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    Van der Waals Gas Equation

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    The equation of state of one mole of a van der Waals gas is given by

    (P+a/(v^2))(V-b) = RT

    with a and b are constants.
    a) Calculate the work W in an isothermal reversible process when volume changes from V1 to V2.
    b) Using the energy equation, show that (du/dV) = a/v^2
    c) Calculate the change in internal energy U in the process (a)
    d) Calculate the heat exchange in this process
    e) Calculate the change in quantity PV.deltaPV
    f) Deduce the change in enthalpy

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    Solution Preview

    I've included in this text an Appendix containing part of a previous problem I did for you which is useful for this problem.

    The equation of state for one mole is:

    (P + a/V^2) (V-b) = R T

    Solving for P yields:

    P = R T/(V - b) - a/V^2

    The work done by the gas when its volume changes from V1 to V2 at constant temperature T is

    W = Integral from V1 to V2 of P dV =

    R T Log[(V2-b)/(V1-b)] +a/V2 - a/V1 (1)

    Then Eq. A9 of the Appendix says that:

    dU = Cv dT +a n^2/V^2 dV

    The coefficient of dV in here is the derivative of U w.r.t. V at constant T:

    (dU/dV)_T = a n^2/V^2 = a/V^2 for n = 1 mole.

    So, when the volume changes from V1 to V2, the change in U is:

    Delta U = Integral from V1 to V2 of a/V^2 dV = -a/V2 + a/V1 (2)

    Then from the First law of thermodynamics we have that:

    Delta U ...

    Solution Summary

    Detailed solution is given where everything is derived from first principles.