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Pressure and Temperature and Heat

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    An ideal gas occupies a volume of 0.3 m3 at the pressure of 400 kPa. If the volume changes to 0.85 m3 and the temperature remains constant, what is the pressure of the gas? Give answer in kPa.


    If 500 cal of heat are added to a gas, and the gas expands doing 394 J of work on its surroundings, what is the change in the internal energy of the gas in Joules.


    150 g of a certain metal, initially at 100o C is dropped into an insulated beaker containing 100 g of water at 20o C. The final temperature of the metal and water in the beaker is measured at 43o C. What is the specific heat capacity of the metal in cal/(g o C)?

    2. Relevant equations

    W = P*deltaV
    Q = mcdeltaT

    3. The attempt at a solution
    I have no idea what's going on with any of these problems... if anyone can help me I would be awesomely grateful.
  2. jcsd
  3. Nov 24, 2008 #2

    Andrew Mason

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    For the first question you need to use the ideal gas equation: PV=nRT

    For the second, you need to use the first law of thermodynamics: dq = dU + dW where dW is the work done by the system.

  4. Nov 24, 2008 #3

    Just one question, and then I'll do the work on my own, I promise -- for the ideal gas equation, where N determines the number of molecules, how do I figure out how many molecules there are?
  5. Nov 24, 2008 #4
    Okay, so, for the second one, I looked through my textbook and found the equation PV=(2/3)K. I tried using that equation (because it doesn't consider temperature or N) but I came up with a ridiculously low number. Is that right, or am I completely off base?
  6. Nov 24, 2008 #5

    Andrew Mason

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    You don't need to calculate n. You just need to know that it does not change. So if T does not change, what can you say about the product PV? Does it change?

  7. Nov 25, 2008 #6
    No, then it doesn't. So if nRT is constant, the two PVs should be the same.

    Thank you so so much for all your help.
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