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Isobaric process

  1. Dec 20, 2009 #1
    NVM I solved it. Thanks for reading anyway :)

    1. The problem statement, all variables and given/known data
    610 J of heat is added to 3.4 mol of a monatomic ideal gas at constant pressure. Find the
    change of temperature of the gas.

    2. Relevant equations
    At isobaric conditions, w = p[tex]\Delta[/tex]V = nR[tex]\Delta[/tex]T

    w = nR[tex]\Delta[/tex]T

    3. The attempt at a solution

    w = 610 J; n = 3.4

    610/(3.4*8.314) = 21.5 K = [tex]\Delta[/tex]T

    Calculations look right to me,but the answer is 8.63 K
    Last edited: Dec 20, 2009
  2. jcsd
  3. Dec 21, 2009 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    You are equating Q with W. That is true only if the internal energy does not change, which of course is not the case when temperature increases.

    The 610 joules of heat flow does two things: causes the gas to do work and increases its internal energy.

    The specific heat at constant pressure is the heat flow per degree of temperature rise (per mole) of a monatomic gas during a constant pressure process:

    [tex]\Delta Q = nC_p\Delta T[/tex]

    Use that to calculate the change in temperature. What is Cp for a monatomic gas?

  4. Dec 21, 2009 #3
    Yea I realized that and used the other equation after some trial and error. Cp for a monoatomic gas is 5/2R :)

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