PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 18. The Laws of Thermodynamics Chapter 12.

PHYSICS 231

INTRODUCTORY PHYSICS I

PHYSICS 231

INTRODUCTORY PHYSICS I

Lecture 18

The Laws of ThermodynamicsThe Laws of Thermodynamics

Chapter 12

Principles of Thermodynamics

• Energy is conserved• FIRST LAW OF THERMODYNAMICS• Examples:

• Engines (Heat -> Mechanical Energy)• Friction (Mechanical Energy -> Heat)

• All processes must increase entropy• SECOND LAW OF THERMODYNAMICS• Entropy is measure of disorder• Engines can not be 100% efficient

• Adding heat Q can:

• Change temperature

• Change state of matter

• Can also change U by doing work on the

Work done on a gas

Change ofInternal Energy U

Work done on the gas

W =r F ⋅Δ

r x = F(−Δy) = (PA )(−Δy)

W = −PΔV

First Law of Thermodynamics

U = Q + W

Note: (Work done by the gas) = - (Work done on the

•Conservation of Energy•Can change internal energy U by

• Adding heat to gas: Q• Doing work on gas:

W = −PΔV

Q = ΔU + Wby the gas

Add heat => Increase Int. Energy & Gas does work€

Wby the gas = +PΔV

Example 12.1

A cylinder of radius 5 cm is kept at pressure with a piston of mass 75 kg. a) What is the pressure inside the cylinder?

b) If the gas expands such that the cylinder rises 12.0 cm, what work was done by the gas?

c) What amount of the work went into changing the gravitational PE of the piston?

d) Where did the rest of the work go?

1.950x105 Pa

183.8 J

88.3 J

Compressing the outside air

Example 12.2aA massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air.

The pressure inside thecylinder is _________ thepressure outside.

a) Greater thanb) Less thanc) Equal to

Example 12.2bA massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air.

The temperature inside the cylinder is __________ the temperatureoutside.

Example 12.2cA massive copper piston traps an ideal gas as shown to the right. The piston is allowed to freely slide up and down and equilibrate with the outside air.

If the gas is heated by a steadyflame, and the piston rises to a new equilibrium position, the new pressure will be _________ thanthe previous pressure.a) Greater than

b) Less thanc) Equal to

Some Vocabulary

• Isobaric• P = constant

• Isovolumetric• V = constant• W = 0

• Isothermal• T = constant• U = 0 (ideal gas)

• Adiabatic• Q = 0

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates atroom temperature (a)Weight is slowly added to the piston, isothermally compressing the gas to half its original volume (b)

Pb is _______ Pa

Outside Air: Room T, Atm. PExample 12.3a

U = Q − PΔV

Wby the gas = PΔV

Tb is ________ Ta

Outside Air: Room T, Atm. PExample 12.3b

U = Q − PΔV

Wby the gas = PΔV

Wab is ________ 0

Outside Air: Room T, Atm. P

Vocabulary: Wab is work done by gas between a and b

Example 12.3c

U = Q − PΔV

Wby the gas = PΔV

Ub is ________ Ua

Outside Air: Room T, Atm. PExample 12.3d

U = Q − PΔV

Wby the gas = PΔV

Qab is ________ 0

Outside Air: Room T, Atm. P

Vocabulary: Qab is heat added to gas between a and b

Example 12.3e

U = Q − PΔV

Wby the gas = PΔV

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates atroom temperature (a)Weight is slowly added to the piston, adiabatically compressing the gas to half its original volume (b)

Pb is _______ Pa

U = Q − PΔV

Wby the gas = PΔV

Wab is ______ 0

U = Q − PΔV

Wby the gas = PΔV

Qab is _______ 0

Outside Air: Room T, Atm. PExample 12.4c

U = Q − PΔV

Wby the gas = PΔV

Ub is _______ Ua

U = Q − PΔV

Wby the gas = PΔV

Tb is _______ Ta

Outside Air: Room T, Atm. PExample 12.4e

U = Q − PΔV

Wby the gas = PΔV

A massive piston traps an amount of Helium gas as shown. The piston freely slides up and down. The system initially equilibrates atroom temperature (a)The gas is cooled, isobarically compressing the gas to half its original volume (b)

Pb is _______ Pa

PV =nRT

U = Q − PΔV

Wby the gas = PΔV

Wab is _______ 0

PV =nRT

U = Q − PΔV

Wby the gas = PΔV

Tb is _______ Ta

Outside Air: Room T, Atm. PExample 12.5c

PV =nRT

U = Q − PΔV

Wby the gas = PΔV

Ub is _______ Ua

PV =nRT

U = Q − PΔV

Wby the gas = PΔV

Qab is _______ 0

Outside Air: Room T, Atm. PExample 12.5e

PV =nRT

U = Q − PΔV

Wby the gas = PΔV

P-V DiagramsP

Path moves to right:

Wby the gas = Area under curve

Path moves to left:

Wby the gas = - Area under curve

(Won the gas = - Wby the gas)

Work from closed cycles

Consider cycle A -> B -> A

WA->B = Area

WB->A = - Area

(work done by gas)

Consider cycle A -> B -> A

WA->B->A= Area

(work done by gas)

Reverse the cycle, make it counter clockwise

WB->A = - Area

WA->B = Area

(work done by gas)

WA->B->A= - Area

Reverse the cycle, make it counter clockwise

(work done by gas)

Internal Energy in closed cycles

in closed cycles

Example 12.6 a) What amount of work is performed by the gas in the cycle IAFI?

b) How much heat was inserted into the gas in the cycle IAFI?

c) What amount of work is performed by the gas in the cycle IBFI?

W=3.04x105 J

W = -3.04x105 J

Q = 3.04x105 J

V (m3)

Example 12.7Consider a monotonic ideal gas.

a) What work was done by the gas from A to B?

b) What heat was added to the gas between A and B?

c) What work was done by the gas from B to C?

d) What heat was added to the gas beween B and C?

e) What work was done by the gas from C to A?

f) What heat was added to the gas from C to A?

V (m3)

P (kPa)

0.2 0.4 0.6

20,000 J

20,000

-10,000 J

-25,000 J

15,000 J

Example Continued

Take solutions from last problem and find:

a) Net work done by gas in the cycleb) Amount of heat added to gas

WAB + WBC + WCA = 10,000 JQAB + QBC + QCA = 10,000 J

This does NOT mean that the engine is 100% efficient!

Example 12.8a

Consider an ideal gas undergoing the trajectory through the PV diagram.In going from A to B to C, the work done BY the gas is _______ 0.

Example 12.8b

In going from A to B to C, the change of the internal energy of the gas is _______ 0.

Example 12.8c

In going from A to B to C, the amount of heat added to the gas is _______ 0.

Example 12.8d

In going from A to B to C to D to A, the work done BY the gas is _______ 0.

Example 12.8e

In going from A to B to C to D to A, the change of the internal energy of the gas is _______ 0.

Example 12.8f

In going from A to B to C to D to A, the heat added to the gas is _______ 0.

PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 18. The Laws of Thermodynamics Chapter 12.

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