Explore BrainMass
Share

# Van der Waals Gas Equation

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

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

© BrainMass Inc. brainmass.com October 10, 2019, 6:15 am ad1c9bdddf
https://brainmass.com/chemistry/energetics-and-thermodynamics/van-der-waals-gas-equation-535865

#### 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.

\$2.19