Lecture 6: Thermo and Entropy

• Reading: Zumdahl 10.2, 10.3

• Outline– Isothermal processes– Isothermal gas expansion and work– Reversible Processes

Weight and Entropy• The connection between weight () and entropy (S)

is given by Boltzmann’s Formula:

S = k ln

k = Boltzmann’s constant = R/Na

= 1.38 x 10-23 J/K

• The dominant configuration will have the largest ; therefore, S is greatest for this configuration

Example: Crystal of CO

• For a mole of CO:

= Na!/(Na/2!)2

= 2Na

• Then, S = k ln() = k ln (2Na)

= Nak ln(2) = R ln(2) = 5.64 J/mol.K

Another Example: Expansion

• What is S for the expansion of an ideal gas from V1 to 2V1?

• Focus on an individual particle. After expansion, each particle will have twice the number of positions available.

Expansion (cont.)

• Note in the previous example that weight was directly proportional to volume.

• Generalizing:

S = k ln (final) - kln(initial)

= k ln(final/initial)

= k ln(final/initial)

= Nk ln(final/initial) for N molec.= Nkln(Vfinal/Vinitial)

Isothermal Processes

• Recall: Isothermal means T = 0.

• Since E = nCvT, then E = 0 for an isothermal process.

• Since E = q + w:

q = -w (isothermal process)

Example: Isothermal Expansion• Consider a mass connected to a ideal gas

contained in a “piston”. Piston is submerged in a constant T bath such that T = 0.

Isothermal Expansion (cont.)

• Initially, V = V1

P = P1

• Pressure of gas is equal to that created by mass:

P1 = force/area = M1g/A

where A = piston area

g = gravitational acceleration (9.8 m/s2)

Isothermal Expansion (cont.)

• One-Step Expansion. We change the weight to M1/4, then

Pext = (M1/4)g/A = P1/4

• The mass will be lifted until the internal pressure equals the external pressure. In this case

Vfinal = 4V1

• w = -PextV = -P1/4 (4V1 - V1) = -3/4 P1V1

Two Step Expansion (cont.)• Graphically, we can envision this two-step process

on a PV diagram:

• Work is given by the area under the “PV” curve.

Two Step Expansion

• In this expansion we go in two steps:

Step 1: M1 to M1/2

Step 2: M1/2 to M1/4• In first step:

Pext = P1/2, Vfinal = 2V1

• w1 = -PextV = -P1/2 (2V1 - V1) = -1/2 P1V1

Two Step Expansion (cont.)

• In Step 2 (M1/2 to M1/4 ):

Pext = P1/4, Vfinal = 4V1

• w2 = -PextV =- P1/4 (4V1 - 2V1) = -1/2 P1V1

• wtotal = w1 + w2 = -P1V1/2 - P1V1/2 = -P1V1

• wtotal,2 step > wtotal,1 step

Infinite Step Expansion (cont.)

• Graphically:

Two Step Reversible

Infinite Step Expansion

• Imagine that we perform a process in which we change the weight “infinitesimally” between expansions.

• Instead of determining the sum of work performed at each step to get wtotal, we integrate:

w PexdVVinitial

V final

PdVVinitial

V final

nRT

VdV

Vinitial

V final

Infinite Step Expansion (cont.)

• If we perform the integration from V1 to V2:

wtotal PdVV1

V2

nRT

VdV

V1

V2

nRT(ln(V )) |V1

V2 nRT lnV2

V1

wrev nRT lnV2

V1

qrev

Two Step Compression• Now we will do the opposite….take the gas and

compress:

Vinit = 4V1

Pinit = P1/4

• Compression in two steps:first place on mass = M1/2second, replace mass with one = M1

Two Step Compression (cont.)• In first step:

w1 = -PextV = -P1/2 (2V1 - 4V1)

= P1V1

• wtotal = w1 + w2 = 2P1V1 (see table 10.3)

• In second step:

w2 = -PextV = -P1 (V1 - 2V1)

= P1V1

Compression/Expansion

• In two step example:

wexpan. = -P1V1

wcomp. = 2P1V1

wtotal = P1V1

qtotal = -P1V1

• We have undergone a “cycle” where the system returns to the starting state.

• Now, E = 0 (state fxn)

• But, q = -w ≠ 0

Defining Entropy

• Let’s consider the four-step cycle illustrated:– 1: Isothermal expansion

– 2: Isochoric cooling

– 3: Isothermal compression

– 4: Isochoric heating

VolumeV1 V2

1

2

3

4

Defining Entropy (cont)

• Step 1: Isothermal Expansion

at T = Thigh from V1 to V2

• NowT = 0; therefore, E = 0 and q = -w

• Do expansion reversibly. Then: Volume

V1 V2

1

2

3

4

q1 w1 nRThigh lnV2

V1

Defining Entropy (cont)

• Step 2: Isochoric Cooling to T = Tlow.

• NowV = 0; therefore, w = 0

• q2 = E = nCvT

= nCv(Tlow-Thigh) Volume

V1 V2

1

2

3

4

Defining Entropy (cont)

• Step 3: Isothermal Compression at T = Tlow from V2 to V1.

• NowT = 0; therefore, E = 0 and q = -w

• Do compression reversibly, thenVolume

V1 V2

1

2

3

4

q3 w3 nRTlow lnV1

V2

Defining Entropy (cont)

• Step 4: Isochoric Heating to T = Thigh.

• NowV = 0; therefore, w = 0

• q4 = E = nCvT

= nCv(Thigh-Tlow) = -q2 Volume

V1 V2

1

2

3

4

Defining Entropy (cont)

VolumeV1 V2

1

2

3

4

qtotal q1 q2 q3 q4

qtotal q1 q3

qtotal nRThigh lnV2

V1

nRTlow ln

V1

V2

qtotal nRThigh lnV2

V1

nRTlow ln

V2

V1

Defining Entropy (end)

VolumeV1 V2

1

2

3

4

qtotal nRThigh lnV2

V1

nRTlow ln

V2

V1

0 q1Thigh

q3

Tlow

The thermodynamic definition of entropy (finally!)

S dqrevTinitial

final

qrevT

Calculating Entropy

S dqrevTinitial

final

T = 0

S dqrevTinitial

final

nR lnV finalVinitial

V = 0

S dqrevTinitial

final

nCv lnTfinalTinitial

P = 0

S dqrevTinitial

final

nCp lnTfinalTinitial

dqrev nRTdV

V

dqrev nCvdT

dqrev nCPdT

Calculating Entropy• Example: What is S for the heating of a mole of a monatomic

gas isochorically from 298 K to 350 K?

S dqrevTinitial

final

nCv lnTfinalTinitial

3/2R

S (1mol) 32R ln 350K

298K

S 2J K

Connecting with Dr Boltzmann

S dqrevTinitial

final

qrevT

nRT

TlnV finalVinitial

nR ln

V finalVinitial

• From this lecture:

• Exactly the same as derived in the previous lecture!

S Nk lnV finalVinitial

nR ln

V finalVinitial