First Law of Thermodynamics

Michael Moats

Met Eng 3620

Chapter 2

Thermodynamic Functions

• To adequately describe the energy state of a system two terms are introduced– Internal Energy, U– Enthalpy, H

• Internal Energy related to the energy of a body.– Example: Potential Energy

• Body has a potential energy related to its mass and height (mgh)

• To move a body from one height to another takes work, w.

w = force * distance = mg * (h2-h1) = mgh2 – mgh1 = Potential Energy at height 2 – Potential Energy at Point 1

Work and Heat

• A system interacts with its environment through work, w, or heat, q.w = -(UB – UA) (if no heat is involved)

• If w < 0, work is done to the system• If w > 0, work is done by the system

q = (UB – UA) (if no work is involved)• If q < 0, work flows out of the body (exothermic)• If q > 0, heat flows into the body (endothermic)

Change in Internal EnergywqU

wqdU U is a state function. Therefore, the path between condition 1 and condition 2 do not affect the differential value. Heat and work are not state functions. The path taken between condition 1 and 2 does matter, hence the use of a partial differential.

Pre

ssur

e

Volume

P1

P2

V2V1

ab

c

1

2

PdVw

Fixing Internal Energy

• For a simple system consisting of a given amount of substance of a fixed composition, U is fixed once any two properties (independent variables) are fixed.

),( TVUU

dTT

UdV

V

UdU vT )()(

Convenient Processes

• Isochore or Isometric - Constant Volume

• Isobaric – Constant Pressure

• Isothermal – Constant Temperature

• Adiabatic – q = 0

• If during a process the system maintains a constant volume, then no work is performed.

Recalland

Thus

where the subscript v means constant volume

• Hence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system.

Constant Volume Processes

PdVwwqdU

vqdU

Constant Pressure Processes

• Again starting with the first law and the definition of work:

• Combining them and integrating gives

where the subscript p means constant pressure

• Solving for qp and rearranging a little gives

)( 1212 VVPqUU p

PdVwwqdU

)()( 1122 PVUPVUq p

Enthalpy

• Since U, P and V are all state function, the expression U+PV is also a state function.

• This state function is termed enthalpy, H

H = U + PV

• Therefore qp = H2 – H1 = ΔH

• In a constant pressure system, the heat absorbed or withdrawn is equal to the change in enthalpy.

Heat Capacity

• Before discussing isothermal or adiabatic processes, a new term is needed to make the calculations easier.

• Heat Capacity, C is equal to the ratio of the heat absorbed or withdrawn from the system to the resultant change in temperature.

• Note: This is only true when phase change does not occur.

T

qC

Defining Thermal Process Path

• To state that the system has changed temperature is not enough to define change in the internal energy.

• It is most convenient to combine change in temperature while holding volume or pressure constant.

• Then calculations can be made as to the work performed and/or heat generated.

Thermal Process at Constant V

• Define heat capacity at a constant volume

• Recalling that at a constant volume

• Leading to

vqdU

vv dT

qC )(

vv dT

dUC )( dTCdU vor

Thermal Process at Constant P

• Define heat capacity at a constant pressure

• Recalling that at a constant pressure

• Leading to

pqdH

pp dT

qC )(

orpp dT

dHC )( dTCdH p

Molar Heat Capacity

• Heat capacity is an extensive property (e.g. dependent on size of system)

• Useful to define molar heat capacity

vv Cnc pp Cnc

where n is the number of moles and cv and cp are the molar heat capacity at constant volume and pressure, respectively.

Molar Heat Capacity - II

• cp > cv

• cv – heat only needed to raise temperature

• cp – heat needed to raise temperature and perform work

• Therefore the difference between cv and cp is the work performed.

pvp T

VPcc )(

Long derivation and further explanation in section 2.6Rcc vp

Reversible Adiabatic Processes

• Adiabatic; q = 0

• Reversible;

• First law;

• Substitution gives us;

• For one mole of ideal gas;

• Recall that

• Leading to

PdVwwqdU

PdVdTCv

V

RTdVdTcv

xx

dxln

)ln()ln(2

1

1

2

V

VR

T

Tcv

Reversible Adiabatic – cont.

• Rearranging gives

• Combining exponents and recalling that cp-cv=R gives

• Defining a term,

• From the ideal gas law

• Finally

RC

V

V

T

Tv )()(

2

1

1

2

v

vp

c

cc

V

V

T

T)(

2

1

1

2 )(

1

2

1

1

2 )(

V

V

T

Tvp cc / gives

)(11

22

1

2

VP

VP

T

T

1122 VPVP constant

Reversible Isothermal P or V Change

• Recall• Isothermal means dT = 0, so dU = 0• Rearranging first law• Substituting reversible work

ideal gas law gives• Integration leaves• Isothermal process occurs at constant

internal energy and work done = heat absorbed.

dTCdU v

wq PdVw and

VRTdvPdVwq /)ln(

1

2

V

VRTwq

Example Calculation

• 10 liters of monatomic ideal gas at 25oC and 10 atm are expanded to 1 atm. The cv = 3/2R. Calculate work done, heat absorbed and the change in internal energy and enthalpy for both a reversible isothermal process and an adiabatic and reversible process.

First Determine Size of System

• Using the ideal gas law

molesK

moleK

atmliterlatm

RT

PVn 09.4

298**

*08206.0

10*10

Isothermal Reversible Process

• Isothermal process; dT = 0, dU = 0

• To calculate work, first we need to know the final volume.

• Then we integrate

litersatm

litersatm

P

VPV 100

1

10*10)(

2

112

PdVw

kJliters

litersK

Kmole

Jmolesw

V

dVnRTPdVw

3.23)10

100ln(*298*

*3144.8*09.4

2

1

2

1

Isothermal Reversible Process –continued

• Since dU = 0, q = w = 23.3 kJ

• Recall definition of enthalpy

H = U + PV

0)(

)(

1212

11221122

TTnRnRTnRTH

VPVPVPVPUH

Isothermal = constant temperature

Reversible Adiabatic Expansion• Adiabatic means q = 0

• Recall

• Since cv = 3/2R, then cp = 5/2R and

• Solve for V3

• Solve for T3

1133 VPVP constant

vp cc / Rcc vp

3/5

litersP

VPV 8.39]

1

10*10[)( 5/3

3/55/3

3

3/511

3

K

Kmole

atmlitermoles

litersatm

nR

VPT 119]

*

*08206.0*09.4

8.39*1[)( 33

3

Reversible Adiabatic Expansion - continued

kJKmoleK

Jmoles

TTncdTncwU vv

13.9)298119(**

3144.8*5.1*09.4

)( 13

3

1

Text shows five examples of path does not matter in determining U.

kJatmliter

JlitersatmlitersatmJ

VPVPUH

2.15*

3.101*)]10*10()8.39*1[(9130

)( 1133