Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 1

Ideal Gas Equation

pV nRT1

constant

pV

pV

Boyles’s Law. At constant

temperature, gas pressure is

inversely proportional to volume.

constant

V T

V

T

Charles’ Law. At constant

pressure, gas volume is

proportional to temperature.

p

V

n

R

T

Pressure in Pascals (Pa)

Volume / cubic metres

Number of moles of gas

Molar gas constant

8.314 Jmol-1K-1

Temperature /Kelvin

Ideal Gases and Heat Engines

The operation, and theoretical efficiency, of combustion driven piston engines (e.g. diesel or petrol fuelled) can

be analysed by considering charges to pressure, temperature and volume of the gaseous components. This

proceeds by accounting for the energy changes in the gas as a result of heat applied and work done

combined with the ideal gas equation, which relates the physical properties of the gas.

An ideal gas assumes a large number of point

particles colliding elastically. It neglects any

short-range intermolecular forces resulting from

repulsion or attraction due to molecular

charges, and the fact that molecules have a

finite volume i.e. are not infinitely small! This

means a real gas is not infinitely compressible

whereas an ideal gas has no such limits.

Special cases of the ideal gas equation:

Jacques Charles

1746-1823

Robert Boyle

1627-1691

Temperature scales

273C

T T

9

532

FT T

/T K

-40o F = -40o C

Fahrenheit is a temperature scale, where 32o F is

the freezing point of water and 212o F is the boiling

point of water, defined at sea level at standard

atmospheric pressure (101,325Pa).

It was proposed in 1724 by Daniel Gabriel Fahrenheit.

0o F corresponded to the lowest temperature he could

cool brine (salt water) and 100o F was the average

human body temperature (37oC).

A more popular scale

is the Celsius scale,

with 0o C and 100o C

representing the

freezing and boiling

points of water at

standard atmospheric

pressure.

The Kelvin temperature scale (or

“absolute” scale) is proportional to the

mean kinetic energy of molecules.

1

2U nRT

Internal energy of n

moles of gas

Number of degrees of freedom of

molecular motion (e.g = 3 for

x,y,z translation)

Anders Celsius

1701-1744

Daniel Fahrenheit

1686-1736

1 mol = 6.02 x 1023 molecules. So energy of a molecule is

100o C is

equivalent to

212o F.

At 1atm = 101,325Pa, one mole of gas at

200C = 293K has volume

V = 2.40 x10-2 m3 = 24 litres

11000

o

101325 8.314atm litres K

/ 273/ atm

12.187 / litres

p V Tn

n T Cp

V

Ideal Gas Equation in practical units

William Thompson (Lord Kelvin)

1824-1907

o/ 273/ atm

12.187 / litres

n T Cp

V

o500T C

o20T C/ litresV

/ atmp

1

2 Bu nk T

Boltzmann's constant 23 23 1/ 6.02 10 1.38 10 JK

Bk R

Heat, work and internal energy

of an ideal gas

12

12

0

V

V

dV

dQ dU

C dT nRdT

C nR

V

V

dQC

dT

dQ C dT

Constant volume process

(isochoric)

1

2U nRT

1

2

pV nRT

U pV

12

1

2

U nRT

dU nRdT

dx

A

F

Consider a cylinder of gas

being compressed by a force F.

The work done on the gas by the force is:

x

dW Fdx

The pressure acting upon the gas is:

Fp

A

and the volume change is:

dV Adx

dVdW pA

A

dW pdV

If heat dQ is supplied to the gas

then the First Law of

Thermodynamics (that Energy in a

closed system is conserved) means

the internal energy change is

dU dQ pdV

dQ dU pdV

The internal energy for n moles of

an ideal gas is

The Ideal Gas Equation is

First Law

First Law

i.e. no work done on gas

Constant volume heat capacity for n moles

So for a constant volume

change, the heat capacity

is a constant for an ideal gas.

12V

Rc

M

Constant volume specific heat capacity.

M is the molar volume /kg

Gas M /gmol-1

Acetylene 26.04

Air 28.966

Ammonia 17.02

Argon 39.948

Benzene 78.11

Butane 58.12

Carbon dioxide 44.01

Carbon

Monoxide

28.011

Chlorine 70.906

Ethyl Alcohol 46.07

Fluorine 37.996

Helium 4.003

Hydrogen

Chloride

36.461

Hydrogen

Sulphide

34.076

Krypton 83.80

Methane 16.044

Natural Gas 19.00

Nitrogen 28.0134

Neon 20.179

Oxygen 31.9988

Ozone 47.998

Propane 44.097

Sulphur dioxide 64.06

Toluene 92.13

Xenon 131.30

Water vapour 18.02

http://www.engineeringtoolbox.com/molecular-weight-gas-vapor-d_1156.html

VQ mc T

Total amount of heat supplied to m kg of

gas is therefore:

Constant pressure process

(isobaric)

P

P

dQC

dT

dQ C dT

12

1

2

U nRT

dU nRdT

pV nRT

pdV nRdT

Since constantp

12

12

P

P

dQ dU pdV

C dT nRdT nRdT

C nR nR

12V

P V

C nR

C C nR

Experimentally it is very hard to maintain a

constant volume as heat is added, so

constant volume heat capacity is difficult to

measure directly.

However, constant pressure heat capacity

is much easier to measure, as one can

allow volumes to change in order to

maintain equilibrium with the ambient

pressure.

The Mayer relationship is therefore very

useful in working out the constant volume

heat capacity from the constant pressure

heat capacity.

Constant pressure heat capacity for n

moles of ideal gas

First Law

Ideal Gas Equation

Mayer Relationship

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 2

Constant temperature

(isothermal) process

Adiabatic (or isentropic) process

i.e. no heat added. Work done

is the sole cause of changes in

internal energy

0

0

0

constant

1

ln

V

V

dU

T

dQ pdV

nRTdQ dV

V

Q nRT dVV

VQ nRT

V

2

2

12

12

1

2

1 12 2

1 12 2

1 1

2 2

2

2

2

1

1

( )

0

1

1

1

1 ln ln

ln

constant

U nRT

pV nRT

U pV

dU d pV

dU Vdp pdV

dQ

dU pdV

dU pdV

Vdp pdV pdV

pdV Vdp

pdV Vdp

dV dp

V p

V p k

pV k

pV

1

2

12

2

1

1

1

P V

P

V V

V

P

V

P P

V V

C C nR

C nR

C C

C nR

C nR

C nR

C c

C c

0 0

constantpV

pV p V

Work done on gas

0ln

dW pdV

W pdV

W dQ

W Q

VW nRT

V

Work done on gas for an adiabatic change

00 0

1 10 0

0

1

0 0

0

2

10 02

0

2

010 02

1

11

1

1

V

V

dW pdV

W pdV

W p V V dV

p VW V V

p V VW

V

VW p V

V

VW p V

V

First Law

Using the Ideal Gas Equation

Heat supplied to gas

The Mayer Relationship and

2 21 1P

V

C

C

0 0

0 0

pV p V

p p V V

This defines an adiabatic change

Heat supplied and work done for an

isobaric process

P

p

dQ C dT

Q mc T

12

12

( 1)

VP

p

p

C nRC nR nRc

nM nM nM

Rc

M

0

0

V

VW pdV

W p V V

Since p is constant

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 3

1

0 0 011

p V VW

V

So work done by gas on the

surroundings is:

P V

VP

P V

C C nR

CC nR

Mn Mn Mn

Rc c

M

1

1

1

1

V V

V

V

P

Rc c

M

Rc

M

Rc

M

Rc

M

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 4

Heat Engines

Carnot Cycle

Positions 1 to 2

Isothermal expansion of an ideal gas at

the hot reservoir temperature. Since gas

temperature, and therefore internal

energy is constant, the work done by the

gas on the surroundings must exactly

equate to the heat absorbed by the gas.

Positions 2 to 3

Isentropic (i.e. adiabatic or ‘no heat

added or lost’) expansion of the gas. The

work done by the gas on the

surroundings is powered by the loss of

internal energy of the gas as it cools

from the temperature of the hot reservoir

to the temperature of the cold reservoir.

Positions 3 to 4

Isothermal compression of the gas. In

order for the temperature, and hence the

internal energy, to remain constant, the

heat lost by the gas to the cold reservoir

must equate to the work done on it by

the surroundings.

Positions 4 to 1

Isentropic compression of the gas,

heating it from the temperature of the

cold reservoir to the temperature of the

cold reservoir.

1

0 0 0

constant

11

pV

p V VW

V

0

lnV

Q nRTV

1

2

pV nRT

U nRT

constantpV

P

V

c

c

Assume we have n moles of ideal gas, none of which

are lost in the process. Input parameters are:

H

C

T

T

Hot reservoir temperature (Kelvin)

Cold reservoir temperature (Kelvin)

Volume of gas at position 1 in the cycle

Volume of gas at position 2 in the cycle

Volume of gas at position 3 in the cycle

1 1 HpV nRT

2 2 Hp V nRT

3 3 Cp V nRT

4 4 Cp V nRT

Between positions 1 and 2

2

in

1

2

1,2

1

ln

ln

H

H

VQ nRT

V

VW nRT

V

1 1 H

H

pV p V nRT

nRTp

V

Between positions 2 and 3

2

2 2 2

1

2 2 2

2,3

3

11

VpV p V p p

V

p V VW

V

Between positions 4 and 1

4

4 4 4

1

4 4 4

4,1

1

11

VpV p V p p

V

p V VW

V

Between positions 3 and 4

3

out

4

3

3,4

4

ln

ln

C

C

VQ nRT

V

VW nRT

V

3 3 C

C

pV p V nRT

nRTp

V

1

2

V V

V V

Volume of gas at position 4 in the cycle

3

4

V V

V V

To complete the cycle

4 4

1 4

1 1 4 1

1

11 1

4 1 4 1

CH

H H

C C

nRTV nRT Vp p

V V V V

T TV V V V

T T

2 3

2 3

2

3 2

3

2

3 2 3

1

11 1

3 2 3 2

, CH

C H

H H

C C

nRTnRTp p

V V

Vp p

V

nRT nRT V

V V V

T TV V V V

T T

Ideal gas equation Mayer relation

Nicolas Léonard

Sadi Carnot

(1796-1832)

Work done by the gas on the

surroundings is the area

enclosed by the cycle

W pdV

1( ,1)p

4

4

1

,V

pV

3

3

1

,V

pV

2

2

1

,V

pV

inQ

outQ

Work done by the ideal

gas on the surroundings

These two volumes are derived from the other inputs:

1

2

3

4

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 5

Carnot engine cont....

Total work done is:

1,2 2,3 3,4 4,1

1 1

32 2 2 2 4 4 4

1 3 4 1

11

12 2

1 1

ln 1 ln 11 1

ln 1 ln1

H C

CH

H C

H

W pdV W W W W

VV p V V p V VW nRT nRT

V V V V

TV nRT VW nRT nRT

V T V

11

1

2 2

1 1

2

1

2

1

11

ln 1 ln 11 1

ln1 1 1 1

ln

C H

C

C CH H

H C

H C

C CH H

H C

H C

nRT T

T

T nRTV nRT V TW nRT nRT

V T V T

nRT nRTV nRT nRTW nR T T

V

VW nR T T

V

1

1

3 2

H

C

TV V

T

1

1

4 1

H

C

TV V

T

1 1 H

pV nRT 2 2 Hp V nRT

3 3 Cp V nRT 4 4 C

p V nRT 2

in

1

lnH

VQ nRT

V

in

2

1

2

1

ln

ln

1

H C

H

H C

H

C

H

W

Q

VnR T T

V

VnRT

V

T T

T

T

T

1

12

3

C

H

TV

V T

1

14

1

H

C

V T

V T

3 2

4 1

V V

V V

Define heat engine efficiency as the ratio

of Work done by the gas to the heat input

The Carnot Engine efficiency

depends only on the reservoir

temperatures.

The efficiency of a Carnot heat engine can be more simply derived by

consideration of Entropy S. This is a measure of disorder in a substance.

The Second Law of Thermodynamics states for any change, the total

amount of Entropy in the Universe must increase.

If heat is added in a reversible process: dQ

dST

For the Carnot cycle, the isentropic stages have no heat change hence they are at constant

Entropy. (Note this applies to the ideal gas, the surroundings will change in entropy due to the

exchange of work with the ideal gas). We cannot create entropy in the cycle for the gas, as the

cycle returns to the original state (p1,V1, TH), and Entropy is a scalar function of state (i.e. like

potential energy – the path does not matter). The (S,T) curve for the cycle is therefore a rectangle.

In the isothermal stages, temperature is a constant, so in both cases outin 2

1

lnH C

QQ VS nR

T T V

From the First Law of Thermodynamics

dU dQ pdV dU TdS pdV

Over the whole cycle the internal energy doesn’t

change, so the work done by the gas is:

2

1

2

1

in 2

1

ln

ln

1

ln

H C H C

H C

C

H

H

W pdV Tds

VW T T S T T nR

V

VT T nR

V TW

Q TVnRT

V

Since the temperature

range cannot go beyond

the range of reservoir

temperatures, the Carnot

cycle represents the

most efficient way of

extracting work given an

amount of heat input.*

Any other process would

occupy less area in the

S,T diagram.

1 2

34

*There is a better justification on the next page!

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 6

General notes regarding the Second Law of Thermodynamics, and the maximum possible efficiency of heat engines

It is possible to bound the efficiency of any heat engine using (i) the law of conservation of energy and (ii) the Second Law of Thermodynamics. That the upper bound of

efficiency equates with the efficiency of the Carnot engine is perhaps an even stronger justification of the statement that a Carnot engine is the most efficient scheme

possible, if indeed it could be practically realized.

Any heat engine is essentially flow of heat from a hot reservoir to a colder one. By a reservoir we mean a thermal mass that is so large that it will not change

temperature when the heat we associate with our engine is taken from or added to it. The difference in heat taken from the hot reservoir, and the heat transferred to the

cold reservoir, is the maximum possible work W done by the engine. This must be true to satisfy the law of energy conservation, or the First Law of Thermodynamics.

Cold reservoir

Heat Engine

Hot reservoir

in outQ Q W

inQ

outQ

W

HT

CT

First Law:

The Second Law of Thermodynamics states that for every change there can never be an overall decrease

in Entropy. For our idealized system, this means the loss of entropy of the hot reservoir must at least be

compensated for by the gain in entropy of the cold reservoir.

0

0

in

H

H

out

C

C

in out

total H C

H C

total

in out

H C

QS

T

QS

T

Q QS S S

T T

S

Q Q

T T

Second Law:

Entropy changes of

hot and cold reservoirs

Combining with the

First Law expression:

0

0

11

0

1 0

in out

H C

out in

in in

H C

in

H C

C

H in

Q Q

T T

Q Q W

Q Q W

T T

W

Q

T T

T W

T Q

Define engine efficiency:

in

1 0

1

C

H

C

H

W

Q

T

T

T

T

So since the Carnot engine has

efficiency

1 C

H

T

T

this is as efficient as

thermodynamics allows, so the

Carnot cycle is (one example*) of

the most efficient heat engine

possible.

*A Brayton Engine (adiabatic compression, isobaric heating, adiabatic expansion, isobaric cooling) has a

similar theoretical efficiency as a Carnot Cycle.

Notes on reversibility

A reversible heat engine is one which is assumed to operate at thermodynamic

equilibrium at all times. The ideal gas equations, and associated relationships, hold

and there are no losses due to friction etc. In other words, the differential form of the

First Law holds at all times; i.e. where changes dU in internal energy are fully accounted

for by heat change dQ and work done dW = -pdV. This means that there is no net

internal energy and indeed entropy change over the complete cycle. This means the

‘ideal’ cycle could be run in reverse without breaking the Second Law, since a zero net

entropy change is permitted.

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 7

Heat Engines - Rectangular p,V cycle

Assume we have n moles of ideal gas, none of which

are lost in the process. Input parameters are:

1p Pressure at position 1 in the cycle

Pressure at position 3 in the cycle

Volume of gas at position 1,4 in the cycle

Volume of gas at position 2,3 in the cycle

Between positions 1 and 2

Between positions 2 and 3

1

2

V V

V V

2

1

1

1 1

1 1 1

1

1,2 2 1

1,2 1 2 1

2

1,2

1

( )

ln

p

T

p pT

p p

p Vp V nRT n

RT

Q nMc T T

W p V V

TdTS nMc nMc

T T

3

2

2

1 2

1 2 2 2

2,3 2 3

2,3

3

2,3

2

0

ln

V

T

V VT

V V

p Vp V nRT T

nR

Q nMc T T

W

TdTS nMc nMc

T T

Between positions 4 and 1

4

3

3

3 2

3 2 3 3

3,4 3 4

3,4 3 2 1

4

3,4

3

( )

ln

p

T

p pT

p p

p Vp V nRT T

nR

Q nMc T T

W p V V

TdTS nMc nMc

T T

1

4

1

3 1

3 1 4 4

4,1 1 4

4,1

1

4,1

4

0

ln

V

T

V VT

V V

p Vp V nRT T

nR

Q nMc T T

W

TdTS nMc nMc

T T

Between positions 3 and 4

Heat input to gas

Work done by gas

Heat output from gas

Work done by gas

Heat input to gas

Work done by gas

Heat output from gas

Work done by gas

Net heat input

in 1,2 4,1

in 2 1 1 4

in 1 2 1 1 1 1 3 1

in 1 2 1 1 1 3

1

in 1 2 1 1 3

( ) ( )

( ) ( )

( ) ( )

( ) ( )1 1

p V

p V

p V

Q Q Q

Q nMc T T nMc T T

nM nMQ c p V p V c p V p V

nR nR

MQ p c V V V c p p

R

VQ p V V p p

Net work done by gas

2 2

1 1

T V

T V

3 3

2 1

T p

T p

4 1

3 2

T V

T V

1 1

4 3

T p

T p

1,2 2,3 3,4 4,1

1 2 1 3 2 1

1 3 2 1

( - ) - ( - )

- -

W W W W W

W p V V p V V

W p p V V

21p

V

c

c

Engine efficiency

1 3 2 1

1in1 2 1 1 3

1

1 1

1 3 2 1

- -

( ) ( )1 1

1

1 - -

p p V VW

VQp V V p p

p V

p p V V

12V

Rc

M

1T Gas temperature at position 1 in the cycle

3p

1 1,p V

1 2,p V

3 2,p V 3 1

,p V

Note Carnot efficiency

for this heat engine

would be

13.5 2731 75%

873 273

1 1( , )p V

1 2( , )p V

3 2( , )p V

3 1( , )p V

1 2

34

1

2

3

4

1

1

1

V

P

Rc

M

Rc

M

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 8

Heat Engines – The Otto cycle

The Otto Cycle is the basis of spark-ignition

piston engine, which is essentially how a typical

petrol driven engine operates.

Positions 0-1:

Air is drawn into piston/cylinder arrangement at

constant pressure.

Process 1–2

Adiabatic (isentropic) compression of the air via a

piston.

Process 2–3

Constant-volume heat transfer to the working gas from

an external source while the piston is at maximum

compression. This process is intended to represent the

ignition of the fuel-air mixture and the subsequent rapid

burning.

Process 3–4

Adiabatic (isentropic) expansion (power stroke).

Process 4–1

Constant-volume process in which heat is rejected from

the air while the piston is at maximum expansion.

Process 1–0

Air is released to the atmosphere at constant pressure.

Assume we have n moles of ideal gas, none of which

are lost in the process. Input parameters are:

1p Pressure of gas at position 1 in the cycle

Volume of gas at position 1 in the cycle

Volume of gas at position 2 in the cycle

Between positions 1 and 2 Between positions 4 and 1

Between positions 3 and 4

Between positions 2 and 3

1

2

V

V

To complete the cycle

Nikolaus Otto

(1832-1891)

Work done by the

gas on the

surroundings is the

area enclosed by

the cycle W pdV

1 1( , )p V

inQ

outQ

1 1

1 1 1

1

1

1 1 1

1

1 1 1

1,2

2

1,2 1,2

11

0, 0

p Vp V nRT n

RT

VpV p V p p

V

p V VW

V

Q S

3

2

2

2 2

2 2 2 2

3 2

3 2 3 3

2,3 3 2

2,3

3

2,3

2

0

ln

V

T

V VT

V V

p Vp V nRT T

nR

p Vp V nRT T

nR

Q nMc T T

W

TdTS nMc nMc

T T

4 1

4 1 4 4

2

3 2 3

1

3 2 2

3,4

1

3,4 3,4

11

0, 0

p Vp V nRT T

nR

VpV p V p p

V

p V VW

V

Q S

1

4

1

4,1 4 1

4,1

1

4,1

4

0

ln

V

T

V VT

V V

Q nMc T T

W

TdTS nMc nMc

T T

2 2 1 1

1

2 1

2

p V p V

Vp p

V

3 2 4 1

2

4 3

1

p V p V

Vp p

V

3p Pressure of gas at position 3 in the cycle

1T Temperature of gas at position 1

in the cycle

-ve since work

is being done

on the gas

2 2( , )p V

3 2( , )p V

4 1( , )p V

1

2

3

4

Heat input to gas

via spark ignition

Work done

by gas

Heat output

via exhaust

in 1,2 3 2

out 4,1 4 1

V

V

Q Q nMc T T

Q Q nMc T T

21p

V

c

c

12V

Rc

M

1

1

1

V

P

Rc

M

Rc

M

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 9

Otto engine cont....

Total work done is:

in

2

3 11

2 3 1

1

11

1

1

11

W

Q

Vp p r

r

V p p r

r

Define heat engine efficiency as the ratio

of Work done by the gas to the heat input

So the Otto Engine efficiency

depends only on the compression

ratio, and the ratio of specific heats

1,2 3,4

1 1

3 21 1 1 2

2 1

12

1 3 1

1

12

1 31 1

2

31

1 11 1

11 1

1

11 1

1 1

11

1

W pdV

W W W

p Vp V V VW

V V

VW p r r p

r

V rW p r r p

r r

VW p

r

1p r

Work done by the

gas on the

surroundings is the

area enclosed by

the cycle W pdV

1 1( , )p V

inQ

outQ2 2

( , )p V

3 2( , )p V

4 1( , )p V

1

2

3

4

Define the

compression ratio

1

2

Vr

V

in 3 2

1

in 3 2 1 2

2

2 3 1

in

1

1

1

1

1

V

V

V

V

Q nMc T T

nMc VQ p V p V

nR V

Rc

M

Mc

R

V p p rQ

3 2

3

p VT

nR2 2

2

p VT

nR

1

2 1

2

Vp p

V

From the previous page:

1

2

3

4

21p

V

c

c

12V

Rc

M

1

1

1

V

P

Rc

M

Rc

M

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 10

Heat Engines – The Diesel cycle

The Diesel Cycle is the basis of a diesel engine, which is

ubiquitous in transport applications. Unlike petrol-driven

engines, diesel variants are more suited to heavy machinery.

They can be found powering most ships as well as trucks,

buses and cars.

Positions 0-1:

Air is drawn into piston/cylinder arrangement at constant

pressure.

Process 1–2

Adiabatic (isentropic) compression of the air via a piston.

Process 2–3

Constant-pressure (isobaric) heat transfer to the working gas

from an external source while the piston is at maximum

compression. This process is intended to represent the

ignition of the fuel-air mixture and the subsequent rapid

burning. This is different from the Otto cycle, which is constant

volume (isochoric) heating during this stage. In the Diesel

cycle, the heat generated from air compression is sufficient to

ignite introduced fuel vapours. In the Otto cycle a spark plug is

used instead to ignite the fuel.

Process 3–4

Adiabatic (isentropic) expansion (power stroke).

Process 4–1

Constant-volume process in which heat is rejected from the air

while the piston is at maximum expansion.

Process 1–0

Air is released to the atmosphere at constant pressure.

Assume we have n moles of ideal gas, none of which

are lost in the process. Input parameters are:

1p Pressure of gas at position 1 in the cycle

Volume of gas at position 1 in the cycle

Volume of gas at position 2 in the cycle

Between positions 1 and 2 Between positions 4 and 1

Between positions 3 and 4

Between positions 2 and 3

1

2

V

V

To complete the cycle

Work done by the gas

on the surroundings is

the area enclosed by

the cycle W pdV

1 1( , )p V

inQ

outQ

1 1

1 1 1

1

1

1 1 1

1

1 1 1

1,2

2

1,2 1,2

11

0, 0

p Vp V nRT n

RT

VpV p V p p

V

p V VW

V

Q S

3

2

2

2 2

2 2 2 2

2 3

2 3 3 3

2,3 3 2

2,3 2 3 2

3

2,3

2

ln

P

T

P PT

p p

p Vp V nRT T

nR

p Vp V nRT T

nR

Q nMc T T

W p V V

TdTS nMc nMc

T T

4 1

4 1 4 4

3

2 3 2

1

2 3 3

3,4

1

3,4 3,4

11

0, 0

p Vp V nRT T

nR

VpV p V p p

V

p V VW

V

Q S

1

4

1

4,1 4 1

4,1

1

4,1

4

0

ln

V

T

V VT

V V

Q nMc T T

W

TdTS nMc nMc

T T

2 2 1 1

1

2 1

2

p V p V

Vp p

V

2 3 4 1

3

4 2

1

p V p V

Vp p

V

1T Temperature of gas at position 1

in the cycle

-ve since work

is being done

on the gas

2 2( , )p V 2 3

( , )p V

4 1( , )p V

1

2 3

4

Heat input to

gas during

combustion

Work done

by gas

Heat output

via exhaust

in 1,2 3 2

out 4,1 4 1

P

V

Q Q nMc T T

Q Q nMc T T

Rudolf Diesel

(1858-1913)

3V Volume of gas at position 3 in the cycle

21p

V

c

c

12V

Rc

M

1

1

1

V

P

Rc

M

Rc

M

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 11

Diesel engine cont....

Total work done is:

in

1 2

1 2

1

(1 ) 11

11

(1 ) 1

1

11

1

1 11

1

W

Q

pVr s r s

r s pV

r s r s

r s

r s

r s

s

r s

Define heat engine efficiency

as the ratio of Work done by the

gas to the heat input

1,2 2,3 3,4

1 1

2 3 31 1 1

2 3 2

2 1

1

11 2 1 2

1 2

1

11 2

1 11 1

1 1 11 1

1 1 1 11

W pdV

W W W W

p V Vp V VW p V V

V V

p V p r V sW r r p r V s s

r

p V sW r r r s r s

r

1 2

1 2

1 2

1 11

(1 ) 1 1 ( 1)1

(1 ) 11

p VW r r r s r s s r

p VW r s r s r s

p VW r s r s

Define the

compression ratios

1

2

Vr

V

in 3 2

1

in 1 3 2

2

2 1 12 2

12

in 1 2

1

1

11

P

P

p

V P V

V

P

Q nMc T T

nMc VQ p V V

nR V

c R Rc c c

c M M

Mc

R

Q r s p V

2 3

3

p VT

nR 2 2

2

p VT

nR

1

2 1 1

2

Vp p p r

V

From the previous page:

3

2

Vs

V

Work done by the gas

on the surroundings is

the area enclosed by

the cycle W pdV

1 1( , )p V

inQ

outQ

2 2( , )p V 2 3

( , )p V

4 1( , )p V

1

2 3

4

The Diesel engine is typically

more efficient than a petrol (Otto)

engine since the former

works on the basis of self ignition

due to high compression. This

‘knocking’ is undesirable for petrol

engines, so a lower r value is

required.

1 3 2V V V

r s

1

2

3

4

Physics topic handout - Ideal gases & heat engines Dr Andrew French. www.eclecticon.info PAGE 12

in

diesel 1

otto 1

1 11

1

11

W

Q

s

r s

r

Work done by the gas

on the surroundings is

the area enclosed by

the cycle W pdV

1 1( , )p V

inQ

outQ

2 2( , )p V

2 3( , )p V

4 1( , )p V

1

2 3

4

Work done by the

gas on the

surroundings is the

area enclosed by

the cycle W pdV

1 1( , )p V

inQ

outQ2 2

( , )p V

3 2( , )p V

4 1( , )p V

1

2

3

4

Comparing Otto and Diesel heat engines

1

2

Vr

V 3

2

Vs

V

1 3 2V V V

r s

1 2 (1 ) 11

pVW r s r s

in 1 21

1Q r s pV

2

3 11

11

1

VW p p r

r

2 3 1

in1

V p p rQ

Diesel

Otto

21p

V

c

c

12V

Rc

M

1

1

1

V

P

Rc

M

Rc

M