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Heated Piston

A cylinder of cross-sectional area 0.0314m^2, filled with argon (a monatomic ideal gas), is sealed with a piston of mass M=25kg that is free to move up and down. The piston is initially at a height of 30cm, and the gas is at 300K. The cylinder is heated from below. After a few minutes, 171J of heat has been added to the system, and the piston has moved upward by 2cm. See figure 1. This problem has multiple ways to calculate the same quantities. This can be a useful check if you are unsure of your answer.

(a) Draw a free body diagram to show all the forces acting on the piston (include the force of the atmosphere pushing down on the piston from outside). (b) What is the initial pressure of the gas in the cylinder? Is this pressure constant? (hint: think about the forces on the piston) (c) How much work is done by the gas in moving the piston 2cm upward? (note: there are two ways to calculate this based on the information you have) (d) Find the change in internal energy (using the first law). (e) Find the change in temperature of the gas using the ideal gas law, and the number of moles of gas present, then find the change in internal energy using !U = nCV !T for an ideal diatomic gas? The numbers from both methods should agree (there may be a small discrepancy due to rounding). If not, review previous steps to find your error

My work so far:

b)

I used the following equations,
F=pA (pressure x area of piston)
F=ma for the force of gravity acting on the piston
I made the assumption that the pressure acting down on the piston was from the force of gravity, and so made these two equations equal to one another.
pA=ma
p=(ma)/A
p=[(25kg)(9.8m/s^2)]/(.0314m^2)
p=7803kg/ms^2
I know that these are the correct units for pressure, and I also felt that the pressure must be fairly large in order to support such a heavy mass.

c)W=pΔV (ΔV=change in volume of cylinder)
ΔV is unknown so I had to solve for it
The volume of a cylinder = πr^2h = area of base x height = Ah
The change in volume here is a result of a change in height of the piston
so, ΔV = AΔh
ΔV= (.0314m^2)(.002m)
ΔV= .0000628m^3
And then I plugged it into W=pΔV
W=(7803kg/ms^2)(.0000628m^3)
W=(0.49kgm^2/s^2)
W=0.49J
Again, I knew I had the right units which was good.

D) I used the equation, ΔU = Q-W, where ΔU is change in internal energy
I plugged in the given value for Q (171J) and the solved value of W (0.49J)
ΔU = 171-0.49= 170.51 J

E) I used the equation for the ideal gas law, PV=nRT
I used the values of the cylinder in it's initial state in order to solve for n.
n=PV/RT
n=[(7803)(.0314)(.032)]/[(8.314)(300)]
n=.00395 moles of gas
I definitely felt that this number was pretty low but I decided to try to find temperature anyway. T=PV/nR
T=[(7803)(.0314)(.032)]/[(.00295)(8.314)]
T=320 K

I knew this answer fit with what my estimation should be, as heat has been added to the system this final temperature should be higher than the initial temperature of 300K. However, the low value I found for moles of gas is throwing me off. I was hoping that somebody could check my work for error, or to work through it and compare the answers.

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closed as off-topic by Floris, ACuriousMind, Danu, Kyle Oman, Ali Oct 8 '14 at 22:27

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ While "check my work" questions are normally considered off topic, I commend you for doing a lot of the checking yourself, and drawing reasonable conclusions about your work ("units seem right", "magnitude feels low"), and because of this I tried to look for possible errors rather than act on the usual "vote to close" reflex that many questions like this would attract. I hope other community members feel the same way. $\endgroup$ – Floris Oct 8 '14 at 4:30
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    $\begingroup$ Thank you! I sincerely apologize, I honestly just didn't see anything about that when I did the tour, so I appreciate that you let me know. I will definitely keep that in mind for the future and post a more suitable question next time! $\endgroup$ – Mary Olson Oct 8 '14 at 16:44
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    $\begingroup$ Welcome to the site! No need to apologize. People who are polite, think things through, and have genuine questions, are always welcome here. People who "can't be bothered to figure this out for themselves" get a less warm reception. That said, since this question ended up hingeing on a simple mathematical error rather than exploring a bit of "deep" physics, it is possible it will get deleted at some point as having "no value to a future visitor". If that happens, please don't take it personally - and do come back with good questions, and good answers! $\endgroup$ – Floris Oct 8 '14 at 20:50
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    $\begingroup$ This question appears to be off-topic because it is a "check my work" question. $\endgroup$ – Floris Oct 8 '14 at 21:27
  • $\begingroup$ Voting to close as per policy, but as Floris says, welcome to the site, and would be happy to see future questions that are similarly well thought out. $\endgroup$ – Kyle Oman Oct 8 '14 at 22:13
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You have a mistake in the line

$$ΔV= (.0314m^2)(.002m)$$

Note that 2 cm = 0.02 m, not 0.002 m...

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