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