1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reversible expansion

  1. Dec 26, 2005 #1
    we have an ideal monatomic gas which intially occupies 27*(10^-3) m^3 at a pressure of 3.2*(10^5) Pa and at a temperature of 400K. We have to find the heat supplied to the gas, the work done on it, and its increase in internal energy when it is compressed isothermally at 400K to a volume of 8*(10^-3) m^3.

    We were given a standard solution to this question which works out work done using W=-P*dV. I think this is incorrect, though as work done is a function of state and therefore is path dependent, so we would need to be informed that the path was reversible before we could figure out the work done in this way. We know the change in T=0, and for ideal gas U=U(T) therefore change in U is zero, therefore by 1st law W=-Q, so we can find a relationship between the two but can't calculate anything explicitly. I think this is right but am not sure.

    Also, I'd like to discuss the fact that W>-PdV for irreversible changes. I can't quite get my head around the inequality. If a gas was to expand by a certain volume reversibly, then it would do a certain amount of work on the surroundings. If it was to expand irreversibly it would have to do the same amount of work on the surroundings to expand in volume, but it would also have to do work against frictional forces. Therefore the amount of work would have greater modulus but -ve sign. Surely then dW <-P*dV??

    Thanks very much.
     
  2. jcsd
  3. Dec 27, 2005 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Using the first law, of course:

    [tex]\Delta Q = \Delta U + \Delta W[/tex]

    As you noted, since there is no temperature change, the internal energy of the gas does not change ([itex]\Delta U = \frac{3}{2}nR\Delta T[/itex] for a monatomic gas). So:

    [tex]\Delta Q = \Delta W = \int_{Vi}^{Vf} P dv[/tex]

    Since PV = nRT, [itex]P = nRT/V = P_iV_i/V[/itex].

    [tex]\Delta W = P_iV_i\int_{Vi}^{Vf} \frac{dv}{V} = P_iV_i ln\frac{V_f}{V_i}[/tex]


    The definition of work is Force x Distance so, by definition, dW = Fds = PAds = PdV (static pressure). So, this is always true whether it is a reversible or irreversible change. If the change is reversible, the work done to or by the gas is recoverable. If it is not recoverable, it is irreversible. But dW is always equal to PdV (assuming thermodynamic equilibrium).

    Hope that helps.

    AM
     
    Last edited: Dec 27, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?