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Homework Help: Work done by Water

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data

    You have a sealed container with 2kg of water (molar mass: 18g/mol, density: 1000kg/m^3). The lid on the container is very light, and the container side walls are very tall, like a tall graduated cylinder. You start with the water at room temperature 25 C and boil all the water until you get to 200 C. What is the work done by he water molecules.

    2. Relevant equations

    3. The attempt at a solution[/b

    I'm not really sure how to go about starting this. If someone could give me a set of equations or explanation it would be much appreciated.
  2. jcsd
  3. Apr 22, 2014 #2
    The given answers are: A) 234 B) 334 C) 434 D) 534 kJ
  4. Apr 22, 2014 #3
    The key to this problem are the words "the lid on the container is very light." What does this mean to you about what the initial pressure is? What does this mean to you about what the final pressure is? If you know the final pressure, then, from the ideal gas law, you know the final volume. What is it? What is the initial volume? What is the equation for the amount of work done?

  5. Apr 22, 2014 #4
    Well, work done is W=Fd. However the lid on the container is very light doesn't really help me understand how to find the initial pressure. Are we working with PV=nRT or P1V1/T2=P2V2/T2 or both? Either way pressure and volume are both unknown. I need a place to start and then I can probably work my way through the rest of the problem. I just don't know where to start.
  6. Apr 22, 2014 #5
    If the lid to on the container is very light (and the system is initially in thermodynamic and mechanical equilibrium), then the initial pressure in the container must be 1 atm. When the system reaches its final equilibrium state, the pressure within the container must still be 1 atm. In fact, during the entire process, the pressure within the container remains at 1 atm. (to match the pressure on the other side of the lid). Does this make sense?

  7. Apr 22, 2014 #6
    Yes, that makes sense. Sorry, I've had to learn momentum, thermodynamics, collision properties, and fluid properties all over the course of a day so my head is cluttered with equations and concepts. Thermodynamics is still very confusing to me.
  8. Apr 22, 2014 #7
    Sorry to hear this. Thermo is confusing enough to most people without also trying to master these other subjects (in one day).

  9. Apr 22, 2014 #8
    Trust me I found out the hard way. Had to do 125 problems, have 5 left, all thermodynamics, this being one of them.
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