Just do first row for chart and show me how to get the answer. For this SLP, you
ID: 2041443 • Letter: J
Question
Just do first row for chart and show me how to get the answer.
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.
Fendt, W. (2003). Special processes of an ideal gas (simulation). Retrieved on March 1, 2008, from http://www.walter-fendt.de:80/ph14e/gaslaw.htm
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.
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
Explanation / Answer
YOUR ANSWER HERE http://ww2.justanswer.com/uploads/akch2002/2011-01-27_095004_thermodynamics_experiment.doc
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.