- #1
littlejon
- 20
- 0
Homework Statement
Homework Equations
pv = nRt
The Attempt at a Solution
The answer is zero because it does not move since no work is being done.
AI Thread Summary
The discussion revolves around solving a problem using the ideal gas law, pv = nRt, and clarifying the concept of work in thermodynamics. Participants debate whether the work done in a process described as "8000 J" is zero, with one asserting that the answer cannot be zero due to the nature of the process. The process is identified as isobaric, indicating that pressure remains constant during gas expansion. There is confusion about the interpretation of the problem and the definition of work in this context. Ultimately, the conversation emphasizes the importance of understanding thermodynamic principles in solving gas-related problems.
pv = nRt
The answer is zero because it does not move since no work is being done.
- #2
BiGyElLoWhAt
Gold Member
- 1,637
- 138
Doesn't the volume change?
- #3
mafagafo
- 188
- 12
This is pretty disturbing. Firstly, the problem says "this process is8000 J.". Secondly, no, the answer is not zero. I am quite confident about that.
- #4
littlejon
- 20
- 0
Okay since we know it not zero then if 8000 work is being done than is the problem reiterating the work that was done so A?
- #5
mafagafo
- 188
- 12
Show us your work.
From where did you get this problem (please tell me)? I don't want to buy the book by accident.
From where did you get this problem (please tell me)? I don't want to buy the book by accident.
- #6
lep11
- 380
- 7
No, it's not A. How is work defined when gas expands? In this case the process is isobaric meaning pressure is constant.littlejon said:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
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