Thermal Processes.

  • Thread starter dalitwil
  • Start date
  • #1
23
0
Ok guys, my last questions of the semester, Yay!

So I have two of them, one pertaining to a figure (attached), which has me terribly confused:

1.) In the figure (attached), if P1 = 133 kPa, V1 = 1123 cm3 and P2 = 200 kPa, V2 = 10000 cm3, what is the heat absorbed (+) or liberated (-), to the nearest joule, in Path 2? Assume the change in internal energy is 8900 J.

My Work (Reasoning):
I am assuming that W=0, simply because the path appears to be only vertical, So using ΔU=Q-W, we can say that ΔU=Q (because W=0) and ΔU is given as 8900 J, so wouldn't this be the answer? (It is not)

2.) 10 moles of a gas in thermal contact with an oil bath at temperature 300 K is compressed isothermally from a volume of 7541 cm3 to a volume of 1561 cm3. To the nearest joule what is the work done by the piston?

My Work:
W would simply equal nRT*ln(Vf/Vi)
Plugging in the knowns leaves me with -131
This is incorrect. (using the reasoning that ΔU=0)

Any ideas guys?
 

Attachments

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
3,021
2
dalitwil said:
I am assuming that W=0, simply because the path appears to be only vertical

Any ideas guys?
I just took a quick look- so this is totally off the cuff, but have you included the whole path? I think the intent is to go from point 1 to point 2, part horizontal and part vertical
 
  • #3
Andrew Mason
Science Advisor
Homework Helper
7,634
368
dalitwil said:
Ok guys, my last questions of the semester, Yay!

So I have two of them, one pertaining to a figure (attached), which has me terribly confused:

1.) In the figure (attached), if P1 = 133 kPa, V1 = 1123 cm3 and P2 = 200 kPa, V2 = 10000 cm3, what is the heat absorbed (+) or liberated (-), to the nearest joule, in Path 2? Assume the change in internal energy is 8900 J.

My Work (Reasoning):
I am assuming that W=0, simply because the path appears to be only vertical, So using ΔU=Q-W, we can say that ΔU=Q (because W=0) and ΔU is given as 8900 J, so wouldn't this be the answer? (It is not)
As Dan says, the path is from 1 to 2 so you have to look at the horizontal part. I am a little confused by the assumption that change in internal energy is 8900 J.

2.) 10 moles of a gas in thermal contact with an oil bath at temperature 300 K is compressed isothermally from a volume of 7541 cm3 to a volume of 1561 cm3. To the nearest joule what is the work done by the piston?

My Work:
W would simply equal nRT*ln(Vf/Vi)
Plugging in the knowns leaves me with -131
This is incorrect. (using the reasoning that ΔU=0)
Your approach is correct. Try the numbers again. I think you forgot to factor in the temperature.[/QUOTE]

AM
 
  • #4
23
0
The first question is extrememly confusing, because it must mean that the temperature is being held constant (and also not given) so the change in internal energy should equal 0, which it doesn't. I cannot find W, without temperature and moles. I am really stuck :frown:
 
  • #5
Andrew Mason
Science Advisor
Homework Helper
7,634
368
dalitwil said:
The first question is extrememly confusing, because it must mean that the temperature is being held constant (and also not given) so the change in internal energy should equal 0, which it doesn't. I cannot find W, without temperature and moles. I am really stuck :frown:
The temperature is not constant:

[tex]T = PV/nR[/tex]

So:[tex]T_2/T_1 = P_2V_2/P_1V_1 \ne 1[/tex]

I think you have to ignore the assumption that the change in internal energy is 8900 J. The change in internal energy is:

[tex]\Delta U = \Delta (PV) = P_2V_2 - P_1V_1 = 1850 J[/tex]

The work done by the gas from 1 to 2 is [itex] W = P_1(V_2 - V_1)[/itex]

The heat added is the sum of the work done + change in internal energy.

AM
 

Related Threads on Thermal Processes.

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
4
Views
1K
Replies
8
Views
2K
  • Last Post
Replies
2
Views
493
Replies
1
Views
6K
Top