MFS First Law of Thermodynamics Created by: Marlon Flores Sacedon Physics section, DMPS June 2010.

MFS

First Law of Thermodynamics

Created by: Marlon Flores Sacedon Physics section, DMPS

June 2010

MFS

The First Law of ThermodynamicsThermodynamic system is a system that can interact (and exchange energy) with its surroundings, or environment, in at least two ways, one of which is heat transfer.

Thermodynamic process is a process in which there are changes in the state thermodynamic system.

Work Done during volume changes

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Work Done by the system

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dVpdspAdsFdW

Where:W = work done by the system [J]p = pressure [pa]dV = differential volume [m3]V1 & V2 = initial and final volume [m3]

dS

F

pdVdW

2

1

V

VpdVW

A

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2

1

V

VpdVW

Where:W = work done by the systemp = pressuredV = differential volumeV1 & V2 = initial and final volume

V1 V2

W

pV-diagram

0

p

V

Area 2

1

V

VpdV

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Signs of work done

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nRTpV If the pressure is constant during thermodynamic process

]VV[pW 12

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Problem

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Paths Between Thermodynamics States

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Internal Energy (U)

Internal Energy of a system is the sum of kinetic energies of all of its constituent particles, plus the sum of all the potential energies of interaction among these particles.

12 UUU Where: = change in internal energy U1 = initial internal energyU2 = final internal energy

U

MFSSystem

Surroundings(environment)

The First Law of Thermodynamics

WQUUU 12

WUQ

= Q-W = +50 JU

System


= Q-W = -50 JU

System


Q = 150J W = 150J

= Q-W = 0U

Q = -150J W = -100J

Where: = change in internal energy (J) W = work done (J)Q = heat quantity (J)

U

Q = 150J W = 100J

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Ex. A gas in a cylinder is held at a constant pressure of 2.30x105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3. The internal energy of the gas decreases by 1.40x105 J. a) Find the work done by the gas. b) Find the absolute value of the heat flow into or out of the gas, and state the direction of heat flow. c) Does it matter whether or not the gas is ideal? J, b) 2.55x105J, out of gas, c) no (Ans. a) -1.15x105

Ex. A gas in a cylinder is held at a constant pressure of 2.30 x 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3. The internal energy of the gas decreases by 1.40 x 105 J. a) Find the work done by the gas, b) Find the absolute value |Q| of the heat flow into or out of the gas, and state the direction of heat flow, c) Does it matter whether or not the gas if ideal? Why or who not?

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Kinds of Thermodynamic Process1. Adiabatic Process (pronounced “ay-dee-ah-bat-ic”) is defined as one with no heat transfer into or out of a system: Q = 0.

WUUU 12(adiabatic process)

2. Isochoric Process (pronounced “eye-so-kor-ic”) is a constant-volume process. When the volume of thermodynamic system is constant W=0.

QUUU 12(isochoric process)

3. Isobaric Process (pronounced “eye-so-bear-ic”) is a constant –pressure process.

)( 12 VVpW (Isobaric process)

4. Isothermal Process (pronounced “eye-so-bear-ic”) is a constant –temperature process.

2

1

1

2 lnln2

1 p

pnRT

V

VnRT

V

dVnRTW

V

V (Isothermal process)

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Kinds of Thermodynamic Process

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Internal Energy of an Ideal Gas

Property of Ideal Gas: The internal energy of an ideal gas depends only on its temperature, and not on its pressure and volume.

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Heat Capacity of an Ideal GasMolar heat capacity at constant volume (CV)

WUQ

RCC Vp

Molar heat capacity at constant pressure (Cp)

dTnCdQ V0dW

dTnCdQ ppdVdW

nRdTdUdTnC p

nRdTdTnCdTnC Vp

(First Law)

At constant volume

dUdQ (from First Law)

dTnCdU V (because dQ=dU)

dWdUdQ or

At constant pressure

nRdTdW ( from pV=nRT )

(Molar heat capacities of an ideal gas)

V

p

C

C

(ratio of heat capacities)

Where:Cp = molar specific at constant pressure (J/mol.K)CV = molar specific at constant volume (J/mol.K)R = ideal gas constant initial and final volume

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Type of Gas

Gas CV (J/mol.K)

Cp (J/mol.K)

Cp-CV (J/mol.K)

(J/mol.K)

Monatomic He 12.47 20.78 8.31 1.67

Ar 12.47 20.78 8.31 1.67

Diatomic H2 20.42 28.74 8.32 1.41

N2 20.76 29.07 8.31 1.40

O2 20.85 29.17 8.31 1.40

CO 20.85 29.16 8.31 1.40

Polyatomic CO2 28.46 36.94 8.48 1.30

SO2 31.39 40.37 8.98 1.29

H2S 25.95 34.60 8.65 1.33

Molar Heat Capacities of Gases

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Heat Capacity of an Ideal Gas

RRRRCC Vp 2

5

2

3

Molar heat capacities for Monatomic ideal gas

67.13

5

23

25

R

R

C

C

V

p

RRRRCC Vp 2

7

2

5 40.1

5

7

25

27

R

R

C

C

V

p

Molar heat capacities for Diatomic ideal gas

RRRRCC Vp 2

9

2

7 29.1

7

9

27

29

R

R

C

C

V

p

Molar heat capacities for Polyatomic ideal gas

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Example. In an experiment to simulate conditions within an automobile engine, 645J of heat is transferred to 0.185 mol of air-conditioned within a cylinder of volume 40.0cm3. Initially the nitrogen is at a pressure of 3.00x106 Pa and a temperature of 780K. a) If the volume of the cylinder is held fixed, what is the final temperature of the air? Assume that the air is essentially nitrogen gas, use the Table. Draw a pV-diagram for this process. b) Find the final temperature of the air if the pressure remains constant. Draw a pV-diagram for this process

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Adiabatic Process for an Ideal Gas

WU No heat transfer, Q = 0

dTnCdU VpdVdW dWdU

pdVdTnCV

dVV

nRTdTnCV

0V

dV

C

R

T

dT

V

Vp CC

11C

C

C

CC

C

R

V

p

p

Vp

V

0V

dV)1(

T

dT

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Adiabatic Process for an Ideal GasttanconsVln)1(Tln

ttanconsVlnTln 1

ttancons)TVln( 1

ttanconsTV 1

122

111 VTVT

ttanconsVnR

pV 1

ttanconspV

221

11 VpVp

Adiabatic process, ideal gas


)TT(nCW 21V

)VpVp(1

1)VpVp(

R

CW 22112211

V


MFS First Law of Thermodynamics Created by: Marlon Flores Sacedon Physics section, DMPS June 2010.

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Transcript of MFS First Law of Thermodynamics Created by: Marlon Flores Sacedon Physics section, DMPS June 2010.