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Announcements 9/14/12 Prayer “Real” thermodynamics (more unified, fewer disjointed topics): a. Today PV diagrams work isothermal contours internal energy First Law of Thermodynamics b. Continues for the next 4 lectures after today. Then one more lecture. Then exam! Pearl s Befor e Swine
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Announcements 9/14/12

Jan 04, 2016

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Announcements 9/14/12. Prayer “Real” thermodynamics (more unified, fewer disjointed topics): Today PV diagrams work isothermal contours internal energy First Law of Thermodynamics Continues for the next 4 lectures after today. Then one more lecture. Then exam!. Pearls Before Swine. - PowerPoint PPT Presentation
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Page 1: Announcements 9/14/12

Announcements 9/14/12 Prayer “Real” thermodynamics (more unified, fewer

disjointed topics):a. Today

– PV diagrams– work – isothermal contours– internal energy– First Law of Thermodynamics

b. Continues for the next 4 lectures after today. Then one more lecture. Then exam!

Pearls Before Swine

Page 2: Announcements 9/14/12

From warmup

Extra time on?a. 9 different answers

Other comments?a. this chapter has a lot of information and quite

confusing. I still don't understand the majority of what this chapter is about...could you simplify this chapter into simple points?

b. Does our reading include all the example problems as well as the text?

Page 3: Announcements 9/14/12

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

Page 4: Announcements 9/14/12

More on Work…

PV diagrams What if pressure

doesn’t stay constant?

Work done on gas vs work done by gas

on gasW PdV

Page 5: Announcements 9/14/12

Clicker question:

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.

Page 6: Announcements 9/14/12

Quick Writing

First: in which path would the gas (pushing against some sort of container) do the most work?

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

Page 7: Announcements 9/14/12

From warmup What is a "state variable"? In your own words,

and without referring to the text if possible, why do things like temperature, internal energy, volume, and pressure call into this category?

a. A state variable is something that helps specify the state of the entire system. They describe macroscopic quantities. State variables are often part of an "equation of state" that describes the dependence of the system's state on these variables. The given quantities fall into this category because individual molecules contribute to temperature, pressure, etc., but T and P measure the contributions from ALL molecules.

State postulate: state is fixed by two independent state variables

Page 8: Announcements 9/14/12

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)

Page 9: Announcements 9/14/12

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

Page 10: Announcements 9/14/12

Clicker question:

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

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

same

Page 11: Announcements 9/14/12

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

Page 12: Announcements 9/14/12

“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)

Page 13: Announcements 9/14/12

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

Page 14: Announcements 9/14/12

From warmup

Are Q, W, and ΔEint +, -, or 0 for the following situations?(A) Rapidly pumping up a bicycle tire (the system in question is the air in the pump)(B) Lukewarm water in a pan on a hot stove (the system in question is the water in the pan)(C) Air quickly leaking out of a balloon (the system in question is the air that was originally in the balloon)

answers: Q W deltaE(A) 0 + +(B) + 0 +(C) 0 - -

(only 3 students correct…for now…by exam ALL students should be correct!)

Page 15: Announcements 9/14/12

Worked Problems

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)

Page 16: Announcements 9/14/12

Quick Answers From Students

Eint will be positive if ______________

Qadded will be positive if ______________

Won system will be positive if ______________

Page 17: Announcements 9/14/12

From warmup Match the letters A-D to the

appropriate path. a. isovolumetric (constant volume)b. adiabatic c. isothermic (constant

temperature) isobaric (constant pressure)

Which processes are most common in typical situations (motors, heaters, calorimeters, refrigerators, leaf blowers, etc.)?

My answer: In many applications (motors) adiabatic and isovolumetric processes are common (see Otto process in Chap. 22). In many science experiments, isothermal processes are common. Isobaric processes are important when the system is open to the atmosphere, such as boiling water on the stove. I guess they are all important! :-)

Student answers:adiabatic (2x)isovol. & isothermalisotherm. & isobaricisovol & isobarisovol