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Thermal Physics Help Needed

  1. Jan 16, 2005 #1
    Hi, I'm a high school physics student, and I'm having problems with two questions:

    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?
    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 tried to solve them myself, and my work is shown below, but I do not think the answers are correct. If anybody could please guide me, or correct me, that would be very helpful. If there's an error, could you please explain why it is wrong? I would really like to know what I am doing, and not just get the right answers. Thank you.

    Here are my solutions (that I am not sure of):
    1. W = PV
    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

    U = Q - W
    Q = cmT
    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
    U = 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

    Gokul43201

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    [tex]W = P \Delta V [/tex]

    You don't do work unless you change the volume. Does the volume change ?
     
  4. Jan 16, 2005 #3

    Gokul43201

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    You're using the wrong formula for Isothermal Work done. Check the formula.
     
  5. Jan 16, 2005 #4
    Oh, really? I'm just using the formulas my teacher provided...

    I would just like to know whether or not these solutions are possible. My physics teacher says she's not too picky about the final answer; she just wants to see how we got to our solutions. So, I would mainly like to know whether or not you guys think that these solutions are possible or if they're completely wrong.
     
  6. Jan 16, 2005 #5
    I think his using correct formula, but first you must find the volume change:
    [itex] \frac {P_1 V_1} {T_1} = \frac {P_2 V_2} {T_2} [/itex]
     
    Last edited: Jan 16, 2005
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