HUKUM PERTAMA TERMODINAMIKA

&&BEBERAPA KONSEP

DASAR

Jenis Besaran/Sifat Termodinamika

• Besaran/Sifat intensif : tidak tergantung jumlah massa zatcontoh : T, P, ρ, konsentrasi , energi per satuan massa, dllsatuan massa, dll

• Besaran/Sifat ekstensif : tergantung jumlah massa zatcontoh : Volume total, panjang, luas, energi total, dll

Pengertian Sistem

• Sistem : bagian dari alam semesta yg dipelajari/ dianalisis/ ditinjau

• Lingkungan (surrounding) : segala sesuatu di luar sistemdi luar sistem

• Batas sistem : pemisah antara sistem dan lingkungan, bisa diam atau bergerak

Jenis Sistem

• Sistem Tertutup : ‘tidak’ mengijinkan perpindahan massa melalui batas sistem

• Sistem Terbuka (volume atur) : mengijinkan perpindahan massa melalui mengijinkan perpindahan massa melalui batas sistem

• Sistem Terisolasi : tidak mengijinkan perpindahan massa maupun panas/energi melalui batas sistem

Kaidah Fase (The Phase Rule)

NF +−= π2

F = derajat kebebasan; jumlah sifat intensif yang bisa ditentukan besarnyabisa ditentukan besarnya

π = jumlah fase

Fase = suatu materi yang homogen

N = jumlah komponen

Thermodynamic state

• Keadaan termodinamika (thermodynamic state) : kondisi termodinamika yang ditentukan oleh sifat-sifat sistemSejumlah sifat sistem bisa menentukan Sejumlah sifat sistem bisa menentukan keadaan sistem. Berapa? (Jika zat homogen murni � cukup 2 sifat) Mengapa?

Fungsi Keadaan (State Functions)

• Fungsi Keadaan adalah sifat termodinamika yang tidak tergantung pada sejarah atau jalan (path) bagaimana mencapai keadaan tertentuContoh : P, T, V, U, H, S dll

• Non state functions (bukan fungsi keadaan) : Q, W• Non state functions (bukan fungsi keadaan) : Q, W

∫∫

∫

==

∆=−=

WdWQdQ

PPPdPP

P

juga

But

2

1

12

Perubahan yg sangat kecil (Infinitesimal change)

Jumlah yg sangat kecil (Infinitesimal amount)

Equilibrium

• Kesetimbangan (equilibrium) : suatu kondisi dimana semua sifat sistem tidak berubah terhadap waktu (tetap)

• Mencakup pula tidak adanya tendensi • Mencakup pula tidak adanya tendensi untuk berubah

• Tendensi untuk berubah dipengaruhi oleh hambatan dan driving forceContoh: Q = ∆T/ RR = sangat besar atau ∆T = 0

Reversible• A process is reversible when its direction can be

reversed at any point by infinitesimal change in external conditions

• A reversible process :� is frictionless� is never more than differentially removed from

equlibrium� traverses a succesion of equilibrium states� traverses a succesion of equilibrium states� is driven by forces whose imbalance is

differensial in magnitude� can be reversed at any point by infinitesimal/

differensial change in external conditions� When reversed, retraces its forward path, and

restores the initial state of system and surroundings

ENERGI DALAM (U)

• Eksperimen Joule?• Energi yang dimiliki benda karena aktivitas

dalam atom, gerakan molekul-molekulnya, interaksi antar molekul.interaksi antar molekul.

• Tidak diketahui berapa besar absolutnya• Hanya diketahui berapa yg keluar maupun

yg masuk sistem

Internal Energy

system U = kinetic + potential

system boundary

The internal energy of a system of particles, U, is the sum of the kineticenergy in the reference frame in which the center of mass is at rest and thepotential energy arising from the forces of the particles on each other.

Difference between the total energy and the internal energy?

The internal energy is a state function – it depends only onB The internal energy is a state function – it depends only onthe values of macroparameters (the state of a system), noton the method of preparation of this state (the “path” in themacroparameter space is irrelevant).

U = U (V, T)In equilibrium [ f (P,V,T)=0 ] :

U depends on the kinetic energy of particles in a system and an average inter-particle distance (~ V-1/3) – interactions.

P

V T A

B

For an ideal gas (no interactions) : U = U (T) - “pure” kinetic

Hukum Pertama

• Energi memiliki banyak bentuk, tetapi kuantitas total energi adalah tetap, jika suatu bentuk energi hilang, maka akan muncul dalam bentuk lainmuncul dalam bentuk lain

• Hukum Pertama � Neraca Energi pada sistem (konservasi energi)∆(Energi pd sistem) + ∆(Energi pd lingkungan) = 0

• ∆(Energi pd sistem) + ∆(Energi pd lingkungan) = 0

lingkungan Q W

• ∆Ut + -(Q + W) = 0

Hukum Pertama (Neraca Energi) pada sistem tertutup dan diam

Sistem

• ∆U + -(Q + W) = 0

atau ∆Ut = Q + Wdengan: ∆Ut = perub. energi dalam (joule, kalori)Q = panas (joule, kalori)W = kerja (joule, kalori)

Sign convention: we consider Q and W to be positive if energy flows into the system.

Work and Heating (“Heat”)

We are often interested in ∆∆∆∆U , not U. ∆∆∆∆U is due to:

Q - energy flow between a system and itsenvironment due to ∆∆∆∆T across a boundary and a finitethermal conductivity of the boundary

– heating (Q > 0) /cooling (Q < 0)(there is no such physical quantity as “heat”; toemphasize this fact, it is better to use the term“heating” rather than “heat”)

HEATING

WORK

W - any other kind of energy transfer acrossboundary - work

Heating/cooling processes :conduction : the energy transfer by molecular contact – fast-moving

molecules transfer energy to slow-moving molecules by collisions;convection : by macroscopic motion of gas or liquidradiation : by emission/absorption of electromagnetic radiation.

Work and Heating are both defined to describe energy transferacross a system boundary.

Work

The sign: if the volume is decreased, W is positive (bycompressing gas, we increase its internal energy); if thevolume is increased, W is negative (the gas decreases

The work done by an external force on a gas enclosed within a cylinder fitted with a piston:

W = (PA) dx = P (Adx ) = - PdV

∆∆∆∆x

A – the piston area

force

volume is increased, W is negative (the gas decreasesits internal energy by doing some work on theenvironment).

∫−=−2

1

),(21

V

VdVVTPW

P

W = - PdV - applies to any shape of system boundary

The work is not necessarily associated with the volume changes – e.g.,in the Joule’s experiments on determining the “mechanical equivalent ofheat”, the system (water) was heated by stirring.

dU = Q – PdV

W and Q are not State Functions

∫−=−2

1

),(21

V

VdVVTPW

- we can bring the system from state 1 tostate 2 along infinite # of paths, and for eachpath P(T,V) will be different.

U is a state function, W - is not ⇒∆∆∆∆U = Q + W

Since the work done on a system depends notonly on the initial and final states, but also on theintermediate states, it is not a state function.

P

V T

1

2

P

V

P2

P1

V1 V2

A B

CD- the work is negative for the “clockwise” cycle; ifthe cyclic process were carried out in the reverseorder (counterclockwise), the net work done onthe gas would be positive.

( ) ( )( )( ) 01212

211122

<−−−=−−−−=+=

VVPP

VVPVVPWWW CDABnet

U is a state function, W - is not ⇒thus, Q is not a state function either .

∆∆∆∆U = Q + W

PV diagram

Comment on State Functions

U, P, T, and V are the state functions, Q andW are not.

Specifying an initial and final states of asystem does not fix the values of Q and W,we need to know the whole process (thewe need to know the whole process (theintermediate states). The initial and finalpositions do not determine the work, theentire path must be specified.

The Enthalpy

Isobaric processes (P = const ):

∆U = Q - P∆∆∆∆V = Q -∆∆∆∆(PV) ⇒⇒⇒⇒ Q = ∆∆∆∆ U + ∆∆∆∆(PV)

The enthalpy is a state function , because U, P,and V are state functions. In isobaric processes,the energy received by a system by heating equals

H ≡≡≡≡ U + PV - the enthalpy⇒⇒⇒⇒

the energy received by a system by heating equalsto the change in enthalpy.

Q = ∆∆∆∆ H

isochoric:

isobaric:

in both cases, Qdoes notdepend on thepath from 1 to 2.

Consequence : the energy released (absorbed) in chemical reactions at constant volume (pressure) depends only on the initial and final states of a system.

The enthalpy of an ideal gas:(depends on T only)

TNkf

TNkTNkf

PVUH BBB

+=+=+= 122

Q = ∆∆∆∆ U

Heat Capacity

T

QC

∆≡ δThe heat capacity of a system - the amount of energy

transfer due to heating required to produce a unittemperature rise in that system

C is NOT a state function (since Q is not a state function) – it depends on the path

T

T1+dTf1 f2 f3

state function) – it depends on the path between two states of a system ⇒

V

T1 i

The specific heat capacitym

Cc ≡

( isothermic – C = ∞, adiabatic – C = 0 )

CV and CP

dT

PdVdU

dT

QC

+== δ VV T

UC

∂∂= the heat capacity at

constant volume

the heat capacity at constant pressure

PP T

HC

∂∂=

Heat capacities at ...

• At constant volume • At constant pressure

∫=∆

∂∂≡

2

)(T

VV

VconstdTCU

T

UC

∫=∆

∂∂≡

2

)(T

PP

PconstdTCH

T

HC

– CV is a state function and is independent of the process.

– CP is a state function and is independent of the process.

∫=∆ 2

1

)(T

T V VconstdTCU ∫=∆ 2

1

)(T

T P PconstdTCH

Calculate ∆U and ∆H for 1 kg of water when it is vaporized at the constant temperature of 100 °C and the constant pressure of 101.33 kPa. The specific volumes of liquid and vapor water at these conditions are 0.00104 and 1.673 m3/kg. For this change, heat in the amount of 2256.9 kJ is added to the water.

Imagine the fluid contained in a cylinder by a frictionless piston which exerts a constant pressure of 101.33 kPa. As heat is added, the water expands from its initial to its final volume. For the 1-kg the water expands from its initial to its final volume. For the 1-kg system:

kJQH 9.2256==∆

kJ

kJkJ

mkPakJ

VPHU

5.2087

4.1699.2256

)001.0673.1(33.1019.2256

)(3

=−=

−×−=∆−∆=∆

Air at 1 bar and 298.15K is compressed to 5 bar and 298.15K by two different mechanically reversible processes: (1) cooling at constant pressure followed by heating at constant volume; (2) heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and ∆U and ∆H of the air for each path. Information: the following heat capacities for air may be assumed independent of temperature: CV = 20.78 and CP = 29.10 J/mol.K. Assuming for air that PV/T is a constant, regardless of the changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m3/mol.

The final volume: 3

2

112 004958.0 m

P

PVV ==

(1) The temperature of the air at the end of the cooling step: KV

VTT 63.59

1

21 ==′

JTCHQ P 6941−=∆=∆= JVPHU 4958−=∆−∆=∆

During the second step: JTCQU V 4958=∆==∆The complete process: JQ 198349586941 −=+−= 0=∆U JQUW 1983=−∆=

0)( =∆+∆=∆ PVUH

(2) The temperature of the air at the end of the heating step: KP

PTT 75.1490

1

21 ==′

JTCUQ V 24788=∆=∆=

During the second step: JTCHQ P 34703−=∆=∆=

The complete process: JQ 99153470324788 −=−= 0=∆U JQUW 9915=−∆=0)( =∆+∆=∆ PVUH

JVPHU 24788−=∆−∆=∆

Calculate the internal-energy and enthalpy changes that occur when air is changed from an initial state of 40°F and 10 atm, where its molar volume is 36.49 ft3/lb-mole, to a final state of 140°F and 1 atm.Assume for air that PV/T is constant and that CV = 5 and CP = 7 Btu/lb-mol.F.

(1) cooled at constant volume to the final pressure;(2) heated at constant pressure to the final temperature.

Independent of paths! Two-step process:

RT 67.49967.459401 =+= RT 67.59967.4591402 =+=

Constant volume

RP

PTT 97.49

2

11 ==′

BtuTCU V 5.2248−=∆=∆

BtuPVUH 6.3141)( −=∆+∆=∆

Constant pressure

BtuVPHU 2.2756=∆−∆=∆

BtuTCH P 9.3847=∆=∆

3

12

2112 93.437 ft

TP

TPVV ==

Intermediate stateBtuHHH 7.50721 =∆+∆=∆BtuUUU 3.70621 =∆+∆=∆

Hukum Pertama (Neraca Energi) pada Proses Alir keadaan tunak (Sistem terbuka)

( ) sWQgzuPVUm &&& +=

+++∆ 2

2

1

WQgzuH s

&&

+=

++∆ 21

mmgzuH

&&+=

++∆2

sWQzgu

H +=∆+∆+∆2

2

scc

WQzg

g

g

uH +=∆+∆+∆

2

2 Jika dengan satuan Inggris

Air at 1 bar and 25 °C enters a compressor at low velocity, discharge at 3 bar, and enters a nozzle in which it expands to a final velocity of 600 m/s at the initial conditions of pressure and temperature. If the work of compression is 240 kJ/kg of air, how much heat must be removed during the compression?

Steady state flow processReturn to initial pressure and temperature, no enthalpy changeNo potential energy change

sWQzgu

H +=∆+∆+∆2

2

sWQu +=2

22

( ) kgkJkgJQ /60/1060102406002

1 332 −=×−=×−=

Initial kinetic energy is negligible

Heat must be removed in the amount of 60 kJ for each kilogram of air compressed.

Water at 200 °F is pumped from a storage tank at the rate of 50 gal/min. The motor for the pump supplies work at the rate of 2 (hp). The water goes through a heat exchanger, giving up heat at the rate of 40000 Btu/min. and is delivered to a second storage tank at an elevation 50 ft above the first tank. What is the temperature of the water delivered to the second tank?

Steady state flow processKinetic energy is negligible

scc

WQzg

g

g

uH +=∆+∆+∆

2

2

mc

s lbBtuzg

gWQH /35.99−=∆−+=∆

mlbBtuHHH /74.6809.16835.9912 =+−=+∆=

From steam table, H1 = 168.09 Btu/lbm

From steam table

Ft o74.100=

Problem

Imagine that an ideal monatomic gas is taken from its initial state A to stateB by an isothermal process, from B to C by an isobaric process, and fromC back to its initial state A by an isochoric process. Fill in the signs of Q,W, and ∆∆∆∆U for each step.

P, 105 Pa

AStep Q W ∆∆∆∆U

→→→→2

+ -- 0

V, m3

BC

A →→→→ B

B →→→→ C

C →→→→ A

2

1

1 2

+ -- 0

-- + --

+ 0 +

T=const