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How much work is done on a gas?

  1. Oct 25, 2007 #1
    First of all...thanks for any help. This is ultrabasic introduction to therm, so I know everyone says that no question is stupid, but I just feel like I'm really missing the boat here.

    1. The problem statement, all variables and given/known data

    The temperature of a monatomic ideal gas remains constant during a process in which 4390 J of heat flows out of the gas. How much work (including the proper + or - sign) is done on the gas?

    2. Relevant equations


    and I think...maybe

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

    3. The attempt at a solution

    Since it says that the temperature remains constant, I was thinking that [tex]\Delta[/tex]E=0

    Q is the heat supplied TO the system, so if 4390 J of heat came out, then it would be -Q.

    So I figured that:
    [tex]\Delta[/tex]E=Q - W
    0=-4390 - W
    W=-4390 Joules

    That's not correct. They must be telling me that it's a monatomic ideal gas for some reason.

  2. jcsd
  3. Oct 25, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Sign convention

    Realize that in this equation:
    Q = the heat energy transferred into the the system
    W = the work done by the system (on the environment)

    So the work done on the system is just the negative of that.

    Many modern books use this form of the first law:
    [tex]\Delta[/tex]E=Q + W, in which W is the work done on the system. This form makes it clearer that this is just a statement of energy conservation.
  4. Oct 25, 2007 #3
    Thank you...obviously I was looking at the meanings incorrectly.

    The answer is W=4390 Joules.

    The way you worded your answer actually helps me with yet another problem I was having! Have a great day!
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