physics 12 giancoli chapter 15. objectives the first law of thermodynamics deduce an expression for...

THERMODYNAMICS

Physics 12Giancoli Chapter 15

ObjectivesThe first law of thermodynamics Deduce an expression for the work involved in a volume

change of a gas at constant pressure. State the first law of thermodynamics. Identify the first law of thermodynamics as a statement of

the principle of energy conservation. Describe the isochoric (isovolumetric), isobaric, isothermal

and adiabatic changes of state of an ideal gas. Draw and annotate thermodynamic processes and cycles on

P-V diagrams. Calculate from a P-V diagram the work done in a

thermodynamic cycle. Solve problems involving state changes of a gas.

ObjectivesSecond law of thermodynamics and entropy State that the second law of thermodynamics

implies that thermal energy cannot spontaneously transfer from a region of low temperature to a region of high temperature.

State that entropy is a system property that expresses the degree of disorder in the system

State the second law of thermodynamics in terms of entropy changes.

Discuss examples of natural processes in terms of entropy changes.

Thermodynamics The study of processes in which energy is

transferred as heat and as work. Important to define the system that we are

dealing with; Everything other than the system shall be

referred to as the “environment” or “surroundings”.

First law of thermodynamics Otherwise known as the law of

conservation of energy. Relates work/heat to the change of the

internal energy of a system.

What are some forms of energy that we have considered in the past? How do they relate to the conservation of energy?

First law of thermodynamics The internal energy is the sum total of all

the energy of the molecules in a system.

What would happen to the internal energy if heat was added to the system?

If heat was taken away from the system?

First law of thermodynamics

The change in internal energy ΔU of a closed system will be equal to the energy Q added to the system minus the work W done by the system on the surroundings.

ΔU = Q + W

First law of thermodynamics Internal energy of an ideal gas

The internal energy U is the sum of the translational kinetic energies of the atoms.

First law of thermodynamicsU = N(KEavg)

where N is the number of molecules of gas orN = nNA

where n is the number of moles of gas andNA is Avogadro’s number

The average KE of an ideal gas is given by the equationKEavg = (3/2)kBT

where kB is Boltzmann’s constant

and T is the temperature of the gas.


Therefore, the internal energy of an ideal gas is given by the expression

U = (3/2)NkBT

whereN is the number of molecules of gas

kB is Boltzmann’s constant, and

T is the temperature of the gas.


ΔU = Q + W

Q is the heat added to or taken away from the system.

W is the work done by or on the system.

What would negative signs for either Q or W imply?


ΔU = Q + W

IN OTHER WORDS…The change in energy in a system is the

total of the heat and work that goes in and out of the system.

Signs will make a difference.


2500 J of heat is added to a system, and 1800 J of work is done on the system. What is the change in internal energy of the system?


4300 J

What would the change in internal energy be if 2500 J of heat was added to the system but 1800 J of work is done by the system?


700 J

First law of thermodynamics The first law of thermodynamics is one of

the great laws of physics that can be proven experimentally and to which no exceptions have been seen.

For simplicity’s sake, we shall discuss these laws in the context of gases.

First law of thermodynamics Consider a fixed mass of an ideal gas

enclosed in a container fixed with a movable piston: (p 410 Fig 15-1).

If we compress the gas (lowering the piston), what happens to its pressure and volume?

First law of thermodynamics Decreasing the volume will increase the

pressure. There is a constant relationship between

pressure and volume when T is constant:PV = nRT

A thermodynamic process in which T is constant is called an isothermal process.

p 410 Fig 15-2

constant

First law of thermodynamics What would you expect the P-V graph of

an isothermal process to look like?

How would the graph change if the process were to occur at a lower temperature?

First law of thermodynamics p 410 Fig 15-2

When the gas is compressed, was work done on the system or by the system?

First law of thermodynamics The act of compressing the gas is work done

on the system +W.

However, in order for T to remain CONSTANT the decrease in V requires an increase in pressure P which is carried out by the gas.

So what is the change in internal energy of the system?


-W = Qsince ΔU = Q + W,

ΔU = 0

This means that there is no change in the internal energy of the system.

Does this necessarily mean that there was no work done on or by the system?

First law of thermodynamics This is an important concept in physics

(wherein the net work being zero does not necessarily mean that no work was done ).

In what other units have we explored this concept?

First law of thermodynamics Another thermodynamic process is one in

which no heat is allowed to flow into or out of the system.

This is called an adiabatic process. In terms of the eq’n for the first law of

thermodynamics, which term is 0?ΔU = Q + W

First law of thermodynamics If Q = 0, what is ΔU?


If Q = 0, and ΔU = Q + W

then ΔU = W

But what does this mean?


Adiabatic expansionQuantity Value / Effect

Q 0

ΔU = Q + W= 0 + W = W

V increases

P ?

T ?


Conceptual exampleWhen you suddenly expand a rubber band,

what happens when you touch it with your lips? Explain this in terms of thermodynamics.


Conceptual exampleThe rubber band will feel warmer than before

it was stretched.When the rubber band is stretched, you do

work on the system (- W). Since it was done suddenly, heat was essentially not allowed to leave the system (Q = 0). Thus, when you touch it with your lips you will feel an increase in temperature.

Try it at home.


Comparison of adiabatic and isothermal compression

Quantity Isothermal Adiabatic

Q

ΔU

V decrease decrease

P

T

First law of thermodynamicsComparison of adiabatic and isothermal compression

In a fuel engine, the fuel and air is compressed so rapidly (adiabatically) in the fuel piston that the T increase causes the mixture to ignite.

Quantity Isothermal Adiabatic

Q W 0

ΔU 0 -W

V decrease decrease

P increase increase

T constant increase

First law of thermodynamics Other thermodynamic processes that may

occur are:isobaric (constant pressure)isochoric or isovolumetric (constant volume)

How would you expect these processes to appear on a P-V graph?

First law of thermodynamics So far, we have discussed situations

qualitatively (which is often insufficient in physics).

How do we calculate the work done on a system?

How have we calculated work before?


Recall that W = Fd

In terms of gases,P = F / A

F = PAthereforeW = PAd

if Ad = ΔVthen W = P ΔV


Dimensional analysis:1 J = 1 Nm

A calculation of PΔV (using SI units) would yield

(Nm-2)(m3)= Nm

= J


Comparison of thermodynamic processes



Constant Important characteristic

Isothermal

Isobaric

Isovolumetric

Adiabatic



Constant Important characteristic

Isothermal T Q = -W

Isobaric P ΔU = Q + W ΔU= Q + PΔV

Isovolumetric V ΔV = 0Q = ΔU

Adiabatic Q = 0 Q = 0U = WΔ


Comparison of adiabatic and isothermal expansion

Look at the Figure 15-3 on p 411.In which process was more work done by the

gas?


Comparison of adiabatic and isothermal expansion

More work was done by the gas in the isothermal process.

The average pressure is higher during the isothermal process.

Work can also be represented graphically by the area under a P-V curve.


An ideal gas is slowly compressed at a constant pressure of 2.0 atm from 10.0 L to 2.0 L (B to C). In this process, some heat flows out of the gas and the temperature drops. Heat is then added to the gas (C to A), holding the volume constant, and the pressure and temperature are allowed to rise until the temperature reaches its original value.

Sketch a P-V graph of the processes involved.Calculate the total work done by the gas in the

process CBA .Calculate the total heat flow into the gas.


Graph:BC: isobaric compression (V = 10.0 L to V = 2.0 L,

P = 2.0 atm )CA: isovolumetric increase in pressure

Work is done only in the compression CB. In CA, ΔV = 0 so W = 0. During CB, W = -1.6 x 103 J.

Because the temperature at the beginning and end of the process is the same, ΔT = 0 so ΔU = 0. Therefore the total heat flow into the gas is -1.6 x 103 J.


In an engine, 0.25 moles of an ideal monatomic gas in the cylinder expands rapidly and adiabatically against the piston. In the process, the temperature of the gas drops from 1150 K to 400 K. How much work does the gas do?


2300 J

First law of thermodynamics Recall the relationship between the heat and the

change in temperature of a substance:Q = mcΔT

In terms of gases, we use the expressionQ = nCΔT

where n is the number of moles of gasC is the molar heat capacity of the gas (this may be

expressed as Cv or Cp at constant volume or pressure)

and ΔT is the change in temperature of the gas

Activity Worksheet – First law of thermodynamics

Second law of thermodynamics

If a hot object is placed in contact with a cold object, in which direction will heat flow?


Experience tells us that the heat will flow from the hot object to the cold object.

Does heat flow from a colder object to a hotter object violate the first law of thermodynamics?


Heat flow from cold to hot does NOT violate the first law of thermodynamics but it does violate the second law of thermodynamics:

Heat will flow spontaneously from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot

object.


This concept is especially relevant in the study of heat engines.

Activity Independent study

Heat engines and the Carnot cycleHandout Sections 15-5 and 15-6 of Giancoli

physics 12 giancoli chapter 15. objectives the first law of thermodynamics deduce an expression for...

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