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Heat Absorbed during a process on an Ideal Gas

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data

    0.21 x 10-3 kmol of ideal gas occupy a volume of 5.0 x 10-3 m3 initially at a pressure of 1.2 x 105 Pa and temperature T.

    The volume of the gas is first decreased to 2.0 x10-3 m3 at a constant pressure of 1.2 x 105 Pa, and then the pressure is increased to 3.0 x105 Pa at a constant volume of 2.0 x10-3 m3.

    For this process, calculate:

    A) The Work done on the gas

    B) The Heat absorbed by the gas

    2. Relevant equations

    ΔU = Q - W
    (where ΔU is change in internal energy, Q is Heat and W is work done by the gas)

    U = n NA ([itex]\frac{1}{2} m \overline{v^2}[/itex]) = [itex]\frac{3}{2} R T[/itex]
    where U is internal energy

    W = p ( V2 - V1 ) (for isobaric process)
    W = 0 (for isovolumetric process)

    3. The attempt at a solution

    A rough P-V diagram would look like:

    [PLAIN]http://img695.imageshack.us/img695/5465/pvdiagram.png [Broken]

    A) Work done on the gas.

    Work done is area under P-V curve.

    W = p (V2 - V1) = - 1.2 x 105 x (2.0 x 10-3 - 5.0 x 10-3) = + 360 J

    B) Heat absorbed by the gas.

    ΔU = Q - W, therefore:
    Q = ΔU + 360

    However I don't know how to find ΔU and seem to be stuck.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 7, 2011 #2

    ehild

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    You can calculate the internal energy from the temperature. Use the ideal gas law. How does the temperature change between the initial and final states?

    ehild
     
  4. Nov 7, 2011 #3
    Using [itex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/itex] (ideal gas law):
    [itex]\frac{1.2 \times 10^5 \times 5.0 \times 10^-3}{T_{initial}} = \frac{3.0 \times 10^5 \times 2.0 \times 10^-3}{T_{final}}[/itex]
    [itex]600 T_{initial} = 600 T_{final}[/itex]
    [itex]T_{initial} = T_{final}[/itex]

    Therefore as an isothermic process (constant temperature), ΔU = 0, Q=W.
    So Q = 360 J.

    Is this correct? The answers I have for this sample exam give -360 J instead.
     
  5. Nov 7, 2011 #4

    ehild

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    W is the work of the gas. As the volume decreases in the first step, the work of the gas is negative, W=-360 J. At the same time, Q is the adsorbed heat. ΔU=Q-W(gas)=Q-(-360J)=0 --->Q=-360 J. Some external force does work on the gas, and the gas transfers heat to the surroundings.


    ehild
     
  6. Nov 7, 2011 #5
    I get it now...
    I was getting confused because the question asked work done on the gas when my formulas are for work done by the gas, and so I used the wrong number/sign in the second part. Your explanation makes sense.

    Thanks, ehild.
     
  7. Nov 7, 2011 #6

    ehild

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    You are welcome.

    The First Law of Thermodynamics states that the internal energy increases by the amount of heat added to and by the work done on the gas: ΔU=Q+W. If the work is done very slowly, so the gas is very nearly in equilibrium with the external force, W=-PdV=-Wgas.

    ehild
     
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