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Piston with heater

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    a piston of mass M is attached to a cylinder made from thermal insulation. There is no friction between the piston and the cylinder. At first, the cylinder is placed horizontally and n mol of an ideal gas of monatomic molecule at the same temperature To and the same pressure Po as the atmosphere is contained in the cylinder. The distance from the bottom of the cylinder to the piston is L as shown in the figure. Let the gas constant be R and let the gravitational acceleration be g.

    piston.jpg

    1. when this cylinder was stood perpendicularly, the piston fell slowly and stopped at a height of L' from the bottom. How much was the increase in the internal energy inside the cylinder at this time?

    2. how much was the increase in temperature inside the cylinder at this time?

    3. after that, when the inside of the cylinder was heated with the heater, the height of the piston returned from L' to L. How much did the temperature in the cylinder increase as compared with the case where the cylinder was placed horizontally?

    2. Relevant equations
    ideal gas
    first law of thermodynamics


    3. The attempt at a solution
    1. done (ans : Mg (L - L') )

    2. done (ans : 2/3 Mg (L-L') / nR )

    3. don't understand to start....

    thanks
     
  2. jcsd
  3. May 26, 2010 #2
    It looks like you have completed a cycle with the final temperature equaling To. Do you have the answer to this problem?
     
  4. May 26, 2010 #3
    yes, the answer is MgL / nR. How can you know that the final temperature is To? the question asks about How much the temperature in the cylinder increased as compared with the case where the cylinder was placed horizontally, so the answer will be zero (because it is the same) ?
     
  5. May 27, 2010 #4
    It looks like part 3 is a constant pressure process where Delta U= Q-W. Delta U=1.5R*n*(Tf-Ti), Q=2.5R*n*(Tf-Ti) and W=mg(L-L'). You might try this approach but I don't see how you eliminate L' from the final expression.
     
  6. May 27, 2010 #5
    How can we know this is an isobaric process?
     
  7. May 30, 2010 #6
    *bump
     
  8. Jun 1, 2010 #7
    The process is isobaric because the pressure of the gas inside the cylinder doesn't change. Pressure before you heat is the atmospheric pressure plus the pressure applied by the piston. This is also the pressure after you heat the gas. If that pressure changes, it'll force the piston to move up and down.

    Now use the first law of thermodynamics and the adiabatic index for monatomic gases. It should work out to be some multiple of your answer to part b.

    Also consider if your answer should be greater than or less than the answer to part b. Consider the pressure and the volume of the system in both cases. Think about which one would require a greater increase in temperature of the system. That would also help you understand what is going on better.
     
  9. Jun 3, 2010 #8
    I don't get it. the pressure could change resulting the piston moving up till it stops when the inner pressure is equal to atmospheric pressure.

    ΔU = ΔQ - ΔW, there is heat supply so there must be certain value of ΔQ and I'm not sure if the pressure is constant so I have problem to find ΔW.

    the change in volume is the same for part (b) and (c) and I have no idea how to compare both answers. the pressure could be not constant (again :smile:) so the comparison in temperature change between part (b) and (c) can't be determined

    Need more hint please. I still don't understand. Thanks
     
  10. Jun 4, 2010 #9
    All right, lets dig into this:

    This is probably too much detail but let's do it anyways. Yes, the pressure inside the cylinder changes very slightly. That is actually what pushes the piston upwards. That very slight change in pressure is what creates the force on the piston and pushes it up. However, when you idealize everything (ideal gas, frictionless piston, no heat loss, etc..) and think about things in the limit, that "very slight" becomes "none". The piston moves to equalize the pressure with "atmospheric pressure + pressure applied by the piston" An infinitesimal change in pressure of the gas moves up the piston infinitesimally. Then the pressure is equalized again. That's why the pressure remains constant during the process.

    Now that you are (possibly :smile:) convinced that this is an isobaric process, you can use the first law of thermodynamics. You don't need to know if the process is isobaric to find the work done by (or on) the system, at least in this case. That has to do with how much the piston moves and in what direction.

    Yes, the volume change is the same for both processes. Now think about what pressure each of these processes occur at. And consider what happens to the PV product. This will tell you if you need to change the temperature of the system more or less (compared to the first process) if you plug these into the ideal gas law.
     
  11. Jun 5, 2010 #10
    just a little question, how can we know that the pressure inside the cylinder changes very slightly? the question never said anything about it.

    so, if the pressure constant :
    ΔW = p ΔV = p A (L - L') = F (L-L') = Mg (L-L')

    then how to find the change in temperature? there should be ΔQ because there is heat supplied.
    ΔU = ΔQ - ΔW
    3/2 nR ΔT = ΔQ - Mg (L-L')...stuck

    these processes (b) and (c) occur at the same pressure. because there is heat supplied in (c), I think the change in temperature should be lower than (b)

    thanks
     
  12. Jun 7, 2010 #11
    I probably shouldn't have said that. The question doesn't say anything about it. It happens because that's what creates the force that pushes the piston upwards. But as I said, that happens in infinitesimal increments. So the pressure changes only for infinitely short time. So you can think of it as it doesn't change. Forget the fact that this is even there. It will only confuse you more. You should get more practice before you think about those details. Cool?

    What is ΔQ for an isobaric process?

    Why do you think so?
     
  13. Jun 9, 2010 #12
    ok

    ΔQ = n Cp ΔT, where Cp= 5/2 nR ΔT

    so, subs that into : 3/2 nR ΔT = ΔQ - Mg (L-L'), I got ΔT = Mg (L-L') / nR

    and now to answer the question : How much did the temperature in the cylinder increase as compared with the case where the cylinder was placed horizontally

    I tried to find the initial temperature.

    T = PV / nR = P AL / nR = FL / nR = MgL / nR

    so the change in temperature is MgL' / nR which is differs from the manual (MgL / nR). Where is my mistake?

    and I'm not so sure about the bold part (changing F to Mg when finding initial temperature) because it's horizontal, not vertical so the force may not be equal to weight.


    because the pressure is constant ?


    thanks
     
  14. Jun 9, 2010 #13
    Yes, that's right.

    That's not correct. The pressure is equal to the atmospheric pressure. Notice that when the cylinder is placed horizontally, the weight of the piston is a vector that points downwards, not sideways. So it doesn't contribute to the pressure inside the cylinder.

    No it isn't. Be careful. Even if your answer in bold was correct, this wouldn't be right. You are subtracting a difference in temperature from an absolute temperature. That gives you another absolute temperature, not a difference. Clear?

    Now, let me ask you this. You have two temperature changes: one that happened when you put the cylinder vertically, and one that happened when you heated it. Those are both increases in temperature. Aren't you asked to find the difference between the initial temperature and the final temperature? (Correct me here if I misunderstood the question). So why don't you just add those two changes that you've calculated?
     
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