THERMODYNAMICS

Department of Physics, Faculty of Science

Jazan University, KSA

Part-3

222 PHYS

The first law of thermodynamics is an extension of the law of conservation of

energy

“The change in internal energy of a system is equal to the difference between heat

added to the system and the work done by the system”

The first law of thermodynamics

the internal energy of a system can be

changed by doing work on it or by

heating/cooling it.

From the microscopic point of view, this

statement is equivalent to a statement of

conservation of energy.

Note: The first law of thermodynamics tells us that in order to change the internal energy of a system we must add (or remove) heat and/or do work on (or have work done by) the system.

ΔU = Q - W

+ = Change in internal energy

Energy supplied to system as heat

Energy supplied to system as work

U= Q (heat) + w (work) q

w

q

w U

Or we can say….

U like reserves of a bank: bank accepts deposits or withdrawals in two currencies (Q & w) but stores them as common fund, U.

System

U

Environment Q W

U– Change in internal energy [J] Q – Heat added or lost by the system [J] W – Work done on or by the system [J]

Heat and work are forms of energy transfer and energy is conserved.

The First Law of Thermodynamics

U = Q + Won

work done on the system

change in total internal

energy

heat added

to system

U = Q - Wby

State Function Path Functions

Positive Q ->heat added to the system Positive W -> work done by the system

statement of energy conservation for a thermodynamic system

The internal energy of a system tends to increase if energy is added via heat (Q) and decrease via work (W) done by the system.

or

The signs on Q and W depend on the way the internal energy is changed. If Q and W are positive the internal energy increases, but if Q and W are negative the internal energy decreases.

heat is random molecular motion while work is force times distanced moved

under its influence

Exothermic Processes release heat and have Q<0

Endothermic Processes absorb heat and have Q>0

Energy: The SI unit is joule (J) although we will frequently use calorie ;

1 cal = 4.2 J

Recap

The work is not necessarily associated with the volume changes – e.g.,

in the Joule’s experiments on determining the “mechanical equivalent of

heat”, the system (water) was heated by stirring.

ΔU = Ufinal - Uinital = Q – W

From the first Law of Thermodynamics: Energy is Conserved

Q = heat absorbed by the system from the surroundings

W = work done by the system on the surroundings

Special Cases of the 1st Law of Thermodynamics

Isolated System The system doesn’t interact with the environment.

What does this mean in terms of the 1st law of thermodynamics?

Q = 0 -------- No heat is transferred into or out of the system. W = 0 ------- No work is done on the system.

U Q W 0 0

0U i fU U

Cyclic Process The system starts and ends in the same state (same internal energy). The system is not necessarily isolated.

The function that describes the changes in the state on a PV - diagram would be a closed curve.

U = 0 ------- No net change in the internal energy.

U Q W 0

WQ 0 WQ

Adiabatic Process

W

This process considers a system where there is no loss or gain of energy through heat.

This can be accomplished by: 1. Thermally insulating the chamber 2. Performing the process very rapidly

– no time for heat to be transferred.

Examples: • Expansion of hot gases in an internal combustion engine • Liquefaction of gases in a cooling system

Adiabatic Free Expansion This is a special case of an adiabatic process, where the gas expands into free space.

U Q W 0

U W

U Q W 0 0

0U

If we do work to compress the gas the temperature of the gas should increase. The increase in the temperature of the gas corresponds to an increase in the internal energy of the system.

Applications of 1st law of thermodynamic

Isobaric Process Constant pressure process.

U Q W 0Q 0W if VVPW

Isovolumetric Process Constant volume process.

U Q W 0Q 0W

If volume doesn’t change work cannot be done to compress the gas.

U Q

Isothermal Process

U Q W 0U WQ

Constant temperature process.

If you do work to compress a gas the energy you put in is released through heat.

On a PV – diagram it is common to use isotherms to show how the temperature changes for a process.

Isotherm – Hyperbolic line of constant temperature on a PV – diagram.

F

W = F y

A

FP

PAF

W= PA y V = A y

W = P V

y

V

Area

Work in Thermodynamic Processes

During a compression:

Work done on a gas is

positive.

As stated previously, pressure, temperature and volume are considered state variables and are used to define the particular state of the system. Work and Heat are called transfer variables. These describe changes in the state. They do not describe the state. We know how to describe the work done on a system. For example let us look at applying a force to a piston in order to compress the gas inside a container.

External force is equal and opposite to force gas exerts on piston. Work done on gas!

dW F d s

PAdy PdVFdydW

f

i

V

V

PdVW -F is parallel to y Total Work done to change the volume of a gas

If the gas is compressed slowly enough for all of the system to remain in thermal equilibrium.

Quasi-static process Work is the transfer of energy that takes place when an object is moved against an opposing force

Work

The sign: if the volume is decreased, W is positive (by compressing gas, we increase its internal

energy); if the volume is increased, W is negative

(the gas decreases its internal energy by doing

some work on the environment).

2

11 2

V

VW P dV

The work done by an external force to compress a

gas enclosed within a cylinder fitted with a piston:

W = (PA) dx = P (Adx) = - PdV

x

P W = PdV applies to any

shape of

system

boundary dU = Q – PdV

A – the

piston area force

Work Done by an Expanding Gas

Gas expands slowly enough to maintain thermodynamic equilibrium.

+dV Positive Work (Work is

done by the gas)

-dV Negative Work (Work is

done on the gas)

Energy leaves the system and goes to the environment.

Energy enters the system from the environment.

PAdyFdydW

Increase in volume, dV

PdVdW

f

i

V

V

PdVW Total Work done to change the volume of a gas

What type of process is described by each of the arrows?

This is a PV – diagram showing several isotherms.

A PV-Diagram is a plot of pressure vs. volume. The work done during the process shown by the PV-diagram can be determined by looking at the area under the curve. Remember this is the same as the integral expression for the work.

Example: In the three figures shown (a), (b) and (c), rank the amount of work done by each of the processes shown from largest to smallest. How much work is done in each case?

(b) > (c) > (a)

The amount of work done during a process depends on the path you take from your initial point to your final point. In other words it depends on how you change your pressure and volume!

(a) W=-Pi(Vf-Vi) (b) W=-Pf(Vf-Vi) (c)

f

i

V

V

dVVPW )(

PV-diagram.

Shaded area is the work done by the system

P depends on V in general

There are

many ways to take the system from i to f.

The work W

done and Q

depends on

the path.

𝑊 = 𝑝𝑑𝑉𝑉𝑓

𝑉𝑖

W and Q are not State Functions

P

V

P2

P1

V1 V2

A B

C D

The work is negative for the “clockwise” cycle; if

the cyclic process were carried out in the reverse

order (counterclockwise), the net work done on the

gas would be positive.

01212

211122

VVPP

VVPVVPWWW CDABnet

2

1

),(21

V

VdVVTPW

We can bring the system from state 1 to state 2 along infinite

number of paths, and for each path P (T,V ) will be different.

U is a state function, W - is not

Q is not a state function either. U = Q + W

Since the work done on a system depends not only on the initial and final states, but also on the intermediate states, it is not a state function.

PV diagram

P

V T

1

2

Work from closed cycles

Consider cycle A -> B -> A

WA->B

-WB->A

Work from closed cycles

Consider cycle A -> B -> A

WA->B->A= Area

Reverse the cycle, make it counter clockwise

-WB->A

WA->B

Change of State: implies one or more properties of the system has

changed.

How these properties would change outside of time is curiously

outside the framework of equilibrium thermodynamics!

The best way to think of them is that the changes are slow relative

to the underlying molecular time scales.

Processes and cycles

• Process: is a succession of changes of state.

Assuming processes are all sufficiently slow such that each stage of

the process is near equilibrium.

Certain common processes are isos, meaning “equal”:

A. Adiabatic: no heat transferred

B. Isothermal: constant temperature,

C. Isobaric: constant pressure, and

D. Isochoric: constant volume.

An important notion in thermodynamics is that of a

• Cycle: series of processes which returns to the original state.

The cycle is a thermodynamic “round trip.”

(1) Adiabatic

(2) Isochoric

(3) Isothermal

(4) Isobaric

Paths on the PV diagram

p

V

2

1

3

4

T1

T2

T3 T4

???? W = 0 ???? W = P V

Ideal gas

A. Adiabatic Process • An adiabatic process transfers no heat

ΔU = Q – W

– But Q = 0 , → ΔU = – W

• When a system expands adiabatically, W is positive (the system does work) so ΔU is negative.

• When a system compresses adiabatically, W is negative (work is done on the system) so ΔU is positive.

Adiabatic processes can occur when the system is well insulated or a very

rapid process occurs so that there is not enough time for a significant heat

to be transferred (e.g., rapid expansion of a gas; a series of compressions

and expansions as a sound wave propagates through air).

•For an ideal gas, and most real gasses, •dQ = dU + PdV •dQ = CVdT + PdV,.

•Then, when dQ = 0,

VC

PdVdT

Adiabatic Processes

In adiabatic process there is no thermal energy transfer to or from a system (Q = 0)

A reversible adiabatic process involves a “worked” expansion in which we can return all of the energy transferred. A Polytropic process is a thermodynamic process that obeys the relation:

p

V

2

1

3

4

T1

T2

T3 T4

Atmospheric processes which lead to changes in

atmospheric pressure often adiabatic: HIGH pressure cell,

falling air is compressed and warmed. LOW pressure cell,

rising air expands and cooled condensation and rain.

A Polytropic process

2 2

2 1 2 1 1 2

1 1

( 1)

p 2 1

1 2

( 1)

2 21 1 2 2 2 1

1 2 1 1 1

( 1)

2

; ln( / ) ln( / ) ln ( / )

; 1

constant

,

;

1

v

RT V

Cv

T Vv v

p v

v v

nRTU W nC dT P dV dV

V

dT R dV RT T V V V V

T C V C

C T VRR C C

C C T V

TV

P VP V T PV V

T T T PV V

Henc

2 1

1 2

cons an, t tP V

eP

VV

P

Q = 0 Adiabatic Processes, Derivation

3 5 5Monatomic: ; ;

2 2 3

5 7 7Diatomic: ; ;

2 2 5

PV P

V

PV P

V

cR Rc c

c

cR Rc c

c

Adiabatic expansion of a perfect gas For a reversible process, CVdT = -pdV along the path. Now, per mole, for an ideal gas, PV = RT

Thermo & Stat Mech - Spring 2006 Class 3

Adiabatic Process

, and

Then,V

PV PdV VdPT dT

nR nR

PdV PdV VdP

C nR

For an ideal gas, PV=nRT, so

, and

1 1Then, 0

0

V V

V

V

pV PdV VdPT dT

nR nR

PdV PdV VdP VdPPdV

C nR C nR nR

nR C VdPPdV

C nR nR

Another way

Thermo & Stat Mech - Spring 2006 Class 3 29

whe

0

0 re,

VV P

V

pP

V V

nR CPdV VdP nR C C

C

CCPdV VdP PdV VdP

C C

but

0

0, which can be integrated,

ln ln constant

ln ln ln constant

constant

PdV VdP

dV dP

V P

V P

V P PV

PV

30

Adiabatic expansion of an ideal gas

From the 1st Law, dU = dq + dw. For an adiabatic process dU = dw and dU = Cv dT, so for any expansion (or compression): TCW V

For an irreversible process, against constant pressure:

ext VW P V C T

ext

V

P VT

C

The gas cools!

1,

iiff

adiabgason

VPVPW

We can use the ideal gas to rewrite the work done on the gas in an adiabatic process in the form

Thermo & Stat Mech - Spring 2006 Class 3 31

constant

constant

as, expressed be alsocan this

of help With the

constant

1

1

P

T

TV

nRTPV

PV

for “Ideal Gasses”

33.16

21 :polyatomic

40.15

21 :diatomic

67.13

21 :monatomic

21

Adiabatic Process in an Ideal Gas

adiabatic (thermally isolated system)

The amount of work needed to change the state of a thermally isolated system

depends only on the initial and final states and not on the intermediate states.

021 Q

2

1

),(21

V

V

dVTVPW

V

P

V1

PV= NkBT1

PV= NkBT2

1

2

V2

PdVdTNkf

dUTNkf

U BB 22

( f – the # of “unfrozen” degrees of freedom )

dTNkVdPPdVTNkPV BB

PVPdVf

VdPPdV 2

to calculate W1-2 , we need to know P (V,T)

for an adiabatic process

fP

dP

fV

dV 21,0

21

0

11

P

P

V

VP

dP

V

dV

constVPPVP

P

V

V

111

1

lnln

An adiabata is “steeper” than an isotherma: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy its temperature will decrease.

2 2

1 1

1 11 2 1 1

2 1

. 1 1( , )

1

V V

V V

PV constW P V T dV dV

V V V

1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic) (again, neglecting the vibrational degrees of freedom)

Non-equilibrium Adiabatic Processes

- applies only to quasi-equilibrium processes !!!

constTV 1

(Joule’s Free-Expansion Experiment)

1.V – increases T – decreases (cooling)

2. On the other hand, U = Q + W = 0

U ~ T T – unchanged

(agrees with experimental finding)

Contradiction – because approach #1 cannot be justified – violent

expansion of gas is not a quasi-static process. T must remain the same.

B. Isothermal Process

• An isothermal process is a constant temperature process.

• Any heat flow into or out of the system must be slow

enough to maintain thermal equilibrium

• For ideal gases, if ΔT is zero, ΔU = 0

• Therefore, Q = W

– Any energy entering the system (Q) must leave as work

(W)

Isothermal processes

• Work done when PV = nRT = constant P = nRT / V

p

V

3 T1

T2

T3 T4

Isothermal change T = 0

U = 0, PV = n R T

22 2

1 11

2 1 11 1 2 2

1 2 2

2 2 2

1 1 1

ln

ln

ln

V V

V V

W W

nRT nRT

VnRTQ W PdV dV nRT

V V

V P PP V P V Q W n R T

V P P

Therefore

V V PWe and e

V nRT V P

Boyle’s Law (1627 -1691)

T1= T2 ------- P1V1 = P2 V2

Isothermal Process in an Ideal Gas

2 22

1 2

1 1 1

lnV V

B BV V

VdVW PdV Nk T Nk T

V V

ln lnf fi f B

i i

V VW Nk T nRT

V V

Wi-f > 0 if Vi >Vf (compression)

Wi-f < 0 if Vi <Vf (expansion)

Work in isothermal process

2121 WQ

0dU

V

P

PV= NkBT

V2 V1

W

D. Isochoric Process

• An isochoric process is a constant volume process. When the volume of a

system doesn’t change, it will do no work on its surroundings. W = 0

ΔU = Q

Heating gas in a closed container is an isochoric process

isochoric ( V = const )

021 W TCTTNkQ VB 02

31221

21 QdU

V

P

V1,2

PV= NkBT1

PV= NkBT2 1

2

Quasistatic Processes in an Ideal Gas

C. Isobaric Process • An isobaric process is a constant pressure process.

• ΔU, W, and Q are generally non-zero, but calculating the work done by an

ideal gas is straightforward

W = P·ΔV

Water boiling in a saucepan is an example of an isobar process

2

1 2 2 11

0W PdV P V V

TCTTNkQ PB 02

51221

2121 QWdUV

P

V1

PV= NkBT1

PV= NkBT2 1

2

V2

Isobaric Process : constant P

2 1W p V V p V

Q U W U p V

isobaric processes PQ n C T

CP = molar specific heat at constant pressure

P Vn C T n C T p V Ideal gas, isobaric :

Vn C T n R T

P VC C R Ideal gas

Isotherms

1P

P

dQC

n dT

Isobaric Processes & Specific Heat

Isochoric (isovolumetric)

Isobaric

)(==12

VVPVPW -

Since the volume of the system in isovolumetric process

remains unchanged, thus

Therefore the work done in the isovolumetric process is

0 PdVW

0dV

The work done during the isobaric process which change of

volume from V1 to V2 is given by

2

1

V

VPdVW and constantP

2

1

V

VdVPW

Work done at constant

volume

Work done at

constant

pressure

Summary, Work and ideal gases

Isochoric

Isobaric

Is

0

ln othermalf

i

f i

Vf

Vi

W PdV

W P dV P V V

VdVW PdV nRT nRT

V V

)(12

constconst2

1

2

1

VVPdVW V

V

V

V V

dV

V

Adiabatic (and

reversible)

𝑊 = 0 Isochoric

2 2

1 12 1 2 1

2 2 1 1 2 1

( )

( ) ( )

constant P, closed system, P-V work only

V V

p pV V

p

p

H U PV

U U q w q PdV q P dV q P V V

q U PV U PV H H

H q

The heat qp absorbed in a constant-pressure

process equals the system’s enthalpy change.

Enthalpy

R C-C

dTCqΔU )dT

dU(

dT

dqC

C

dTCqΔH)dT

dH(

dT

dqC

C

TnCUE T nCH

vp

T

Tvvv

vv

v

T

Tppp

p

p

p

VP

2

1

2

1

pressureconstant at capacity heat

pressureconstant at capacity heat

ΔΔΔΔΔ

),(:n

CC

n

CCnote v

v

p

p

Thermodynamics of Ideal Gases

RCCn

C

n

C

nRdT

dUnRTUd

dT

dU

dT

dHCC

VpVp

Vp

)(

The Enthalpy

Isobaric processes (P = const):

dU = Q - PV = Q -(PV) Q = U + (PV)

H U + PV - -- the enthalpy

The enthalpy is a state function, because U, P,

and V are state functions. In isobaric processes,

the energy received by a system by heating equals

to the change in enthalpy.

Q = H

isochoric:

isobaric:

in both cases, Q

does not depend on

the path from 1 to 2.

Q = U

Consequence: the energy released (absorbed) in chemical reactions at constant

volume (pressure) depends only on the initial and final states of a system.

The enthalpy of an ideal gas: (depends on T only)

TNkf

TNkTNkf

PVUH BBB

1

22

The heat supplied is equal to the change in another thermodynamic property called enthalpy (H)

i.e. H = Qp [only valid at constant pressure]

As most reactions in chemistry take place at constant pressure we can say that:

A change in enthalpy = heat supplied

Enthalpy and heat capacity •Enthalpy, H = U + PV, turns out to be a useful quantity for calculating the heat capacity at constant pressure

P

P P

dQ HC

dT T

Always true

0

0 0 and T

P P PT

dHC H H H C dT C T T

dT

•For an ideal gas, it can be shown that the enthalpy depends only on the temperature of the gas T. Therefore,

dH = dU + PdV + VdP = dQ + VdP

Enthalpy (H)

2

1

2

1

2

1

dVPdQdU

121212 VVPQQUU

1212 PVPVQUU P PQPVUPVU 1122

PVUH

On integrating between the limits initial and final stages, we can write from 1st Law,

Or,

PQQQ 12

For constant pressure,

Therefore, and

Replacing by Hi, ii PVU PQHHH 12

This H is called enthalpy

At constant pressure, dTCdq PP

From the definition of Enthalpy, dHdqP

So after rearranging, dTCdH P

On integration between the limits, T0 (standard temperature) to

T (any other temperature),

T

T

T

T

P dTCdH0 0

00 TTCHH PTT

The equation enables us to calculate enthalpy of formation

of compound at temperature T.

EXOTHERMIC & ENDOTHERMIC REACTIONS

Exothermic process: a change (e.g. a chemical reaction) that releases heat.

A release of heat corresponds to a decrease in enthalpy.

Exothermic process: H < 0 (at constant pressure).

Endothermic process: a change (e.g. a chemical reaction) that requires (or absorbs) heat.

Absorption of heat corresponds to an increase in enthalpy.

Endothermic process: H > 0 (at constant pressure).

Evaporation, fusion, melting of solids

Condensation, crystallization of liquids

Vaporisation

Energy has to be supplied to a liquid to enable it to overcome forces that hold molecules together

• endothermic process (H positive)

Melting

Energy is supplied to a solid to enable it to vibrate more vigorously until molecules can move past each other and flow as a liquid

• endothermic process (H positive)

Freezing

Liquid releases energy and allows molecules to settle into a lower energy state and form a solid

• exothermic process (H negative)

(we remove heat from water when making ice in freezer)

CV and CP

dT

PdVdU

dT

QC

V

VT

UC

the heat capacity at

constant volume

the heat capacity at constant pressure

P

PT

HC

To find CP and CV, we need f (P,V,T) = 0

and U = U (V,T)

dT

dVP

V

U

T

U

dVV

UdT

T

UdU

dT

PdVdUC

TV

TV

PT

VPdT

dVP

V

UCC

Note: for an ideal gas, U = U(q ), so W = Q for isothermal processes.

It is also always true that, for an ideal gas,

andV f i P f iU C T T H C T T

Adiabatic processes: dQ = 0, so W = U, also PV = constant.

1

1V f i f f i iW C T T P V PV

Changes in Thermal Systems

Example of a Reversible Process:

Cylinder must be pulled or pushed slowly enough that the system remains in thermal equilibrium

Changes in Thermal Systems

Example of an Irreversible Process:

The gas expands freely when

the valve is opened.