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Heat Engines. Coal fired steam engines. Petrol engines Diesel engines Jet engines Power station turbines. DECC http://www.decc.gov.uk/assets/decc/statistics/publications/flow/193-energy-flow-chart-2009.pdf. Combined Cycle. THE LAWS OF THERMODYNAMICS

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Heat engines
Heat Engines

  • Coal fired steam engines.

  • Petrol engines

  • Diesel engines

  • Jet engines

  • Power station turbines


Decc http www decc gov uk assets decc statistics publications flow 193 energy flow chart 2009 pdf
DECChttp://www.decc.gov.uk/assets/decc/statistics/publications/flow/193-energy-flow-chart-2009.pdf



THE LAWS OF THERMODYNAMICS

1. You cannot win you can only break even.

2. You can only break even at absolute zero.

3. You can never achieve absolute zero.




Boyle’s Law

p

p  1/V

1/V


Charles’s Law

V

V  T

T


Pressure Law

p

p  T

T


Common sense Law

p

p  N

Number of molecules, N


Isotherms

(constant temperature)

p

p

Isochors

(constant volume)

1/V

T

Isobars (constant pressure)

V

T


pV

= constant

T

For ideal gases only

A gas that obeys Boyles law

In summary…

p  1/V

p  T

V  T


Ideal gas?

Most gases approximate ideal behaviour

Ideal gases assume:-

  • No intermolecular forces

Not true - gases form liquids then solids as temperature decreases

  • Volume of molecules is negligible

Not true - do have a size


pV

= constant

T

p2V2

p1V1

T1

T2

Only useful if dealing with same gas before (1) and after (2) an event

=


Ideal Gas Law

pV = nRT

p = pressure, Pa

V = volume, m3

n = number of moles

R = Molar Gas constant (8.31 J K-1 mol-1 )

T = temperature, K

Macroscopic model of gases


Which can also be written as …

pV = NkT

N = number of molecules

k = Boltzmann’s constant (1.38 x 10-23 J K-1)


z

v

y

x

Kinetic Theory

First there was a box and one molecule…

Molecule:- mass = m

velocity = v


2mv

-2mv

-v

v

mv - mu

pmol

Molecule

Remember p = F so a force is felt by the box

t

Molecule hits side of box…(elastic collision)

= -mv - mv = -2mv

pbox = -pmol = 2mv

Box


z

y

x

s

2x

=

=

v

v

Molecule collides with side of box, rebounds, hits other side and rebounds back again.

Time between hitting same side, t


F =

p

= p v

= 2mv v

= mv2

t

2x

2x

Force exerted on box

x

Time

Actual force during collision

Average force, exerted by 1 molecule on box

Average Force


Consider more molecules

-v6

z

v2

v5

v1

-v8

vN

v3

v4

-v7

y

x

All molecules travelling at slightly different velocities so v2 varies - take mean - v2


F = Nmv2

Mean square velocity

x

Nmv2

Nmv2

Pressure = Force

=

=

Area

xyz

V

Nmv2

1

p=

V

3

Force created by N molecules hitting the box…

But, molecules move in 3D

Kinetic Theory equation


Brownian Motion

Why does it support the Kinetic Theory?

  • confirms pressure of a gas is the result of randomly moving molecules bombarding container walls

  • rate of movement of molecules increases with temperature

  • confirms a range of speeds of molecules

  • continual motion - justifies elastic collision


1

Nmv2

pV=

3

pV

= NkT

Nmv2

=

NkT

1

3

Microscopic

Macroscopic

(In terms of molecules)

(In terms of physical observations)


mv2

=3kT

(1)

Already commented that looks a bit like K.E.

K.E. = ½mv2

3

K.E. =

kT

2

Average K.E. of one molecule

Rearrange (and remove N)

Substitute into (1)


3

K.E.Total =

NkT

2

3

NkT

U =

2

Total K.E. of gas (with N molecules)

This is translational energy only

- not rotational, or vibrational

And generally referred to as internal energy, U


3

NkT

U =

2

Physically hit molecules

Energy and gases

Internal Energy of a gas

Sum of the K.E. of all molecules

How can the internal energy (K.E.) of a gas be increased?

1) Heat it - K.E.  T

2) Do work on the gas


Change in Internal Energy

Work done on material

Energy transferred thermally

=

+

U = W + Q

Basically conservation of energy

Also known as the First Law of Thermodynamics


Heat, Q – energy transferred between two areas because of a temperature difference

+ve when energy added

-ve when energy removed

Work, W – energy transferred by means that is independent of temperature

i.e. change in volume

+ve when work done on gas - compression

-ve when work done by gas - expansion


Bonds between atoms a temperature difference

Jiggling around

(vibrational energy)

Einstein’s Model of a solid

Atom requires energy to break them

U  kT


Mechanical properties change with temperature a temperature difference

T = high

can break and make bonds quickly – atoms slide easily over each other

Liquid: less viscous

Solid: more ductile

T = low

difficult to break bonds – atoms don’t slide over each other easily

Solid: more brittle

Liquid: more viscous


Activation energy, a temperature difference 

Can think of bonds as potential wells in which atoms live

Activation energy,  - energy required for an event to happen i.e. get out of a potential well


The magic a temperature difference /kT ratio

 - energy needed to do something

kT - average energy of a molecule

/kT = 1

Already happened

/kT = 10 - 30

Probably will happen

/kT > 100

Won’t happen


Probability a temperature difference

1

Exponential

0

Energy

Probability of molecule having a specific energy


e a temperature difference -/kT

/kT

Boltzmann Factor

e-/kT

Probability of molecules achieving an event characterised by activation energy, 

1

0.37

10 - 30

4.5 x 10-6 - 9.36 x 10-14

3.7 x 10-44

> 100

Nb. 109 to 1013 opportunities per second to gain energy


Amongst particles a temperature difference

Entropy

Number of ways quanta of energy can be distributed in a system

Lots of energy – lots of ways

Not much energy – very few ways

An “event” will only happen if entropy increases or remains constant


2 a temperature difference nd law of thermodynamics

S = k ln W

S = entropy

k = Boltzmann’s constant

W = number of ways


Δ a temperature difference S = ΔQ

T


At a thermal boundary a temperature difference

Energy will go from hot to cold

Hot

Number of ways decreases – a bit

Cold

Number of ways increases – significantly

Result - entropy increase


  • Efficiency = W/Q a temperature difference H = (QH – QC ) / QH

  • BUT Δ S = Q/T

  • SOEfficiency =(TH – TC)/ TH

  • = 1 – TC/TH



Symbol = c a temperature difference

Unit = J kg-1 K-1

Energy and solids (& liquids)

Supplying energy to a material causes an increase in temperature

Specific Thermal Capacity

Energy required to raise 1kg of a material by 1K


E = mc a temperature difference 

E = Energy needed to change temperature of substance / J

m = Mass of substance / kg

c = Specific thermal capacity of substance

/ J kg-1 K-1

 = Change in temperature / K


Energy Units a temperature difference

From E = qV

Energy gained by an electron when accelerated by a 1V potential difference

E = 1.6 x 10-19 x 1

= 1.6 x 10-19J

= 1eV

From E = NAkT

Energy of 1 mole’s worth of particles

kJ mol-1


Latent Heat a temperature difference

Extra energy required to change phase

Solid liquid

Latent Heat of fusion

Latent Heat of vaporisation

Liquid gas

At a phase boundary there is no change in temperature - energy used just to break bonds


N a temperature difference A = 6.02 x 1023

Molar masswater = 18g

mass evaporated

NA

molar mass

energy used

no of molecules evaporated

1kg

NA Energy to vaporise one molecule

molar mass

SLHV - water

Calculate

1) Number of molecules of water lost

2) Energy used per molecule to evaporate

3) Energy used to vaporise 1kg of water


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