Lecture 7: Thermo and Entropy

• Reading: Zumdahl 10.2, 10.3

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

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)kg m-1 s-2 = 1 Pa

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

• 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

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.

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.)

• Graphically:

Two Step Reversible

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

• Now T = 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.

• Now V = 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.

• Now T = 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.

• Now V = 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 Lecture 6

€

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

⎛

⎝ ⎜

⎞

⎠ ⎟