Thermodynamics : Temperature, Heat Transfer, and

First Law of Thermodynamics,

Yohanes Edi GunantoDept. of Math. Educ.

UPH

Temperature

Apakah Suhu = Panas ?

• Apa yang dimaksud dengan Suhu ?

• Apa yang dimaksud dengan Panas ?

Apa yang terjadi ketika benda padat, cair dan gas menerima panas ?

Ketika menerima panas, molekul-molekul bergerak makin lama makin cepat !

• Inilah yang disebut dengan pemuaian

• Jadi apakah pemuaian itu ?

Mengapa termoemeter dapat digunakan untuk mengetahui panas

suatu benda ?

Bandingkan skala-skala termometer di bawah ini

Bagaimana cara panas berpindah?

Panas berpindah dari benda bersuhu panas ke benda bersuhu dinginPanas berpindah dengan cara konveksi, konduksi dan radiasi

Konduksi

Perpindahan panas tanpa memindahkan penghantarnya

Konveksi

• Perpindahan panas dengan memindahkan perantaranya

Radiasi

• Perpindahan panas tanpa memerlukan perantara

THERMODYNAMICS BASICS

Zeroth Law

A B C

If A and B and B and C are inthermal equil, then A and C arein thermal equil. [ie. At same T]

First Law of Thermodynamics

Conservation of Energy for Thermal Systems

Joule Equivalent of Heat

• James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy.

• He showed that 1 calorie of heat was equivalent to 4.184 J of work.

1 cal = 4.184 J

Energy

• Mechanical Energy: KE, PE, E • Work is done by energy transfer.• Heat is another form of energy.

Need to expand the conservation of energy principle to accommodate thermal systems.

1st Law of Thermodynamics

• Consider an example system of a piston and cylinder with an enclosed dilute gas characterized by P,V,T & n.

11stst Law of Thermodynamics Law of Thermodynamics

• What happens to the gas if the piston is moved inwards ?

11stst Law of Thermodynamics Law of Thermodynamics

• If the container is insulated the temperature will rise, the atoms move faster and the pressure rises.

• Is there more internal energy in the gas?

11stst Law of Thermodynamics Law of Thermodynamics

• External agent did work in pushing the piston inward.

• W = Fd

• =(PA)x

• W =PV

x

11stst Law of Thermodynamics Law of Thermodynamics

• Work done on the gas equals the change in the gases internal energy,

W = U

x

1st Law of TD

• Let’s change the situation:

• Keep the piston fixed at its original location.

• Place the cylinder on a hot plate.

• What happens to gas?

Heat flows into the gas.

Atoms move faster, internal energy increases.

Q = heat in Joules

U = change in internal energy in Joules.

Q = U

1st Law of TD

• What if we added heat and pushed the piston in at the same time?

F

1st Law of TD• Work is done on the

gas, heat is added to the gas and the internal energy of the gas increases!

Q = W + U

F

1st Law of TD

Some conventions:For the gases perspective: • heat added is positive, heat removed is

negative.• Work done on the gas is positive, work done

by the gas is negative.• Temperature increase means internal energy

change is positive.

First Law of Thermodynamics

“Energy cannot be created or destroyed. It can only be changed from one form into

another.”

Rudolf Clausius 1850

First Law of Thermodynamics

• Conservation of Energy• Says Nothing About Direction of Energy

Transfer

1st Law of TD

• Example: 25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?

• Po = 202,600 Pa, Vo = 0.025 m3, To = 300 K, Pf = 202,600 Pa, Vf=0.020 m3, Tf=

n = PV/RT.W = -PV

U = 3/2 nRTQ = W + U

W =-PV = -202,600 Pa (0.020 – 0.025)m3

=1013 J energy added to the gas.U =3/2 nRT=1.5(2.03)(8.31)(-60)=-1518 JQ = W + U = 1013 – 1518 = -505 J heat out

Quasistatic Processes in an Ideal Gas

isochoric ( V = const )

isobaric ( P = const )

021 W

TCTTNkQ VB 02

31221

0),( 12

2

1

21 VVPdVTVPW

TCTTNkQ PB 02

51221

21QdU

2121 QWdU

V

P

V1,2

PV= NkBT1

PV= NkBT21

2

V

P

V1

PV= NkBT1

PV= NkBT212

V2

(see the last slide)

Isothermal Process in an Ideal Gas

1

221 ln),(

2

1

2

1V

VTNk

V

dVTNkdVTVPW B

V

V

B

V

V

f

iBfi V

VTNkW ln

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

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

isothermal ( T = const ) :

V

P

PV= NkBT

V1V2

W

2121 WQ

0dU

Adiabatic Process in an Ideal Gas

adiabatic (thermally isolated system)

PdVdTNkf

dUTNkf

U BB 22

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

dTNkVdPPdVTNkPV BB PVPdVf

VdPPdV 2

fP

dP

fV

dV 21,0

21

constVPPVP

P

V

V

111

1

lnln

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 21WdU

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

for an adiabatic process

2

1

),(21

V

V

dVTVPW

011

P

P

V

V P

dP

V

dV

V

P

V1

PV= NkBT1

PV= NkBT21

2

V2

Adiabaticexponent

Adiabatic Process in an Ideal Gas (cont.)

V

P

V1

PV= NkBT1

PV= NkBT21

2

22 2

1 1 1

11 11 2 1 1

1 1 1 12 1

1( , )

1

1 1 1

1

VV V

V V V

PVW P V T dV dV PV V

V

PVV 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)

constVPPV 11

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.

V2

Prove 1 2 2 2 B

f fW PV Nk T U

Summary of quasi-static processes of ideal gas

Quasi-Static process

U Q WIdeal gas

law

isobaric (P=0)

isochoric (V=0)

0

isothermal (T=0)

0

adiabatic (Q=0)

0

fi

i f

VV

T T

2 2B

f fU Nk T P V

f iU U U

2

2

fP V

P V

2 2B

f fU Nk T P V

2

fP V

fi

i f

PP

T T

i i f fPV P Vln fB

i

VNk T

VW

i i f fPV P V 2 2B

f fU Nk T PV U

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