Conducted byEfa Mai Inanignsih

Objective to be able to formulate work, heat and internal

energy based on the principal law of thermodynamics and apply it in problem solving

to be able to apply general equation of ideal gases in isothermal, isochoric and isobaric processes

to be able to analyze ideal gases process based on pressure against volume graph

to be able to describe the principle of Carnot engine

The phenomenonWhat do you think of

Balloon explosion?? This is it!!If a balloon is inflated

and kept in a hot place, it will finally explode, because the particles of gas in the balloon continue to expand and press the balloon wall so that when the wall is unable to stand the gas pressure,

the balloon will explode

IntroductionThermodynamics discuss the relationship

between heat and mechanical work also the basic knowledge about temperature and

heat, the influence of temperature and heat to the characteristics of substances, and kinetic theory of gases

System and Environment

is the everything becoming our observation object

is the everything which is outside the system

External Work

∆ s

The figure is a tube containing ideal gas closed with a piston

If the gas in is heated at a constant pressure, then the gas will expand and push the piston with force F so that the piston will shift as far as s

External Work (continued)

W = F.sBecause F = pA

Then W = p.A.s Where

W = mechanical work of gas (J) p = gas pressure

A = section area of tube (m2)s = displacement of piston (m)

 

Because A s = V, then the work equation above becomes

W = p.VWhere V is change of volume (m3)

Work done by the gas is directly proportional to pressure (p) and change of volume (V) of the gas.

Sample ProblemA gas is

compressed so that the volume decreases from 5.0 L to 3.5 L at a constant pressure of 1.0 x 105 Pa. Calculate the external work applied to the gas!

Solution

V1 = 5.0 LV2 = 3.5 L p = 1.0 x 105 Pa

because the gas is compressed, then W = p V

= p (V2 – V1)= 1.0 x 105 Pa (3.5 L –

0.5 L) = 1.0 x 105 Pa (-1.5 x 10-3

m3) = -150 J

Thus, the external work applied to the gas is -150 J

ExercisesA number of ideal gases are heated at a

constant pressure of 2 x 105 N/m2 so that the volume changes from 20 liters to 30 liters. Calculate the external work done by the gas during expansion!

A gas is compressed at constant pressure 2.00 x 105 Pa from a volume of 2.00 m3 to a volume of 0.500 m3. What is the work done to the gas? If the temperature initially was 40oC, what is the final temperature of the gas?

Thermodynamics Processes

Isothermal Process Isobaric Process Isochoric Process Adiabatic Process

Isothermal ProcessIsothermal process is the change process

of gas state at constant temperatureFrom the ideal gas state equation, pV =

nRT obtained pV = constant, because nRT has a constant value.So pV = constant then p1V1 = p2V2

Where p1 = initial pressure

V1 = initial volume

p2 = final pressure

V2 = final volume

complies with the Boyle’s law

The external work done by gas in the isothermal process can be determined from the equation pV = nRT and W = p V.

External work in Isothermal Process

V

ΔVnRT W

ΔVV

nRT W

VpΔ W

1

2

12

VV

V

V nRT W

V - V nRT W

V nRT W 2

1

ln

lnln

ln

Isobaric ProcessIsobaric process is the change process of gas

state at constant pressure

because p is constant and nR is always constant, then

p

nR

T

V

nRT pV

2

2

1

1

1

1

T

V

T

V

T

V or constant

the Gay-Lussac’s law

Graph of isobaric process

Isobaric Process(continued)

The external work done by gas in isobaric process can be determined by equation

W = p V = p (V1 – V2)

it can also be determined by the area under the p – V graph (shaded region)

External work in Isobaric Process

Isochoric ProcessIsochoric process is the

change process of gas state at constant volume

because V is constant and nR is always constant, then

This process complies the Gay-Lussac’s law.

V

nR

T

p

nRT pV

2

2

1

1

T

p

T

p

T

p or constant

graph of isochoric process

The external work done by gas in isochoric process:

In other words, in isochoric process (constant volume), the gas does not do any external work. So the area under the p – V graph only forms a point.

External work in Isochoric Process

0

W

ΔV p WThe area under the p – V graph for isochoric process is zero

Adiabatic process is the process change of a gas state which does not experience any transfer of heat or there is no heat entering or coming out of the system (gas)

This process complies with the Poisson’s formula

Where = Laplace constant = Cp/Cv

Cp = specific heat of gas at constant pressure

CV = specific heat of gas at constant volume

Adiabatic Process

2211 Vp Vpor constant V p

The equation above can also be expressed in another equation as follows

1γ-22

1γ-11

γ2

2

2γ1

1

1

γ22

γ11

VT VT

V V

nRT V

V

nRT

Vp Vp

Adiabatic Process(contd.)

Curvature of p – V adiabatic graph is steeper than isothermal

The external work done by gas in adiabatic process is expressed as follows.

)Vp - V(p - γ

W 22111

1

)T-(T nR W 212

3or

External work in Adiabatic Process

The external work done by gas is equal to the area of shaded region under the p – V graph

Sample Problem Two moles of

gas is compressed at a constant temperature of -23oC so that its volume becomes half of the initial. Calculate the external work done by the gas! (R = 8.31 J/mol K, ln 1 = 0, ln 2 = 0.69)

Solution  

Given n = 2 mol R = 8.31 J/mol K T = (-23 + 273) K = 250 K

V2 = ½ V1 so V2/V1 = 1/2

The external work done in isothermal process W =

= (2 mol) (8.31 J/mol K) (250 K) (ln )

= 4155 J (ln 1 – ln 2)= 4155 J (0 – 0.69) = -2866.95 J

Thus, the external work done is -2866.95 J (-) sign indicates that at the gas is applied work.

1

2

V

VnRT ln

A monoatomic ideal gas ( = 5/3) is compressed adiabatically and the volume decreases to its half. Determine the ratio of the final pressure to the initial pressure!

Sample Problem

Solution

Because = 5/3 and V2/V1 = ½then

p1V1 = p2V2

p2 /p1 = V1 /V2

= 2 (5/3)

= Thus, the ratio of the final to the initial pressure is

3 42

3 42

Exercise A gas occupying a room of 40 cm3 is heated at a

constant pressure so that the volume becomes twice the initial. The gas pressure is 105 Pa. Calculate the external work done by the gas!

Two moles of ideal gas initially has a temperature of 27oC, a volume V1 and pressure p1 = 6.0 atm. The gas expands isothermally to V2 volume and pressure p2 = 3.0 atm. Calculate the external work done by the gas! (R = 8.31 J/mole K)

The First Law of ThermodynamicIf an external work is applied on a system,

then the temperature of the system will increase. This happens because the system receives energy from the environment. This increase of temperature relates to the increase of the internal energy.

In adiabatic process, the external work applied on the ideal gas will be equal to the change of internal energy of the ideal gas. 0 or ΔU - WΔU WWhere

W = external work (J) U = the change of internal energy (J)

in non-adiabatic process, the gas will not only receive external work but also heat.

Where Q = heat (J)

This is it!!.....The first law of thermodynamics

That states

“Though heat energy has turned into the change of internal energy and external work, the amount of

all energy is always constant”.

W - Q ΔUW ΔU Q or

The First Law

If a system receives (absorbs) heat from the environment

Q = U + (+W) = U + WIf a system receives heat from

the environment Q = U + (-W) = U – W

The First Law

The rules of W and Q values

In isothermal process (constant temperature), the change of internal energy U = 0, because the change of temperature T = 0. So that the first law of thermodynamics becomes

The First Law in isothermal process

1

2

V

V nRT W Q

W - Q 0

W - Q ΔU

ln

In isochoric process (constant volume), the work applied by gas W = 0 because the change of volume U = 0. So that the first law of thermodynamics becomes

The First Law in isochoric process

Q ΔU

0 - Q ΔU

W - Q ΔU

In isobaric process (constant pressure), the work done by gas W = p V = p (V2 – V1). So, the first law of thermodynamics becomes

The First Law in isobaric process

)V - (V p - Q ΔU

W - Q ΔU

12

In adiabatic process, the system does not receive heat or release heat, so Q = 0. Therefore, the first law of thermodynamics becomes

The First Law in adiabatic process

W- ΔU

W - 0 ΔU

W- Q ΔU

Heat Capacity Heat capacity of gas is the amount of heat

energy needed to increase gas temperature by one Kelvin (1 K) or one degree Celsius (1oC).

ΔT

Q C Where

C = heat capacity (J/K) Q = absorbed heat (J)

T = the change of temperature (K)

processisochoricforΔTC Q or Δ

Q C VV

T

VV

processisobaricforΔTC Q or ΔT

Q C pp

pp

DerivationThe first law of thermodynamics can be

derived as

ΔT

W

ΔT

ΔU C

ΔT

W

ΔT

ΔU

ΔT

Q

W ΔU Q

For isobaric process (p = constant)

Monatomic gas: Diatomic gas:

nRΔR ΔU nRT U

nRΔR ΔV p W

2

3

2

3

nR C

nR nR C

ΔT

nRΔR

ΔT

nRΔR C

p

p

p

2

52

3

23

nR C p 2

7

For isochoric process (V = constant).

Monatomic gas: Diatomic gas:

Relationship between Cp and CV :nR C

ΔT

0

ΔT

nRΔR C

W

V

V

2

3

23

0

nR C V 2

5

nR C - C

nR - nR C - C

Vp

Vp

2

5

2

7

Other parametersMolar heat capacity of gas

ΔT n

Q Cm

process isochoricΔTC n Q or ΔT n

Q C m,VV

VmV,

process isobaricΔTC n Q or ΔT n

Q C mp,p

pmp,

Specific heat of gas

process isochoricforTmC Qor T m

Q C VV

VV

ΔT m

Q C

gas monatomicfor 2

3

M

R CV

gas diatomicfor 2

5

M

R CV

process isobaricfor or ΔTmC Q ΔT m

Q C pp

pp

M

R C p 2

5

M

R C p 2

7

Sample ProblemOne mole of gas is

compressed at a constant temperature of -23oC so that its volume decreases to half of its initial volume. Calculate the work done by the gas! (R = 8.31 J/mole K; ln 1 = 0, ln 2 = 0.69)

Solution  Because

n = 1 mol T = (-23 + 273) K = 250 K V2 = ½ V1

R = 8.31 J/mol K

Then W = nRT ln V2/V1

= 1 x 8.31 x 250 ln ½ = -1.4 x 103 joule

Thus, the work done by the gas is -1.4 x 103 joule.

ExerciseTwo moles of ideal gas initially has temperature of

27oC, volume V1 and pressure p1 = 6.0 atm. The gas expands in isothermic process and reaches volume of V2 and pressure p2 = 3.0 atm. Calculate the external work done by the gas! (R = 8.3 J/mole K) (Answer : 11.5 J)

2.5 m3 of neon gas with temperature 52oC is heated in isobaric process to 91oC. If the pressure of the gas is4.0 x 105 N/m2, determine the work done by the gas!

56 x 10-3 kg of nitrogen is heated from -3oC to 27oC. If it is heated in a free expanding vessel, then required heat of 2.33 kJ. If the nitrogen is heated in a stiff vessel (cannot expand), then the heat required is 1.66 kJ. If the relative mass of nitrogen molecules is 28 g/mole, calculated (a) the heat capacity of nitrogen, (b) the general gas constant!

Thermodynamic Cycle Cycle means the process

which runs from the initial state and returns to that initial state after gas does work

Random cycle in p – V diagram

In the process a – b, the gas expands in adiabatic process and the work done by the gas is the area of plane abV2V1, its value is negative. In process b – c the compressed gas in isothermal process is the area of plane bcV1V2, its value is positive. In process c – a the gas does not do any work because its volume is constant. The process c – a is an isochoric process which is done to make the gas returns to its initial state.

The total external work done by the gas in one cycle a – b – c – a is the area of abca.

A thermodynamics cycle can occurs in a heat engine, such as otto engine (Otto cycle), diesel engine (diesel cycle), steam engine (Rankine cycle), and carnot engine

Wabca =Wab + Wbc + Wca

area of abV2V1 + (-area of bcV1V2) + 0

Wabca =area of abca

Thermodynamic Cycle: Carnot engineCarnot engine is assumed

as an ideal heat engine which works cyclically and reversible between two temperatures without any loss of energy.

imaginary Carnot engine

Within one cycle, the gas returns to its initial state, so there is no change of internal energy (U = 0).

Q = U + WQ1 – Q2 = 0 + WW = Q1 – Q2

Work process of Carnot engine to produce Carnot cycle

Entire of process in the Carnot can be represented in pressure (P) against volume (V) graph:

Carnot cycle

EfficiencyEfficiency of engine

In the Carnot engine, holds W = Q1 – Q2

%Q

Wη

1

100

100%1

100%

1

2

1

21

Q

Qη

Q

Q-Qη

%T

T

1

2 1001

Sample ProblemThe figure below

indicates the thermodynamics change of system from iitial state A to B and C and back to A. If VA = 0, VB = 30 joule and heat given to the system in process B C = 50 J

Determine

a. the system internal energy in state C,

b. the heat given to the system in A B process, and

c. the heat given to the system or taken from in C A process.

Solution a. The heat which is given away to the system in B

C process, is QBC = +50 joule.

QBC = +50 joule

WBC = 0, because B C process is isochoric

Use the first law of thermodynamics QBC = UBC + WBC UBC = QBC – WBC

UBC = 50 – 0 = 50 J

UBC = UC – UB UC = UBC + UB

UC = 50 + 30 = 80 J  

Thus, the system internal energy in state C is 80 J.

b. Process from A BWAB = area of ABED

= AB x BE = 2 x 30 = 60 J

UAB = UB – UA

= 30 – 0 = 30 J QAB is calculated by using the first law of thermodynamics

QAB = UAB + WAB

QAB = 30 + 60 = 90 J

Thus, the heat given in A B process is 90 J.

c. Process from C AWCA = -area of ACED

= -(area of ABED + area ABC)

= -120 JUCA = UA – UC = 0 – 80 = 80 J

QCA is calculated by using the first law of thermodynamics. QCA = UCA + WCA = -80 + (-120)

= -200 J Thus, the energy taken in C A process is -200 J.

2

BC AB60

2

60260

The Second Law of Thermodynamics

is a restriction of the first law of thermodynamics which expresses the energy conservation

It is states: “energy cannot be created or destroyed but can only change from one form to another”

Rudolf Clausius: Heat flows spontaneously from an object of high temperature to an object of lower temperature and it does not flow spontaneously in the opposite direction without external work.

EntropyThe total entropy of the universe does not change

when a reversible process occurs (Suniverse = 0) and increase when the irreversible process occurs (Suniverse > 0).

entropy is a measurement of the amount of energy or heat which cannot changed into work

the total change of entropy of the Carnot engine is

01

1

2

212 T

Q -

T

Q ΔS ΔS

T

Q ΔS with S = the change of entropy (J/K)

Kelvin and Planck formulate the second law of thermodynamics about heat engine that it is impossible to make an engine with 100% efficiency

Principle of cooler engines: is flowing heat from the cool reservoir T2 to the hot reservoir T1 by exerting external effort on the system.

The magnitude of external work needed in a cooler engine is formulated as

Where :

Q1 = heat absorbed from low temperature

Q2 = heat given at high temperature

21 Q - Q W

Sample ProblemA motor operates a cooler engine for

producing ice. Q2 heat is taken from a cooling room which contains an amount of water at 0oC and Q1 heat is given away to the air around it at 15oC. Suppose the cooler engine has a coefficient of performance of 20% of the coefficient of performance of an ideal cooler engine. a. Calculate the work done by the motor to make 1

kg of ice? (ice latent heat is 3.4 x 105 J/kg)b. What is time required to make 1 kg of ice if the

power of the motor is 50 W?

Solution The scheme of the cooler engine

m = 1 kgLice = 3.4 x 105 J/kg T1 = 15 + 273 = 298 K T2 = 0 + 273 = 273 K Cp engine = 20% x Cp ideal p = 50 W

21

2idealp

T-T

TC

15

273

K 273 -K 288

K 273

Cp engine = 20% x Cp ideal

The work done by electric motor (W) is

Thus, the work done by the electric motor (W) is 9.3x104 J.

b. The time required to make 1 kg of ice is

Thus, the time required to make 1 kg of ice is 31 minutes.

25

91

15

273

100

20

W

QC 2

p engine

J 104 x 9.3 enginepice

engine p

2 C/LmC

QW

minutes 31seconds 1860 power

work

P

Wt

ExercisesAn ideal refrigerator has coefficient of

performance of 5.0. If the room temperature outside the refrigerator is 27oC, what is the lowest temperature in the refrigerator which can be obtained?

The coefficient of performance of a refrigerator is 4.0. What is the electric energy used to transfer 4000 joule of heat from the food in the refrigerator?

That’s all!!!Thanks, ……bye bye