Chapter 4: Applications of the First Law

Different types of work:

• Configuration work: (reversible process)• Dissipative work: (irreversible process)• Adiabatic work: (independent of path)• Work: not a system property, i.e. not a state

variable• Conventional sign of work: on (-) or by (+) the

system• Expansivity and isothermal compressibility

4.2 Mayer’s equation

• Heat capacity: limiting ratio of …… ( not an exact differential !!)

• Heat capacity depends on the conditions at which heat transfer takes place

• Specific heat capacity cv and cp

• For the ideal gas system: cp – cv = R

• Calculate the internal energy of an ideal gas system based on the definition of cv

• If cv is independent of temperature, then

u = u0 + cv(T – T0)

dTcuu

dTcdudT

duc

T

T

uc

T

T

v

vv

v

vv

0

0

so

on dependent

only isenergy internal thesystem gas idealan for since

4.3 Enthalpy and Heats of Transformation

• The heat of transformation is the heat transfer accompanying a phase change.

• Phase change is an isothermal and isobaric process.

• Phase change only entails a change of volume.

thus the work (only configuration work) equals:

w = P(v2 – v1)

• Variation in the internal energy is expressed as:

du = dq - Pdv

for a finite change: u2 – u1 = l – P(v2-v1)

where l represents the latent heat of transformation.

thus l = (u2 + Pv2) – (u1 + Pv1)

At this point, introducing h = u + Pv (the small h denotes the specific enthalpy)

• Therefore, the latent heat of transformation is equal to the difference in enthalpies of the two phases.

• Conventional notation: 1 denotes a solid, 2 a liquid and 3 a vapor, i.e. h’ represents the enthalpy of solid, h’’ is the enthalpy of liquid, ….

• l12 = h’’ – h’ represents solid to liquid transformation (fusion).

l23 = h’’’ – h’’ represents liquid to vapor transformation (evaporation).

• Enthalpy is a state function, i.e. integration around a closed cycle produces 0!! (see Fig. 4.2)

,0321 hhh

,0122313 lll .122313 lll

4.4 Relationships involving enthalpy

• The natural choice in the variable h is

h = h(T, P)

• The analysis can be proceeded in the same way as to the internal energy u

.Pddudq

dTT

udu )(

.)(

du

T

.)( dPu

dTT

udq

T

• As

Thus:

Since:

then

.puh

,dPPddudh

.dPdhdq

,dPP

hdT

T

hdq

Tp

pp dT

dqc

pp T

hc

• For an ideal gas

• Then

• Since for ideal gas h depends on T only,

0

TP

h

,vdPdTcdq p

)(

)( :constant is Provided

00

00

00

TTchh

TTchhc

dTchh

dT

dh

T

hc

p

pp

T

T p

pp

4.5 Comparison of u And h

• See Table 4.2 in the textbook

4.6 Work done in an adiabatic process

• In adiabatic process: dq = 0.

• The equation dq = cpdT – vdP

can be rearranged into

vdP = cpdT

• Similarly, one gets Pdv = -cvdT

• Dividing the above two equation:

v

dV

c

c

P

dPthus

c

c

PdV

vdP

p

p

• Assuming

• The integration of the above equation leads to

where K is a constant

• Similarly, one gets

c

c p

,KP

constant1

TP

constant 1 T

• The work done in the adiabatic process is

• For a reversible adiabatic process:

w = u1 – u2 = cv(T1 – T2)

][1

11122 vPvPw

Pdvw

Example: An ideal monatomic gas is enclosed in an insulated chamber with a movable piston. The initial values of the state variables are P1 = 8atm, V1 = 4 m3 and T1 = 400K. The final pressure after the expansion is P2 = 1 atm. Calculate V2, T2, W and ∆U.

• Solution:For an ideal monatomic gas, the ratio of specific heats γ = 5/3

since P1V1γ = P2V2

γ V2 = V1(P1/P2)1/γ

thus V2 = 13.9m3

According to ideal gas law: PV = nRT

T2 can be easily calculated as 174K

The work done by the system is

w = 2.74 x 106 J

For the adiabatic process: ∆U is equal to the work done on the system

thus is -2.74 x 106 J

][1

11122 vPvPw

Chapter 5: Consequences of the First Law

5.1 The Gay-Lussac-Joule Experiments