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Entropy Change of an Expanding Gas

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Two moles of an ideal gas undergo a reversible isothermal expansion from 3.34E-2 m^3 to 4.80E-2 m^3 at a temperature of 25.7C

    What is the change in entropy (ΔS) of the gas?

    2. Relevant equations

    ΔS = Q/T

    3. The attempt at a solution
    first I think I need to calculate the work done by the gas.
    and then Q = ΔEint - W
    but I am confused.. because this formula requires pressure. and the pressure is not given.

    pls.. help me..
    thanks in advance..
     
  2. jcsd
  3. Dec 13, 2007 #2

    dynamicsolo

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    The thing to keep in mind here is that the expansion is isothermal, so there is no change in E_int; that means that Q = -W (W being the work done on the gas in this version of the First Law).

    So you'll need to compute the work done on the gas in such an expansion. You need to integrate -P dV from V_initial to V_final to get W, then take the negative of that result to find Q. Since the gas will be at constant temperature, using the ideal gas law gives

    P = nRT/V , with everything in the numerator being positive. So the integral should be pretty easy to do.

    We want the change in entropy, though. Since the gas remains isothermal, we get to take a shortcut. We need to integrate

    dS = dQ / T over the expansion.

    Since in this situation, dQ = -dW = P dV = (nRT/V) dV , with T constant,
    you can go directly to your entropy integral, with limits from V_initial to V_final. The statement of the problem gives you enough information to work out delta_S in J/K .
     
  4. Dec 13, 2007 #3
    the delta_S should be in a negative value, right?
    since Q = -W,
    but when I type it, the feedback said that since the gas is expanding, the entropy will increase and it should be positive..

    anyway..
    thank you for your help.. :)
     
  5. Dec 13, 2007 #4

    dynamicsolo

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    ...except that W is "work done on the gas" in the definition of the First Law that we're using. So W is negative in an expansion (work done by the gas is positive), making Q positive and thus delta_S as well.

    (You should also have found that the temperature given in the problem is irrelevant.)
     
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