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THERMODYNAMIC EQUILIBRIUM

1.Mechanical Equilibrium

2.Thermal Equilibrium

3.Chemical Equilibrium

QUASISTATIC PROCESS

In which the deviation from thermodynamic

equilibrium is infinitesimal and all the states

through which the system passes during a

quasistatic process can be considered as

equilibrium states.

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Internal Energy and Temperature

“internal energy” : the energy content of a system.

1. kinetic energy of molecules (due to vibrational,

rotational or electronic energy)

2. Potential energy due to intermolecular forces ,

depends upon separation .

3. the energy of electrons and nucleiThe relationship between energy and temperature (for monatomic ideal gas)

2B

1 3av K E /m o lecu le m v k T

2 2= =

Internal Energy U = number of molecules av KE/molecule

= N (3/2) kBT

= (3/2) n RT = (3/2) P V (ideal gas)Internal energy can change by thermodynamic process

====== ZERO’th LAW of Thermodynamics =====

If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

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• The work done is the area under the curve in a PV diagram.

• The work done depends on the path by which the system goes from the initial to the final state.

• The work done depends on the initial, final and intermediate states of the system.

Work : A path dependent function

V

P

1 2

3

V1 V2

P1

P3

Area = (V2-V1)x(P1-P3)/2

What is the total work done by system when going from state 1 to state 2 to state 3 and back to state 1 ?

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V

P 1 2

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V

PW = PV (>0)1 2

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V > 0 V

PW = PV = 01 2

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V = 0

V

PW = PV (<0)1 2

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V < 0 V

W = PV = 01 2

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P

V = 0 V

P 1 2

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If we gothe other way thenWtot < 0

V

P 1 2

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Wtot > 0

Wtot = ??Isobaric Process: P=constantIsochoric Process: V=constant

Work done during volume Work done during volume changechangeWork can be performed by gas system by changing

volume dW=Fdx=PAdxat constant pressure ‘P’ and Area ‘A’

A

2

1

V

V

W P dV=

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Work done is the area under the P-V curve.

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The amount of heat supplied to the system is equal to the

sum of the increase in the internal energy (U) of the

system and the external work done (W) by the system.

Q=(U2-U1)+WFor a very small change in the state of the system

Q=dU+W or dU = Q - PdV

The First Law of Thermodynamics

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Significance of the first law

•It is applicable to any process by which a system

undergoes a physical or chemical change

•It introduces the concept of internal energy, and

•It provides method for determining the change in

internal energy

Limitations of the first law

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Path between thermodynamic Path between thermodynamic statesstates

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Path between thermodynamic Path between thermodynamic statesstates

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First Law of ThermodynamicsExample-1

V

P

1 2

V1 V2

P2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant pressure

P=1000 Pa, where V1 =2m3 and V2

=3m3. Find

T1, T2, dU, W, Q.1. P V1 = n R T1 T1 = P V1/(nR) = 120K

2. P V2 = n RT2 T2 = P V2/(nR) = 180K

3. U = (3/2) n R T = 1500 J or U = (3/2) P V = 1500 J

4. W = P V = 1000 J > 0 Work done by the system (gas)

5. Q = U + W = 1500 J + 1000 J = 2500 J > 0 heat gained by system (gas)

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2 moles of monatomic ideal gas is taken

from state 1 to state 2 at constant volume

V=2m3, where T1=120K and T2 =180K. Find Q.

1. P V1 = n R T1 P1 = n R T1/V = 1000 Pascal

2. P V2 = n R T1 P2 = n R T2/V = 1500 Pa

3. U = (3/2) n R T = 1500 J

4. W = P V = 0 J

5. Q = U + W = 0 + 1500 = 1500 J

=> It requires less heat to raise T at const. volume than at const. pressure.

V

P

2

1

V

P2

P1

First Law of ThermodynamicsExample-2

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Thermodynamic Processes(for gases)• Isothermic: constant temperature

T = 0 means U = 0 (U is constant!)• Adiabatic: no gain/loss of heat energy (Q)

Q = 0 means U = W

Note: Isothermal and Adiabatic are important for Carnot Engine (Ideal heat Engine)

• Isochoric (or isovolumetric): constant volumeV = 0 means W = 0 (no work performed on/by system)

• Isobaric: constant pressureP = 0 means W = PV

Note: Isochoric and Isobaric are important for specific heat of Gases.

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Specific Heat Capacity & 1st Law of Thermodynamics (for monatomic

gases)• The heat (Q) absorbed by a gas can

be expressed as:

Q = CnT

where C is the molar heat capacity (J/mol.K)

• For gases it is necessary to distinguish between molar heat capacity at constant pressure (CP) and constant volume (CV)

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Let’s begin with the 1st Law of Thermodynamics:

δQ = U + δW

At constant pressure: {U = (3/2)nRT & W = PV = nRT}

Q = (3/2)nRT + nRT = (5/2)nRT CP = (5/2)R

At constant volume: {U = (3/2)nRT & W = 0}

Q = (3/2)nRT + 0 = (3/2)nRT CV = (3/2)R

Specific Heat Capacity & 1st Law of Thermodynamics (for monatomic

gases)

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P-V relation for Adiabatic Processes

• When a gas undergoes an adiabatic expansion/contraction the relationship between pressure and volume is

PiVi = PfVf

where is CP/CV

For monoatomic gases, is 5/3

For Diatomic gases, is 7/5

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Selected Specific Heats

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An ideal gas is taken through a cycle abc consisting of the

following processess:

ab, isothermal expansion at temperature T1 from volume

V1 tp 2V1,

bc, compression at constant pressure P1 to volume V1,

and ca, change of pressure at constant volume. Find the

expressions for the work done, the heat transferred and

the change in the internal energy for each part of the

cycle, and show that the sum of the three is zero for the

internal energy only.

P a

P1 c b

V1 2V1

V

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• Book: Basic Thermodynamics - Evelyn Guha (Cahpter-3,4 & 5)

• Worked Examples on first Law: Page-58,59,60 &61 (chapter-3)

Study material on Thermodynamics