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What is the final volume of container B?

  1. May 24, 2007 #1
    Two 800\; cm^3 containers hold identical amounts of a monatomic gas at 20^\circ C. Container A is rigid. Container B has a 100\; cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
    Suppose the heaters have 25 W of power and are turned on for 15 s. What is the final volume of container B?


    I found the initial pressures of both container B, I then plugged that into the initial pV=nRT to find n, then plugged that into
    E=[nc(constant pressure)deltaT] using 25x15=375 J as the value for E. This determined delta T, I then plugged that (with T = delta T + 293) into the final pV=nRT to get V. This seems like it's probably too complicated, and I didn't get the right answer...can someone please help??
     
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  3. May 25, 2007 #2

    Pythagorean

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    I'm somewhat confused with your notation 100\; and I don't know what the significance of container A is.
     
  4. May 25, 2007 #3
    that should just be 100 cm squared. I'm not sure what the significance is of container A is either...it may have to do with question (a) which asked which container would have the higher final temp. Do you think container A should have some part in the answer?
     
  5. May 25, 2007 #4

    Pythagorean

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    ahh, ok, A seems to be only for part a).

    I'll poke at b) and give you hints

    to start you off, they give you power (in watts) and the length of time that power is exposed to the system. Power has energy of units over time and time has units of time. Using those two quantities, you should be able to calculate the energy put into the system by the heaters.

    furthermore, Boltzmann's constant and temperature are equatable to energy.
     
    Last edited: May 25, 2007
  6. May 25, 2007 #5
    I think... that container A and B share the energy from the heater.
     
  7. May 25, 2007 #6
    that would be the thermal energy, right? 375 J...did I use the right equation above with E=ncdeltaT with c=20.8 ? or should it be another equation, since that equation is assuming that the pressure in the system stays constant
     
  8. May 25, 2007 #7

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    Also, for constant pressure, W = P(del V)

    (work equals pressure times the change in volume)

    which can be represented as W = P(V2 - V1)
     
  9. May 25, 2007 #8

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    What is c? That looks an awful lot like part of the phase change equation which shouldn't be used here, that's for when a material changes form (from liquid to gas and such)
     
  10. May 25, 2007 #9
    I tried another equation to get the final temp...thermal energy = 375 = (3/2)NkBT...then plugged it into pv=nRT but this didn't work either
    For the work equation above, does work = 375??? I feel like I'm getting more confused the further I go
     
  11. May 25, 2007 #10

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    I'm not quite sure about this. Let me play with it more and get back to it
     
  12. May 25, 2007 #11

    Pythagorean

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    This may be a good point too, I'm not sure. What's the answer so I know when I get it?
     
  13. May 25, 2007 #12
    I'm not sure what the answer is...sorry
     
  14. May 25, 2007 #13
    but I know the initial pressure is 1.097 atm
     
  15. May 25, 2007 #14

    Pythagorean

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    note that in B, some of the work is going to go into lifting the piston:

    W=mgh
     
  16. May 25, 2007 #15

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    then why wouldn't W = P(V2-V1) work then? pressure stays constant if volume is allowed to change (i.e. the piston)

    You just have to make sure you use the right W (that is, whether it's the full energy delivered by the heaters or only half.
     
  17. May 27, 2007 #16

    Andrew Mason

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    How did you determine the pressure in A and/or the volume in B?

    AM
     
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