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Thermodynamics Help

  1. Jan 16, 2005 #1
    Hi, I need some help with my thermal physics. I posted these questions on the high school forum (I'm just a junior in high school), but my physics teacher takes questions out of college-level books, so I thought it would be more appropriate if I posted here.

    The two questions are:
    1. An aluminum cube 20 cm on a side is heated from 50° C to 150° C in a chamber at atmospheric pressure. Determine the work done by the cube and the change in its internal energy. If the same process was carried out in a vacuum, what would be the change in internal energy?
    and
    2. Determine the amount of work needed to compress 4.0 g of oxygen at STP down to 1/3 its original volume, keeping the temperature constant. Assume it behaves as an ideal gas.

    I gave the it a shot and attempted to solve the questions. My work is shown below. The problem is I have no idea whether or not it is correct. If anybody could guide me through and tell me where I went wrong (and why), that would be very helpful. Thank you.

    My work:
    W = PDV
    1 atm = 1.01x10^5 Pa (process is isobaric; pressure is constant.)
    Volume = 20 cm^3 = 20 cm^3 x (1 m^3/100 cm^3) = 8x10^-3 m^3
    W = (1.01x10^5 Pa)(8x10^-3 m^3)
    = 808 J
    Work is 808 Joules

    DU = Q - W
    Q = cmDT
    Specific heat capacity (c) of aluminum is 900 J/(kg)(K)
    Mass (m) is 0.0269 kg (26.9 g/1000)
    150 degrees - 50 degrees = 100 degrees = 373 K
    Q = (900 J/(kg)(K))(.0269kg)(373 K)
    = 9030 J
    DU = Q - W
    = 9030 J - 808 J
    = 8222 J
    Change in internal pressure is 8222 J

    If the process was carried out in a vacuum, the work (w) would be zero. Thus, the internal energy would be the Q - 0, which is equal to 9030 J.

    2. STP constants: V = 22.4 L and P = 1.01x10^5 Pa
    W = nRTV(f)/V(I) or PV(V(f))/(V(I))
    N = number of moles = 4 g * (1 mole/16 g) = .25 mole
    Volume = .25 mole * (22.4 L/1 mole) = 5.6 L
    W = PV(F)/(I)
    = (1.01x10^5 Pa)(5.6 L)(1)/3
    = 1.8x10^5 J
     
  2. jcsd
  3. Jan 16, 2005 #2

    dextercioby

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    1.For the fisrt part,check again the work "W".Are u sure the volume does not vary??

    Daniel.

    PS.Are u sure with the mass of aluminum??I think it's wrong...
     
  4. Jan 16, 2005 #3
    My teacher gave us a hint that the process was only isothermal... My guess then is that the volume does not vary.

    And the problem with the mass doesn't surprise me. I took the value from teh Periodic Table and divided it by 1000. I didn't take into account the volume. I'm not sure how I would include the mass, so I tried something. Do you know?
     
  5. Jan 16, 2005 #4

    dextercioby

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    Are u kidding me???What kind of a sick teacher u have??The body dilates.Which means that as the temperature rises,the volume rises as well.The process is ISOBAR,which means the pressure in the chamber does not vary.

    The density of aluminum is [itex] 2700Kgm^{-3} [/itex] at 20°.The problem with this densoty is that it varies with temperature.However,i think u can use this number for finding the mass.Afer all,at any other temperature,the mass would be the same...The trick is that the volume varies as well.

    Daniel.
     
  6. Jan 16, 2005 #5

    Andrew Mason

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    The teacher is correct. The volume of air/gas in the room decreases only by the amount that the aluminum expands which is negligible if the pressure change is negligible. So you can treat this as an isothermic and isobaric process.

    I think Vitaly's approach is correct except that the work done is not P x volume of block but P x change in volume of block. Apart from that the thermodynamics is correct. The change in internal energy is the heat added less the work done in expanding the block. In a vacuum, Vitaly is correct that this work would be 0.

    AM
     
  7. Jan 16, 2005 #6

    dextercioby

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    The volume of air decreases,because the solid is dilating.Period.I agree that the dilation is small,yet i also think it is calculable,as long as you can assume the linear approximation for each if the cube's sides.Besides,is it just me who sees problems more complicated than they are??However,i still think the transformation is only isobar,not isothermal...
    Besides,we actually have a nonequlibrium problem.The gas is in contact with a solid and it takes heat.That heat is transmitted through conduction to the lower layer and by convection to the rest of the gas...

    But sure,let's agree with the teacher,though she's obviously wrong... :yuck:


    Daniel.
     
  8. Jan 16, 2005 #7
    For the first problem:

    You must calculate in fact the work done on the air. You need only the definition of work in a isobar process: W=p*DV. The air pressure is p_0 so you have W=p_0*DV. Now you have to estimate DV (you need the linear dilatation coefficient of Al ......the density of Al is useless in this problem!) :

    DV=3*alfa*V_0*Dt where V_0=a^3 (a=0.2m) and Dt=100 K

    In the second problem, the work done by the gas is

    W=niu*R*T*ln(V_2/V_1)

    where V_2=V_1/3

    and the external work will be W'=-W.


    PS:

    In all equations above
    R=8.31E3 J/Kmol/K
    niu=m/(32 Kg/Kmol) (if O2 obvious) with m=4e-3 Kg
    For the second problem you need the temperature: if 0 Celsius then T=273 K
     
    Last edited: Jan 16, 2005
  9. Jan 16, 2005 #8

    dextercioby

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    Am i seeing things,or to you "air=oxygen"???

    Daniel.
     
  10. Jan 16, 2005 #9
    2. Determine the amount of work needed to compress 4.0 g of oxygen at STP down to 1/3 its original volume, keeping the temperature constant. Assume it behaves as an ideal gas.
     
  11. Jan 16, 2005 #10

    Andrew Mason

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    Of course it does, but it is negligible when compared to the room size. So the temperature change is negligible. If the volume decrease was not negligible, the room pressure would increase and that is not the case.

    It is not calculable in this problem since we don't know the volume of the room. We can just determine the change in volume.

    If it is Isobar but an increase in temperature there would have to be an increase in volume of the room. P = nRT/V
    It is assumed that the process happens so quickly that there is a negligible temperature change to the room. Otherwise there would be a pressure increase.

    AM
     
  12. Jan 16, 2005 #11

    dextercioby

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    Yes,you're right...You were talking about oxygen and air in the same post and i thought u mixed problems... :tongue2:

    Sorry.

    Daniel.
     
  13. Jan 16, 2005 #12
    Thank you for all of your comments. I'm just glad to know I'm on the right track.
     
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