Air Conditioners, and Heat Pumps

Refrigerators

If we compare the heat engine and the refrigerator, we see that the refrigerator is basically a heat engine running backwards – it uses work to extract heat from the cold

reservoir (the inside of the refrigerator) and exhausts to the kitchen. Note that

- more heat is exhausted to the kitchen than is removed from the refrigerator.

Refrigerators

An ideal refrigerator would remove the most heat from the interior while requiring the smallest amount of work. This ratio is called the coefficient of performance, COP:

Typical refrigerators have COP values between 2 and 6. Bigger is better!

An air conditioner is essentially identical to a refrigerator; the cold reservoir is the interior of the house, and the hot reservoir is outdoors.

Heat PumpsFinally, a heat pump is the same as an air conditioner, except with the reservoirs reversed.

Heat is removed from the cold reservoir outside, and exhausted into the house, keeping it warm.

Note that the work the pump does actually contributes to the desired result (a warmer house) in this case.

Heat Pump EfficiencyIn an ideal heat pump with two operating temperatures (cold and hot), the Carnot relationship holds; the work needed to add heat Qh to a room is:

The COP for a heat pump:

a) get warmer

b) get cooler

c) stay the same

Room TemperatureRoom TemperatureYou haven’t had time to install your new air condition in the window yet, so as a short-term measure you decide to place it on the dining-room table and turn it on to cool off a bit. As a result, does the air in the dining room:

a) get warmer

b) get cooler

c) stay the same

Room TemperatureRoom TemperatureYou haven’t had time to install your new air condition in the window yet, so as a short-term measure you decide to place it on the dining-room table and turn it on to cool off a bit. As a result, does the air in the dining room:

The AC motor must do work to pull heat from one side to the other. The heat that is exhausted is the heat drawn from the room plus the work done by the motor. The net effect is that the motor of the AC is adding heat to the room.

AB(O)A BC CD

Gasoline engines

DA

Idealized Diesel cycle

Approaching absolute zero

So a reversible engine has the following relation between the heat transferred and the reservoir temperatures:

so... how cold can we make something?

The efficiency of a reversible engine:

(irreversible engines )

As a system approaches absolute zero, heat becomes harder to extract

The Third Law of Thermodynamics

Absolute zero is a temperature that an object can get arbitrarily close to, but never attain.

Temperatures as low as 2.0 x 10-8 K have been achieved in the laboratory, but absolute zero will remain ever elusive – there is simply nowhere to “put” that last little bit of energy.

This is the third law of thermodynamics:

It is impossible to lower the temperature of an object to absolute zero in a finite number of steps.

The Laws of Thermodynamics

I) ΔU = Q - WA continuous system (which is not consuming internal energy) cannot output more work than it takes in heat energy

YOU CAN’T WIN!

YOU CAN’T BREAK EVEN!

III) TC = 0 is not achievable

II)

Entropy

A reversible engine has the following relation between the heat transferred and the reservoir temperatures:

Rewriting,

This quantity, Q/T, is the same for both reservoirs. This conserved quantity is defined as the change in entropy.

Entropy

Like internal energy, entropy is a state function

In a reversible heat engine, the entropy does not

change.

Unlike energy, entropy is NOT conserved

Entropy

A real engine will operate at a lower efficiency than a reversible engine; this means that less heat is converted to work.

Any irreversible process results in an increase of entropy.

for irreversible processes

Entropy

To generalize:

• The total entropy of the universe increases whenever an irreversible process occurs.

• The total entropy of the universe is unchanged whenever a reversible process occurs.

Since all real processes are irreversible, the entropy of the universe continually increases. If entropy decreases in a system due to work being done on it, a greater increase in entropy occurs outside the system.

Order, Disorder, and EntropyEntropy can be thought of as the increase in

disorder in the universe.

In this diagram, the end state is less ordered than the initial state – the separation between low and high temperature areas has been lost.

Entropy

Entropy

As the total entropy of the universe increases, its ability to do work decreases.

The excess heat exhausted during an irreversible process cannot be recovered into a more organized form of energy, or temperature difference.

Doing that would require a net decrease in entropy, which is not possible.

So far, we have focused on the rather gloomy prospect of the universe constantly evolving toward greater disorder. Is it possible, however, that life is an exception to this rule?