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Thermodynamics: Work, Heat, Internal Energy, and the First Law, Lecture notes of Thermodynamics

Indian Institute of Science Thermodynamics

Work-Heat-Internal energy and First law

Typology: Lecture notes

2018/2019

Uploaded on 08/07/2019

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Work-Heat-Internal energy and First law

State of a system in thermodynamic equilibrium is defined by its state

variables, e.g. p,V,Tfor gas in a container. If the values of the state

variables stay put, state of the system does not change either.

When the values of the state variables change, the system has undergone

aprocess. Processes can be initiated internally i.e. from within the

system (say, explosion) or externally (say, change in volume).

Quantities defining a process are not state variables, e.g. heat, work etc.

Processes considered are quasi-static, meaning the system at all times is

infinitesimally near a state of thermodynamic equilibrium.

Our processes are not only quasi-static but also reversible, meaning every

infinitesimal steps can be reversed (no dissipation) without violating any

physical principles. It can be represented by a continuous line.

p

V

(p1, V1)

(p2, V2)

reversible

irrevesible

Partial preview of the text

Download Thermodynamics: Work, Heat, Internal Energy, and the First Law and more Lecture notes Thermodynamics in PDF only on Docsity!

Work-Heat-Internal energy and First law

State of a system in thermodynamic equilibrium is defined by its state

variables, e.g. p, V , T for gas in a container. If the values of the state

variables stay put, state of the system does not change either.

When the values of the state variables change, the system has undergone

a process. Processes can be initiated internally i.e. from within the

system (say, explosion) or externally (say, change in volume).

Quantities defining a process are not state variables, e.g. heat, work etc.

Processes considered are quasi-static, meaning the system at all times is

infinitesimally near a state of thermodynamic equilibrium.

Our processes are not only quasi-static but also reversible, meaning every

infinitesimal steps can be reversed (no dissipation) without violating any

physical principles. It can be represented by a continuous line.

p

V

(p 1 , V 1 )

(p 2 , V 2 )

reversible

irrevesible

Work is done when a system changes from state 1 to state 2

δW = F~ · d~x → W 12 =

1

~F · d~x

Work integral is path-dependent and the work differential is inexact ⇒

work is not a state variable for a system.

If z = z(x, y ) for every x and y , then

dz =

∂z

∂x

y

dx +

∂z

∂y

x

dy ⇒

1

dz =

C

∂z

∂x

y

dx +

∂z

∂y

x

dy

The integral is said to be path-independent. Conversely, if we are given

dz = M(x, y ) dx + N(x, y ) dy then the associated integrals are

path-independent iff z(x, y ) can be found by solving

M =

∂z

∂x

y

, N =

∂z

∂y

x

∂M

∂y

x

∂N

∂x

y

Exact differential is a differential which yields a path-independent

integral, otherwise it is an inexact differential written as δz.

Isothermal : Wif =

∫ V

f

Vi

p dV = nRT

∫ V

f

Vi

dV

V

= nRT ln

Vf

Vi

Isobaric : Wif =

∫ V

f

Vi

p dV = p

∫ V

f

Vi

dV = p (Vf − Vi )

Isochoric : Wif =

Vf

Vi

p dV = 0

Examples of other forms of work

I Wire stretched by tensile force F through length change dL

δW = −F dL

I Changing surface area by dA with surface tension S

δW = −S dA

I Moving charge q around a circuit by dx by an electrochemical cell

of emf E

δW = −q E dx

I Changing magnetization by dm of a paramagnetic substance in with

electromagnetic of field H

δW = −μ H dm

Example. A spherical balloon, of diameter D 1 = 0. 3 m, contains air at

p 1 = 150 kPa and is placed in a vacuum. The balloon is heated until its

diameter is D 2 = 0. 4 m. It is given that the pressure in the balloon is

proportional to its diameter. Calculate the work of expansion.

The expression of pressure inside the balloon is

V =

π

D

→ D =

6 V

π

⇒ p = kD = k

6 V

π

and k = p 1

π

6 V 1

where k is the proportionality constant. Hence the work of expansion,

W 12 =

2

1

p dV = k

π

V 2

V 1

V

1 / 3 dV

= k

π

[

V

4 / 3

]V

2

V 1

3 k

π

[

V

4 / 3 2

− V

4 / 3 1

]

3 k

π

V

4 / 3 1

[

V 2

V 1

]

p 1

π

6 V 1

π

V

4 / 3 1

[(

D

3 2

D^3 1

]

= π p 1

D 1

) 3 [(

D 2

D 1

]

= 3. 436 kJ.

The work done by gas on piston,

W =

1

p dV =

∫ V

2

V 1

[

patm +

k

A

2

(V − V 1 )

]

dV

[

patmV +

k

2 A

2

(V − V 1 )

2

]V

2

V 1

= patm (V 2 − V 1 ) +

k

2 A^2

(V 2 − V 1 )

2

= patm (V 2 − V 1 ) ︸︷︷︸

work done on atmosphere

k

V 2 − V 1

A

work done on spring

= 0 .125 kJ

The examples considered here are all isothermal, T = const. But it need

not have to be so. Temperature can change and works can be performed.

Let us look into the role of temperature somewhat more closely.

Problems. (a) What can differentiate ice and water at 0

◦

C or (b) can

explain ice at, say, − 20

◦

C causes frostbite but air doesn’t? (c) What can

account for achieving thermal equilibrium?

The answer is heat.

Heat : transfer of energy across boundary of a system to another system

(or surroundings) by virtue of temperature difference between the two.

I Bodies do not contain heat but has relevance as a type of energy

that crosses system boundaries. It is same with work but heat

cannot be identified as work.

I Heat flows from system at higher temperature to the system at

lower temperature. If the systems are in thermal equilibrium with

each other, then there is no transfer of heat.

I Heat is not a state variable and hence infinitesimal exchange of heat

is expressed as inexact differential δQ.

I Amount of heat or thermal energy transferred or consumed is

defined through heat capacity of the system C = δQ/dT.

Depending on heat capacity C of a system, different amount of Q

will be exchanged for the same difference in T.

There is a special kind of process in which δQ = 0 – adiabatic process.

Adiabatic processes usually happen in systems enclosed by adiabatic wall

and undergoing change in pressure or volume or both. When it happens

it is accompanied by change of temperature as well.

First law of Thermodynamics

Also known as principle of conservation of energy.

I During any cyclic process, the cyclic integral of heat added to a

system is proportional to cyclic integral of work done by the system.

I In all cases in which work is produced by agency of heat, a quantity

of heat is consumed which is proportional to work done.

The mathematical representation of this law is

δW = J

δQ

where

denotes cyclic integral and J = 4.2 J/cal is proportionality

constant known as mechanical equivalent of heat.

Remembered... how Count Rumford used boring a cannon immersed in

water and boiling the water, thereby demonstrating that the work of

boring was converted into heat i.e. first law of thermodynamics.

A practical simplification – when Q is measured in units of work then

J = 1 and first law reads

δW =

δQ.

Consider two cyclic processes, each passing through points 1 and 2

following different paths

I Cycle I : 1 to 2 on path A followed by 2 to 1 on path B.

I Cycle II : 1 to 2 on path A followed by 2 to 1 on path C.

p

V

A

B

C

2

1

The first law

δQ =

δW for the two cycles are

Cycle I :

1

δQA +

2

δQB =

1

δWA +

2

δWB

Cycle II :

2

1

δQA +

1

2

δQC =

2

1

δWA +

1

2

δWC

Heat Capacity

Previously defined heat capacity as C = δQ/dT , the amount of heat

needed to raise the temperature of an object by one degree. We get

specific heat capacity by dividing heat capacity by mass, c = C /m.

It turns out that since δQ is path-dependent, so is C. Two common

paths are isochoric and isobaric,

isochoric : CV =

δQ

dT

V

isobaric : CP =

δQ

dT

p

Expressing U = U(T , V ), from the first law we get,

dU =

∂U

∂T

V

dT +

∂U

∂V

T

dV

δQ = dU + δW =

∂U

∂T

V

dT +

[

∂U

∂V

T

p

]

dV

For isochoric process, we get

CV =

δQ

dT

V

∂U

∂T

V

For isobaric process, first we make use of V = V (T , p) to write δQ,

dV =

∂V

∂T

p

dT +

∂V

∂p

T

dp

δQ = CV dT +

[

∂U

∂V

T

p

] [

∂V

∂T

p

dT +

∂V

∂p

T

dp

]

CP =

δQ

dT

p

= CV +

[

∂U

∂V

T

p

]

∂V

∂T

p

This is as far we can go without assuming any specific system. If we

consider free expansion of ideal gas dU = 0 then we get

∂U

∂V

T

∂V

∂T

p

nR

p

⇒ CP = CV + nR

For constant pressure processes, it is useful to define another state

variable that simplifies the description of energy transfer. The new state

variable is called Enthalpy H,

H = U + pV

p=const −−−−−→ dH = dU + pdV = δQ − δW + pdV = δQ

Hence, the heat capacity at constant pressure CP is

CP =

δQ

dT

P

∂H

∂T

p

Free expansion : Consider an isolated container having an adiabatic

partition separating some gas in one half from vacuum in the other. On

removing the partition, gas flows to the vacuum half irreversibly and

occupy the entire container.

The system does no work on its surroundings and no heat enters the

system from surroundings either. Therefore, dU = 0. Assume dT = 0,

U = U(T , V ) → dU =

∂U

∂T

V

dT +

∂U

∂V

T

dV ⇒

∂U

∂V

T

U = U(T , p) → dU =

∂U

∂T

p

dT +

∂U

∂p

T

dp ⇒

∂U

∂p

T

Hence, in free expansion of gas, U is a function of T only, U = U(T ).

For real gas, however, U is a function of both T , V.

dT =

∂T

∂U

V

dU +

∂T

∂V

U

dV =

∂T

∂V

U

dV = μJ dV

where μJ = Joule coefficient, whose value depends on the gas being

used. So in free expansion, temperature change is

∆T =

∫ V

f

Vi

μJ dV

Continuous flow process : In contrast with free (or Joule) expansion, the

Joule-Kelvin expansion is a more practical expansion process, called

throttling process. Gas is forced steadily and irreversibly through a

porous plug, (pi , Vi , Ti ) → (pf , Vf , Tf ).

piston piston

porous

pi, Vi, Ti pf , Vf , Tf

plug

The process is adiabatic δQ. The work done is,

Uf − Ui = pi (Vi − 0) − pf (Vf − 0) = pi Vi − pf Vf → Ui + pi Vi = Uf + pf Vf

i.e. Hi = Hf the enthalpy is conserved (isenthalpic). Writing T = T (H, p),

dT =

∂T

∂H

p

dH +

∂T

∂p

H

dp = μJK dp,

μJK is Joule-Kelvin coefficient. Temperature change is ∆T =

pf pi

Thermodynamics: Work, Heat, Internal Energy, and the First Law, Lecture notes of Thermodynamics

Related documents

Partial preview of the text

Download Thermodynamics: Work, Heat, Internal Energy, and the First Law and more Lecture notes Thermodynamics in PDF only on Docsity!

Work-Heat-Internal energy and First law

State of a system in thermodynamic equilibrium is defined by its state

variables, e.g. p, V , T for gas in a container. If the values of the state

variables stay put, state of the system does not change either.

When the values of the state variables change, the system has undergone

a process. Processes can be initiated internally i.e. from within the

system (say, explosion) or externally (say, change in volume).

Quantities defining a process are not state variables, e.g. heat, work etc.

Processes considered are quasi-static, meaning the system at all times is

infinitesimally near a state of thermodynamic equilibrium.

Our processes are not only quasi-static but also reversible, meaning every

infinitesimal steps can be reversed (no dissipation) without violating any

physical principles. It can be represented by a continuous line.

Work is done when a system changes from state 1 to state 2

Work integral is path-dependent and the work differential is inexact ⇒

work is not a state variable for a system.

If z = z(x, y ) for every x and y , then

The integral is said to be path-independent. Conversely, if we are given

dz = M(x, y ) dx + N(x, y ) dy then the associated integrals are

path-independent iff z(x, y ) can be found by solving

M =

, N =

∂M

∂N

Exact differential is a differential which yields a path-independent

integral, otherwise it is an inexact differential written as δz.

∫ V

∫ V

V

∫ V

∫ V

Examples of other forms of work

I Wire stretched by tensile force F through length change dL

δW = −F dL

I Changing surface area by dA with surface tension S

δW = −S dA

I Moving charge q around a circuit by dx by an electrochemical cell

of emf E

δW = −q E dx

I Changing magnetization by dm of a paramagnetic substance in with

electromagnetic of field H

δW = −μ H dm

Example. A spherical balloon, of diameter D 1 = 0. 3 m, contains air at

p 1 = 150 kPa and is placed in a vacuum. The balloon is heated until its

diameter is D 2 = 0. 4 m. It is given that the pressure in the balloon is

proportional to its diameter. Calculate the work of expansion.

The expression of pressure inside the balloon is

V =

D

→ D =

6 V

6 V

6 V 1

where k is the proportionality constant. Hence the work of expansion,

W 12 =

V

[

V

]V

[

V

− V

]

V

[

V 2

V 1

]

6 V 1

V

[(

D

]

D 1

) 3 [(

D 2