$25.00 Heat and Thermodynamics 2
- From Physics: General-Physics
- Closed, but you can still post tutorials
- Due on Mar. 03, 2012
- Asked on Feb. 29, 2012 at 09:24:21AM
For this SLP, you will perform a simulation investigating three special processes - isobaric (constant pressure), isochoric (constant volume), and isothermal (constant temperature). - that can be derived from the ideal gas law. The simulation is found at Background Info 1. The simulation allows you to vary the pressure for an isobaric process, the volume for an isochoric process, and the temperature for an isothermal process. The simulation shows how the system goes from one equilibrium state to another and the work that is done on or by the system, and how heat flows into or out of the system.
The variables are:
V1 = initial volume, dm3 (dm = decimeter = 10 cm)
V2 = final volume, dm3
P1 = initial pressure, kPa (absolute; that is, measured against vacuum)
P2 = final pressure, kPa
T1 = initial temperature, K
T2 = final temperature, K
Heat flow: into system (red arrow points right)
out of system (red arrow points left)
Work done: on system (blue arrow points up)
by system (blue arrow points down)
Experiment 1:
Isobaric (constant pressure) Process
P1 = P2 = 100 kPa
Fill in the blanks in the following table.
|
|
|
|
|
|
|
|
|
|
|
V1 |
T1 |
V2 |
T2 |
Heat |
Work |
|
|
|
1.00 |
300 |
0.5 |
150 |
out |
on |
|
|
|
1.00 |
300 |
1.25 |
|
|
|
|
|
|
1.00 |
300 |
2.00 |
|
|
|
|
|
|
1.00 |
300 |
|
200 |
|
|
|
|
|
1.00 |
300 |
|
350 |
|
|
|
|
|
1.00 |
300 |
|
400 |
|
|
|
|
|
|
|
|
|
|
|
|
Show how these results illustrate the Ideal Gas Law for the special case of constant pressure. (Hint: Show that V1/T1 = V2/T2 for all the test conditions.)
__________
Experiment 2:
Isochoric (constant volume) process
V1 = V2 = 1.00 dm3
Fill in the blanks in the following table.
|
|
|
|
|
|
|
|
|
|
|
P1 |
T1 |
P2 |
T2 |
Heat |
Work |
|
|
|
100 |
300 |
66.7 |
200 |
|
|
|
|
|
100 |
300 |
|
400 |
|
|
|
|
|
100 |
300 |
|
500 |
|
|
|
|
|
100 |
300 |
75 |
|
|
|
|
|
|
100 |
300 |
150 |
|
|
|
|
|
|
100 |
300 |
200 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Following a procedure similar to the one in Experiment 1, show how these results illustrate the Ideal Gas Law for the special case of constant volume.
__________
Experiment 3:
Isothermal (constant temperature) process
T1 = T2 = 300 K
Fill in the blanks in the following table:
|
|
|
|
|
|
|
|
|
|
|
P1 |
V1 |
P2 |
V2 |
Heat |
Work |
|
|
|
100 |
1.00 |
75 |
|
|
|
|
|
|
100 |
1.00 |
150 |
|
|
|
|
|
|
100 |
1.00 |
200 |
|
|
|
|
|
|
100 |
1.00 |
|
0.75 |
|
|
|
|
|
100 |
1.00 |
|
1.25 |
|
|
|
|
|
100 |
1.00 |
|
2.00 |
|
|
|
|
|
|
|
|
|
|
|
|
Following a procedure similar to the one in Experiment 1, show how these results illustrate the Ideal Gas Law for the special case of constant temperature.
The isothermal case deserves a special mention, because at first glance it seems to be just plain wrong. Everybody knows that when you squeeze a quantity of gas, the volume goes down and the pressure goes up; but the temperature ALSO goes up. Ask any diesel mechanic! So what's going on here?
Answer: Either the volume is decreased (pressure increased) so slowly that the heat can escape, and the temperature remains a constant, OR the system is compressed to its new, smaller volume, and allowed to cool to its original temperature, before the new pressure is measured. Which is definitely a lab experiment, and not an industrial process.
- This tutorial was purchased 1 time and rated No Rating by students like you.
- Posted on Feb. 29, 2012 at 09:27:59AM
Experiment.docx (32K) - This tutorial hasn't been purchased yet.
- Posted on Feb. 29, 2012 at 09:33:22AM
- This tutorial hasn't been purchased yet.
- Posted on Feb. 29, 2012 at 09:36:22AM
- This tutorial hasn't been purchased yet.
- Posted on Feb. 29, 2012 at 09:45:29AM