Lect# 7: Thermodynamics and Entropy

• Reading: Zumdahl 10.2, 10.3

• Outline:Isothermal processes (∆T = 0)Isothermal gas expansion and work(w)

Reversible and irreversible processes

Isothermal Processes

• Recall: Isothermal means T = 0.

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

• Since E = q + w, then q = -w (isothermal process)

Example: Isothermal Expansion

Consider a mass (M) connected to a ideal gas

confined by a piston. The piston is submerged

in a constant T bath, so T = 0.

Initially, V = V1

P = P1

Pressure of gas is equal to that created by the pull of the hanging mass:

P1 = force/area = M1g/A

A = piston areag = gravitational

acceleration

(9.8 m/s2)

Note: kg m-1 s-2 = 1 Pa

One-Step Expansion: If we change the weight to M1/4, then the pressure becomes

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

The mass will be lifted until the internal pressure

equals the external pressure, at which point

Vfinal = 4V1

What work is done in this expansion?

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

What if we expand in two steps?

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

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

Note: wtotal,2 step > wtotal,1 stepMore work was done in the two-step

expansion

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 decrease the weight very slightly (∆M) between an infinite number of (reversible) expansions.

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

€

w = − PexdVVinitial

V final

∫ = − PdVVinitial

V final

∫ = −nRT

VdV

Vinitial

V final

∫

Graphically, see how much more work is done in the infinite-step expansion (red area)

Two Step

Reversible

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 CompressionNow we will do the opposite, compress

the fully expanded gas:Vinit = 4V1

Pinit = P1/4

Compress in two steps: first put on mass = M1/2, Then, in step two, replace mass M1/2 with a bigger mass M1

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.

Since isothermal (T = 0) then, E = 0

but, q = -w ≠ 0

Entropy: A Thermodynamic Definition

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

2: Const V cooling3: Isothermal compression

4: Const V heatingVolumeV1 V2

1

2

3

4

Step 1: Isothermal Expansion

at T = Thigh from V1 to V2

T = 0, so E = 0 and q = -w

Do this expansion reversibly,

so that

VolumeV1 V2

1

2

3

4

€

q1 = −w1 = nRThigh lnV2

V1

⎛

⎝ ⎜

⎞

⎠ ⎟

Step 2: Const V cooling to T = Tlow.

V = 0, therefore, w = 0

q2 = E = nCvT = nCv(Tlow-Thigh)

VolumeV1 V2

1

2

3

4

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

Since T = 0, then E = 0 and q = -w

Do compression reversibly, then

VolumeV1 V2

1

2

3

4

€

q3 = −w3 = nRTlow lnV1

V2

⎛

⎝ ⎜

⎞

⎠ ⎟

Step 4: Const-V heating to T = Thigh.

V = 0, so, w = 0, and

q4 = E = nCvT = nCv(Thigh-Tlow) = -q2

VolumeV1 V2

1

2

3

4

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

⎛

⎝ ⎜

⎞

⎠ ⎟

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: summary

€

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: simple example

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

gas @constant volume 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

⎛

⎝ ⎜

⎞

⎠ ⎟