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

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.

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