Thermodynamic Processes

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# Thermodynamic Processes - PowerPoint PPT Presentation

Thermodynamic Processes. Thermodynamic Processes.ppt. F. D x. D V. D V = A D x. Area. Work Done on a Gas. W = F D x. W= PA D x. W = P D V. Thermodynamic Processes.ppt. 1. Isotherms (lines of constant temperature). Pressure. T 4. T 3. T 2. T 1. Volume.

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## Thermodynamic Processes

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Thermodynamic

Processes

Thermodynamic Processes.ppt

F

Dx

DV

DV = A Dx

Area

Work Done on a Gas

W = F Dx

W= PA Dx

W = P DV

Thermodynamic Processes.ppt

1

Isotherms

(lines of constant

temperature)

Pressure

T4

T3

T2

T1

Volume

Pressure - Volume Graph

P

Area under curve represents work

Internal energy

is proportional

to temperature

V

Thermodynamic Processes.ppt

W

Q

First Law of Thermodynamics

System

U

Environment

DU = ±Q ±W

Thermodynamic Processes.ppt

Thermodynamic Processes

A. Isobaric

B. Isovolumetric

C. Isothermal

Thermodynamic Processes.ppt

Po

DV

Isobaric (Constant Pressure)

Expansion

P

a) W = -PΔV  negative W

b) DU increases

T4

T3

c) Q = DU + W

T2

T1

V

DU = Q - W

Thermodynamic Processes.ppt

Po

DV

Isobaric (Constant Pressure)

Compression

a) W = -PDV  positive W

P

b) U decreases ΔU<0

T4

T3

c) Q = -DU - W

T2

T1

V

-DU = Q + W

Thermodynamic Processes.ppt

Po

Pf

Isovolumetric (Constant Volume)

Decrease in Pressure

P

a) W = 0 (ΔV = 0)

b) DU<0 (U decreases)

T3

c) Q = -DU

T2

T1

V

DU = Q + W

Thermodynamic Processes.ppt

Pf

Po

Isovolumetric (Constant Volume)

Increase in Pressure

P

a) W = 0

b) DU increases

T3

c) Q = DU

T2

T1

V

DU = Q + W

Thermodynamic Processes.ppt

Po

Pf

Vo

Vf

Isothermal (Constant Temperature)

Expansion

P

W< 0

b) DU = 0

T3

c) Q = W

T2

T1

V

DU = Q + W

Thermodynamic Processes.ppt

Pf

Po

Vo

Vf

Isothermal (Constant Temperature)

Compression

P

W> 0

b) DU = 0

T3

c) Q = -W

T2

T1

Note: ΔU=0 when:

Isothermic process

Cyclic process

V

DU = Q + W

Thermodynamic Processes.ppt

Po

Pf

Vo

Vf

Expansion

P

a) W = -PΔV negative W

b) DU = decreases

T2

c) Q = 0

W= -DU

T1

V

DU = Q + W

Thermodynamic Processes.ppt

Pf

Po

Vf

Vo

Compression

P

a) W = -PΔV  positive W

b) DU = increases

T2

c) Q = 0

W= DU

T1

V

DU = Q +W

Thermodynamic Processes.ppt

The internal energy of a gas increases during

a thermodynamic process. Which of the following

are possible processes?

Single

Concept

II. isothermal

III. isobaric

no

Illustration

(A) I and II only

(B) I only

(C) 1 and III only

(D) III only

(E) II and III only

Multi-

Response

Type

2003 AP Physics Workshop, Ted Vittitoe

An ideal gas expands to 10 times its original volume,

maintaining a constant 440 K temperature. If the gas

does 3.3 kJ of work on its surroundings, how much heat

does it absorb?

W = -3.3 kJ

In an isothermal process DU = 0

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

One mole of a monatomic ideal gas, initially at point A is taken through a 3-process cycle as shown in the the PV diagram.

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

The change in internal energy in each process

DU = Q + W

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Heat added and exhausted in each process

DU = Q + W

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Work done by the system in each process

DU = Q + W

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Temperature

Determine the temperature of

1 mol of gas at A, B, and C

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Internal Energy

Determine the internal energy

of the gas at A, B, and C

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Process AB

Work:

Heat:

QAB = ΔUAB = UB – UA = 7480 J – 3740 J=3740J

Internal Energy:

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Process BC

Work:

Internal Energy:

Heat:

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Process CA

Work:

Internal Energy:

Heat:

QCA = ΔU - W

QCA = -3740 J -2500 J

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Determine the net work done

on the gas in one cycle

Net work is represented by the

enclosed area of the curve

WAB = 0 J

WBC = -3750 J

WCA = 2500 J

Thermodynamic Processes.ppt

Pa

B

2.0

1.0

A

C

50

25

Determine the net heat absorbed

by the gas in one cycle

QAB = 3740 J

QBC = 3750 J

QCA = -6240 J

Qnet = 1250 J

Thermodynamic Processes.ppt

2P0

Po

V0

2Vo

3Vo

II

An ideal gas initially has pressure Po, at volume Vo and absolute temperature To. It then undergoes the following series of processes:

III

I

IV

I. Heated, at constant volume to pressure 2Po

II. Heated, at constant pressure to volume 3Vo

III. Cooled, at constant volume to pressure Po

IV. Cooled, at constant pressure to volume Vo

Thermodynamic Processes.ppt

2P0

P0

V0

2Vo

3Vo

II

2To

6To

III

I

To

IV

3To

Find the temperature at each end point in terms of To

Thermodynamic Processes.ppt

II

2P0

III

I

Find the net work done by the gas in terms of Po and Vo

P0

IV

V0

2Vo

3Vo

Net work equals net area under curve

Thermodynamic Processes.ppt

2Po

Po

Vo

2Vo

3Vo

II

III

I

Find the net change in internal energy

IV

Thermodynamic Processes.ppt

2Po

Po

Vo

2Vo

3Vo

II

III

I

Find the net heat absorbed in terms of Po and Vo

IV

Thermodynamic Processes.ppt

2Po

Po

Vo

2Vo

3Vo

II

2To

6To

III

I

Find the heat transferred during process II in terms

of Po and Vo

IV

Thermodynamic Processes.ppt

Thermodynamic

Processes

END

Thermodynamic Processes.ppt