Announcements 1/24/11 Prayer Substitute today: Dr. Stokes Today’s topics: a. a.PV diagrams b. b.Work c. c.Isothermal contours d. d.Internal energy e. e.First.

Announcements 1/24/11

Prayer Substitute today: Dr. Stokes Today’s topics:

a. PV diagramsb. Workc. Isothermal contoursd. Internal energye. First Law of Thermodynamicsf. Some specific state changes

Reading quiz (graded)

Which of the following is NOT true of the work done on a gas as it goes from one point on a PV diagram to another?

a. It cannot be calculated without knowing n and T.

b. It depends on the path taken.c. It equals minus the integral under the

curve.d. It has units of Joules.e. It is one of the terms in the First Law of

Thermodynamics.

Work done by an expanding gas

1 m3 of an ideal gas at 300 K supports a weight in a piston such that the pressure in the gas is 200,000 Pa (about 2 atm). The gas is heated up. It expands to 3 m3.

Plot the change on a graph of pressure vs. volume (a P-V diagram)

How much work did the gas do as it expanded?

a. How do you know it did work?

W F distance

P Area distance

P V

= 400,000 J

More on Work…

PV diagrams What if pressure

doesn’t stay constant?

Work done on gas vs work done by gas

on gasW PdV

Thought question (ungraded)

A gas in a piston expands from point A to point B on the P-V plot, via either path 1 or path 2. Path 2 is a “combo path,” going down first, then over. The gas does the most work in:

a. path 1b. path 2c. same work

Quick Writing

Describe with words how you could actually make a gas (in some sort of container) change as in path 2.

Internal Energy, Eint (aka U)

Eint = Sum of all of the microscopic kinetic energies. (Also frequently called “U”.)

Return to Equipartition Theorem:a. “The total kinetic energy of a system is shared

equally among all of its independent parts, on the average, once the system has reached thermal equilibrium.”

b. Each “degree of freedom” of a molecule has kinetic energy of kBT/2

c. Monatomic molecules 3 d.o.f. d. At room temperatures, diatomic 5 d.o.f.

(3 translational, 2 rotational)

Internal Energy

Monatomic: Eint = N 3 kBT/2

= (3/2)nRT

Diatomic: (around room temperature) Eint = N 5 kBT/2

= (5/2)nRT

32intE nR T

52intE nR T

Thought question (ungraded)

The process in which Eint is the greatest (magnitude) is:

a. path 1b. path 2c. neither; it’s the

same

Isothermal Contours

A gas changes its volume and pressure simultaneously to keep the temperature constant the whole time as it expands to twice the initial volume. What does this look like on a PV diagram?

What if the temperature is higher? Lower?

PV nRT xy constant

“First Law”

Eint = Qadded + Won system

What does that mean? You can add internal energy, by…

a. …adding heatb. …compressing the gas

Possibly more intuitive version:Qadded = Eint + Wby system

When you add heat, it can either…increase internal energy (temperature)…be used to do work (expand the gas)

Three Specific Cases

Constant pressure, “isobaric”a. Work on = ?

Constant volume, “isovolumetric”a. Work on = ?

Constant temperature, “isothermal”a. Work on = ?

0

–PV

nRT dVPdV dV nRT

V V

2 1lnnRT V V

Worked Problems (as time permits)

For each problem, draw the process on a P-V diagram, state what happens to the temperature (by visualizing contours), and calculate how much heat is added/removed from gas via the First Law.

a. A monatomic gas (1.3 moles, 300K) expands from 0.1 m3 to 0.2 m3 in a constant pressure process.

b. A diatomic gas (0.5 moles, 300K) has its pressure increased from 100,000 Pa to 200,000 Pa in a constant volume process.

c. A diatomic gas (0.7 moles, 300K) gets compressed from 0.4 m3 to 0.2 m3 in a constant temperature process.

T increases, Q = Eint + PV = 8102 J added

T increases, Q = Eint = 3116 J added

T stays constant, Q = –Won gas = –1210 J (i.e., 1210 J of heat removed from gas)