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Homework Help: Isothermal compression vs thermally isolated compression

  1. Jun 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Two equal masses of an ideal gas initially at the same temperature and pressure are compressed to half of their initial volumes, one of them isothermally, and the other while thermally isolated from its surroundings.
    Which one of the following is the same for both samples of the gas?
    A) the heat given during compression
    B) the internal energy of the compressed gas
    C) the density of the compressed gas
    D) the work done on the gas during compression
    2. Relevant equations
    3. The attempt at a solution
    My initial thought was "is there a diff between 'isothermally' and 'thermally isolated from surroundings' ???". Or is it just a trick statement? i gave it a try and chose B as the answer cause i thought internal energy of ideal gas depends on its temperature. Since both are (i assumed) isothermally compressed, internal energy at the end must be the same, regardless of change in pressure or volume. Alas, my physics teacher posted answers on the net and it came out as C. WHY??????
  2. jcsd
  3. Jun 25, 2011 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Thermally isolated from surroundings = adiabatic: no heat flow into or out of the gas.

    In an adiabatic compression, dQ = 0 so dU = -dW where dW is the incremental work done BY the gas. Since work is done ON the gas to compress it, dU>0, so its temperature will increase. So adiabatic compression cannot be isothermal.

    Heat must flow out of the gas to keep temperature constant in an isothermal compression. Heat cannot flow out in an adiabatic compression. Internal energy increases in the adiabatic compression but is constant in the isothermal compression.

    So A and B are out. Since the work done on the gas depends on the pressure of the gas which is proportional to the temperature at a given volume, the adiabatic compression requires more work at the gas heats up. So D is out. That leaves C. Since [itex]\rho = n/V[/itex], density does not depend on how the gas is compressed.

  4. Jun 25, 2011 #3
    thanks alot man!
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