Chapter 17

Heat & The First Law of Thermodynamics

Thermal processes

2

Breaking of equilibrium

If the process is extremely slow, or quasi-statical

system always at equilibrium state in the process

changing of state

2

1P

V

Shown in PV diagram

Point: equilibrium state

Curve: quasi-statical process

Heat as energy transfer

3

Heat is not a fluid substance

and not even a form of energy

calorie: amount of heat to raise 1g water by 1℃

Joule’s theory & mechanical equivalent of heat

1 cal 4.186 J

Heat is energy that transferred from one body to

another because of a difference in temperature

Internal energy

4

—— thermal energy / internal energy

The sum total of all the energy of all molecules

Temperature, heat and internal energy

Internal energy of monatomic (1-atom) ideal gases:

kU N E3

2 NkT

3 3

2 2 nRT PV

n-atoms molecule, real gases, liquids & solids

The first law of thermodynamics

5

The change in internal energy of a closed system, will be equal to the heat added to the system minus the work done by the system.

This is the first law of thermodynamics

U Q W

where Q is the net heat added to the system

and W is the net work done by the system

U is a state variable, but Q and W are not

Calculating the work

6

Consider the gas in a cylinder with a piston

.

.

.

.

.

.

.

.

.P

S

dx

dV

Work done by the gas to move the piston dx :

dW Fdx

PSdx PdV

For a finite change in volume from VA to VB :

B

A

V

VW PdV

B

AP

V

Heat in process

7

Example1: In process abc, 800J heat flow into the system, and 500J work done by system. In process cda, 300J work done to system, What’s the heat?

Solution: First law P

V

a

b

c

d

U Q W

800 500 300

abc cda

cdaU Q

800 500 abc c aU U U

300 cda a c cdaU U U Q600

cdaQ J

ΔU is different!

3 simple processes

8

Isothermal process: ( constant T )3

02

U nR T

B

A

V

VQ W PdV ln B

A

VnRT

V

B

AP

VVA VB

Isobaric process: ( constant P )P

VVA VB

A B

B

A

V

VW PdV P V

Isochoric process: ( constant V ) C

0 B

A

V

VW PdV

Adiabatic process

9

Adiabatic process: ( Q = 0 )

No heat is allowed to flow into or out of system

Well insulated or process happens too quickly

B

AP

V

isothermal

Cadiabatic

Adiabatic curve is steeper

than an isothermal curve

0 U Q W

Temperature decreases as well

Cyclic process

10

Example2: An monatomic (1-atom) gas system goes through processes ab, bc, ca. Determine Q, W and ΔU in each process.

Solution: In process ab:

c

P (105Pa)

4 V ( l )2o

ab1

3200 W P V J

3( )

2 b b a aU PV PV

3300

2 P V J

500 Q U W J

11

In process bc:c

P (105Pa)

4 V ( l )2o

ab1

3

0W

3( )

2 c c b bU PV PV

3600

2 V P J

600 Q U W J

In process ca: 1( )( )

2 a c a cW P P V V

3( )

2 a a c cU PV PV 300 J

400 J

100 Q U W J

Q, W and ΔU

in process abca?

Specific heat

12

Heat transfer in → temperature rises

c is called the specific heat of material

For water at 15 and 1atm: ℃

Q mc T

one of the highest specific heats of all substance

4186 / c J kg C

c as constant (P407, T17-1)

except for gases

Molar specific heat

13

Heat required to raise 1mol gas by 1 ℃ (conditions)

Isochoric (constant V) CV:

For CV of monatomic (1-atom) ideal gas, W = 0

VQ nC T

Isobaric (constant P) CP: PQ nC T

c for gases depend on how the process goes on

3

2 nR T VnC T Q U

3

2 VC R

Degrees of freedom

14

What is CV of diatomic or triatomic gas?

Degrees of freedom: number of independent ways molecules can possess energy.

diatomic:i = 5

monatomic:i = 3

triatomic:i = 6

Equipartition of energy

15

Energy is shared equally among the active degrees of freedom, each degree of freedom of a molecule has on the average energy equal to kT/2.

Average energy of a molecule:2

i

E kT

Internal energy:2 2

i i

U NE nRT PV

3

2kE kT 2 2 21 1 1 1

2 2 2 2 x y zmv mv mv kT

Principle of Equipartition of energy:

Active DoF & CV

16

CV of diatomic gases

by experiments:

Active degrees of freedom at different T

3 / 2 Low

5 / 2 Ordinary

7 / 2 High

V

R T

C R T

R T

Translational motion; Rotation; Vibration

i = 3, 5, 6 for 1, 2, n-atoms

2V

iC RIsochoric molar specific heat:

Energy in gas system

17

Example3: Determine the internal energy of (a) 2l

O2 gas system at 1atm; (b) same system at same T

but O2 is dissociated to 2O.

Solution: (a) Active degrees of freedom: i = 5

5

2U PV 5 35

1.013 10 2 10 5072

J

(b) O2 dissociate to 2O: i = 3

1

5507

2 U nRT J 2

3 2 608

2 U n RT J

Isobaric molar specific heat CP

18

( )PV nRT Q U W

2

2

P V

iC C R R

In an isobaric process (constant P): W P V

VnC T P V

VnC T nR T ( ) Vn C R T

Isobaric molar specific heat:

Adiabatic coefficient

VP

V V

C RC

C C

Adiabatic equation

19

0 dQ dU dW

Equation for adiabatic curve?

0 VnC dT PdV

B

AP

V

isothermal

Cadiabatic

( ) PV nRT PdV VdP nRdT

( ) 0 V VC R PdV C VdP

0 dV dP

V P consant PV

Equation of quasi-static adiabatic process

( )

V

V

C R

C

Some questions

20

1) Isothermal / adiabatic C ?

1 ? P

V

C

C2) Why

3) Monatomic / diatomic / triatomic gas ?

adiabaticB

AP

V

CD

4) Process AC is adiabatic

then does heat flow in or out in process AB and AD?

P

V

Isobaric & adiabatic

21

Example4: N2 system compresses isobarically in

process AB, and then expands adiabatically to C. (a) Q in process AB; (b) PC ; (c) W in process BC.

Solution: (a)

P

V

C

AB P

V 3V

PQ nC T7

2 nR T

7

2 P V 7 PV

(b) (3 ) CPV P V

1.43 CP P 0.215 P

(c) 3

V

VW PdV U

5( 3 )

2 CPV P V 0.89 PV

isothermal ?

Free expansion

22

Example5: 2 well-insulated container. A is filled with gas and B is empty. Open the valve, there is an adiabatic free expansion. What is the final P, T ?

Solution: No heat flows in or out → Q = 0

PA , TA , V V

No work is done → W = 0

U doesn’t change in free expansion!

0 U Q W

, AT T / 2 AP P

Not quasi-static, no PV diagram

*Heat transfer

23

Conduction:

Convection:

Radiation:

dQ dT

kSdt dx

4dQ

e STdt