18. Heat, Work, & First Law of Thermodynamics 1.

2.

3.

The 1

st

Law of Thermodynamics Thermodynamic Processes Specific Heats of an Ideal Gas

A jet aircraft engine converts the energy of burning fuel into mechanical energy. How does energy conservation apply in this process?

E combustion = E mech + Q waste

18.1. The 1

st

Law of Thermodynamics

PE of falling weight  KE of paddle  Heat in water Either heating or stirring can raise

T

of the water.

1 st Law of Thermodynamics : Increase in internal energy = Heat added  Work done

Q W dU d t



d Q d t



dW d t

Thermodynamic state variable = variable independent of history.

e.g.,

U

,

T

,

P

,

V

, … Not

Q

,

W

, … Joule’s apparatus

Example 18.1

. Thermal Pollution

The reactor in a nuclear power plant supplied energy at the rate of 3.0 GW, boiling water to produce steam that turns a turbine-generator.

The spent steam is then condensed through thermal contact with water taken from a river.

If the power plant produces electrical energy at the rate of 1.0 GW, at what rate is heat transferred to the river?

1 st law 

dU d t



d Q d t



dW d t

From standpoint of power plant:

d Q

  2.0

GW d t

 3.0

GW



d Q d t

 1.0

GW

( loses heat to river )

18.2. Thermodynamic Processes

Quasi-static process : Arbitrarily slow process such that system always stays arbitrarily close to thermodynamic equilibrium.

Reversible process : Any changes induced by the process in the universe (system + environment) can be removed by retracing its path.

T water = T gas & rises slowly Reversible processes must be quasi-static.

Irreversible process : Part or whole of process is not reversible.

e.g., any processes involving friction, free expansion of gas ….

system always in thermodynamic equilibrium

Work & Volume Changes

面積

F

F

ext



x

x



W

 

d W

 

V

1

V

2

p dV

Work done by gas on piston

GOT IT? 18.1

.

Two identical gas-cylinder systems are taken from the same initial state to the same final state, but by different processes.

Which are the same in both cases: (a) the work done on or by the gas, (b) the heat added or removed, or (c) the change in internal energy?

Isothermal Processes

Isothermal process :

T

= constant.

W

 

V

1

V

2

p dV

 

V

1

V

2

n R T V dV



n R T

ln

V V

2

V

1

W



n R T

ln

V

2

V

1

U

 3 2

N k T

 0

Q



W



n R T

ln

V

2

V

1 Isothermal processes on ideal gas

Example 18.2

. Bubbles!

A scuba diver is 25 m down, where the pressure is 3.5 atm ( 350 kPa ).

The air she exhales forms bubbles 8.0 mm in radius.

How much work does each bubble do as it arises to the surface, assuming the bubbles remain at 300 K.

W V

2

V

1  

n R T P

1

P

2  ln

V

2

V

1   3.5

1

atm atm

   3.5

P V



n R T T

= const 

P V

1 1 

P V

2 2

W



n R T

ln 3.5



p V

1 1 ln 3.5

  350, 000

Pa

 4   0.008

m

 3 3  0.94

J

ln 3.5

Constant-Volume Processes & Specific Heat

Constant-volume process ( isometric, isochoric, isovolumic ) :

V

= constant  0 

W

0

Q C V

= molar specific heat at constant volume

Q n C V



T

isometric processes

C V



dQ

 

V

Ideal gas:

U = U

(

T

)  

U ideal gas



n C V



T

for all processes Non-ideal gas:

Q



n C V



T

only for isometric processes

Isotherms

Isobaric Processes & Specific Heat

Isobaric Process : constant

P

W

  2 

V

1 

Q W p V C P

= molar specific heat at constant pressure

C P



dQ

 

P Q



n C P



T

isobaric processes Ideal gas, isobaric : 

n C P n C V



n C V C P



C V



R p V

Ideal gas

Adiabatic Processes

Adiabatic process :

Q

= constant e.g., insulated system, quick changes like combustion, … Tactics 18.1. 

W p V

 

const T V

  1 

const

adiabat , ideal gas  

C P C V

 1 Prob. 66

W



p V

1 1  

p V

2 2  1 Prob. 62 Adiabatic: larger 

p

cdf

TACTIC 18.1

. Adiabatic Equation

Ideal gas, any process:

dU



n C dT V

Adiabatic process:

p V



n R T



dU

 

dW

 

p dV



n C dT V



p d V



V d p



n R dT



p C V dV



p dV



V dp R

 

V



p dV



C V d p

V

 0

C p dV p



C V d p V

 0 

dV V



d p p

 0  ln

V

 ln

p



const

 ln

p V

   

C C V p p V

 

const

Conceptual Example 18.1

. Ideal-Gas Law vs Adiabatic Equation

The ideal gas law says

p V = n R T

, but the adiabatic equation says

p V



= const

.

Which is right ?

Both are right.

The adiabatic equation is a special case where

T



V

 +1

Making the Connection

Suppose you halve the volume of an ideal gas with  = 1.4.

What happens to the pressure if the process is (a) isothermal and (b) adiabatic?

Ans:

(a) pV

= const  (b)

p V

 = const 

p =

2

p 0

(doubles)

p = p 0

2 

=

2.64

p 0

Example 18.3

. Diesel Power

Fuel ignites in a diesel engine from the heat of compression (no spark plug needed).

Compression is fast enough to be adiabatic.

If the ignit temperature is 500  C, what compression ratio V max / V min is needed?

Air’s specific heat ratio is  = 1.4, & before the compression the air is at 20  C.

T V

  1 

const V

max

V

min

T

min

T

max  1 /    1   273 273

K K

 500  20

K K

    11 

Application : Smog Alert!

Air is poor heat conductor  convection is adiabatic.

If rising air cools slower than surrounding air, pollution rises high & can be dispersed.

Otherwise, smog.

GOT IT? 18.2

.

Name the basic thermodynamic process involved when each of the following is done to a piston-cylinder system containing ideal gas, & tell also whether

T

,

p

,

V

, &

U

increase or decrease.

(a) the piston is lock in place& a flame is applied to the bottom of the cylinder, (b) the cylinder is completely insulated & the piston is pushed downward, (c) the piston is exposed to atmospheric pressure & is free to move, while the cylinder is cooled by placing it on a block of ice.

(a) isometric;

T

 ,

p

 ,

V

=const,

U

 .

(b) adiabatic;

T

 ,

p

 ,

V

 ,

U

 .

(c) isobaric;

T

 ,

p

=const,

V

 ,

U

 .

Ideal Gas Processes

Cyclic Processes

Cyclic Process : system returns to same thermodynamic state periodically.

Example 18.4

. Finding the Work

An ideal gas with  = 1.4 occupies 4.0 L at 300 K & 100 kPa pressure.

It’s compressed adiabatically to ¼ of original volume, then cooled at constant

V

back to 300 K, & finally allowed to expand isothermally to its original

V

.

How much work is done on the gas?

AB (adiabatic):

W AB



p V A A

 

p V B B

 1

W AB



p V A

  1

A

  1

p B



p A

 

V A V B

  

V A V B

   1     100

kPa

 4.0

L

  BC (isometric): CA (isothermal): work done by gas:

W BC

 0

W CA



n R T V

ln

V C A



A

ln 4  555

J W ABCA



W AB



W BC



W CA

  186

J

   741

J

18.3. Specific Heats of an Ideal Gas

Ideal gas:

K

 1 2

m v

2  3 2

k T U ideal gas



C P

3 2 

N k T C V



R

 3 2

n R T

 5 2

R

  

C V

 1

n



U



T



C C V P

 5 3  3 2

R

 1.67

Experimental values ( room

T

): For monatomic gases,   5/3, e.g., He, Ne, Ar, ….

For diatomic gases,   7/5 = 1.4, C V = 5R/2, e.g., H 2 , O 2 , N 2 , ….

For tri-atomic gases,   1.3, C V = 3.4R, e.g., SO 2 , NO 2 , ….

Degrees of freedom (DoF) = number of independent coordinates required to describe the system Single atom: DoF = 3 (transl) For low

T

( vib modes not active ) : Rigid diatomic molecule : DoF = 5 (3 transl + 2 rot) Rigid triatomic molecule : DoF = 6 (3 transl + 3 rot)

The Equipartition Theorem

Equipartition theorem ( kinetic energy version): For a system in thermodynamic equilibrium, each degree of freedom of a rigid molecule contributes ½

kT

to its average energy.

Equipartition theorem ( general version): For a system in thermodynamic equilibrium, each degree of freedom described by a quadratic term in the energy contributes ½

kT

to its average energy.

U



n N A f

2

k T



f

2

n R T

Monatomic Diatomic Triatomic

DoF ( f )

3 5 6

C V

3/2 5/2 3

C V



f

2

R

C P

5/2 7/2 4

C P



f

2  1

R

 5/3 7/5 4/3  

f

 2

f

Example 18.5

. Gas Mixture

A gas mixture consists of 2.0 mol of oxygen (O 2 ) & 1.0 mol of Argon (Ar).

Find the volume specific heat of the mixture.

U O

2  5 2

n O

2

R T U Ar

 3 2

n Ar R T U mix

 5 2

n O

2  3 2

n Ar R T C mix

 1

n



U



T mix

 5 2

n O

2

n O

2  3 2

n Ar



n Ar R

 2.2

R

 5 2   2.0

2.0

mol



mol

2  1.0

 1.0

mol mol



R

Quantum Effects

Quantum effect : Each mechanism has a threshold energy.

E transl < E rot < E vib rotation+Translation+vibration rotation+Translation Translation

C V

of H 2 gas as function of

T

. Below 20 K hydrogen is liquid, above 3200 K it dissociates into individual atoms.

Reprise

Quasi-static process : Arbitrarily slow process such that system always stays arbitrarily close to thermodynamic equilibrium.

Reversible process : Any changes induced by the process in the universe (system + environment) can be removed by retracing its path.

Dissipative work : Work done on system without changing its configuration, irreversible.

Insulated gas a  c  c : Free expansion with no dissipative work.

b : Adiabatic.

a  d  d : Adiabatic.

b : Free expansion with no dissipative work.

a  e  e : Adiabatic.

b : Adiabatic dissipative work.

1 st law: The net adiabatic work done in all 3 processes are equal (shaded areas are equal).