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Lect# 7: Thermodynamics and Entropy

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- Reading: Zumdahl 10.2, 10.3
- Outline:
Isothermal processes (∆T = 0)

Isothermal gas expansion and work(w)

Reversible and irreversible processes

- Recall: Isothermal means DT = 0.
- Since DE = nCvDT, then DE = 0 for an isothermal process.
- Since DE = q + w, then
q = -w (isothermal process)

Consider a mass (M) connected to a ideal gas

confined by a piston. The piston is submerged

in a constant T bath, so DT = 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 area

g = 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 = -PextDV = -P1/4 (4V1 - V1) = -3/4 P1V1

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 = -PextDV = -P1/2 (2V1 - V1) = -1/2 P1V1

In Step 2 (M1/2 to M1/4 ):

Pext = P1/4, Vfinal = 4V1

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

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

Note: wtotal,2 step > wtotal,1 step

More 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.

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:

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

Reversible

Two Step

If we perform the integration from

V1 to V2:

Now 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 = -PextDV = -P1/2 (2V1 - 4V1)

= P1V1

In second step:

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

= P1V1

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

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 (DT = 0)

then, DE = 0

but, q = -w ≠ 0

Let’s consider the four- step cycle illustrated:

1: Isothermal expansion

2: Const V cooling

3: Isothermal compression

4: Const V heating

Step 1: Isothermal Expansion

at T = Thigh from V1 to V2

DT = 0, so DE = 0 and q = -w

Do this expansion reversibly,

so that

Step 2: Const V cooling to

T = Tlow.

DV = 0, therefore, w = 0

q2 = DE = nCvDT

= nCv(Tlow-Thigh)

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

Since DT = 0, then DE = 0

and q = -w

Do compression reversibly, then

Step 4: Const-V heating to

T = Thigh.

DV = 0, so, w = 0, and

q4 = DE = nCvDT

= nCv(Thigh-Tlow) = -q2

The thermodynamic definition of entropy

(finally!)

DT = 0

DV = 0

DP = 0

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

gas @constant volume from 298 K to 350 K?

3/2R

Connecting with Lecture 6

• From this lecture:

• Exactly the same as derived in the previous lecture!