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Change in internal energy for an isobaric process?

  1. Nov 22, 2005 #1

    When a quantity of monatomic ideal gas expands at a constant pressure of [tex]4.00 \times 10^{4} {\rm Pa}[/tex], the volume of the gas increases from [tex]2.00 \times 10^{ - 3} {\rm m}^{3}[/tex] to [tex]8.00 \times 10^{ - 3} {\rm m}^{3}[/tex].


    What is the change in the internal energy of the gas?

    It's isobaric, so the pressure is constant.

    I know the work is [tex]P\Delta V = (4.00 * 10^4)(6.00 * 10^{-3})[/tex].

    But, I don't know how to get [tex]\Delta U[/tex] from this.
  2. jcsd
  3. Nov 22, 2005 #2

    Physics Monkey

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    What variables does the internal energy of the ideal gas depend on? How do these variables change in the aforementioned process?
  4. Nov 22, 2005 #3
    OK, the internal energy depends only on temperature.

    For a monatomic gas, [tex]\Delta U = \frac{3}{2}nR\Delta T[/tex].

    I don't know the number of moles or the change in temperature.
  5. Nov 22, 2005 #4

    Physics Monkey

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    Progress! Ok, so now you need to know the change in temperature times [tex] n R [/tex], right? You know the pressure and volume of the gas at two different points in P,V space. Can you use this information to find the unknown? Hint: ideal gas law.
  6. Nov 22, 2005 #5


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    This relationship should be helpful as well.

    [tex]\Delta U = Q - W [/tex]

    Apparently you have the equation for the W right. Now use the ideal gas law and a bit of calorimetry.
  7. Nov 22, 2005 #6
    Thanks a lot!
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