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1st Law of Thermodynamics

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Three moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2114 J of heat are added to the gas, and 811 J of work are done on it. What is the final temperature of the gas?


    2. Relevant equations



    3. The attempt at a solution
    I have a feeling I'm missing an equation. I thought the equation was U = Q - W but that won't help me here.
     
  2. jcsd
  3. Jan 19, 2009 #2

    Andrew Mason

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    How is U related to temperature? Hint: what is the specific heat of a monatomic gas at constant volume (where W = 0 so Q = U)?

    AM
     
  4. Jan 20, 2009 #3
    Doesn't it depend upon the gas?
     
  5. Jan 20, 2009 #4

    Andrew Mason

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    Not if it is an ideal gas, which this is. The specific heat is Cv = 3R/2 where R is the universal gas constant (units are J/mole K). Since, by definition, [itex]\Delta Q = nC_v\Delta T[/itex] and [itex]\Delta U = \Delta Q[/itex] for a constant volume process and since U is a function only of T, it follows that

    [tex]\Delta U = nC_v\Delta T[/itex]

    regardless of the type of process.

    AM
     
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