Thermodynamics: Work Under Isothermal Condition

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Homework Statement


[/B]
One mole of 2-D ideal gas is confined in an isotropic cone potential:

[itex]U = \lambda |r|[/itex]

where [itex]\lambda[/itex] is a positive parameter and [itex]r[/itex] is the displacement vector (2-dimensional) from the origin. The mass of each molecule is [itex]m[/itex].

(1) Determine the Helmholtz free energy [itex]A[/itex] of this gas confined in the potential at temperature [itex]T[/itex].

(2) If you want to vary [itex]\lambda → \lambda + \delta \lambda[/itex], you must do some work [itex]\delta W[/itex]. Compute [itex]\delta W[/itex], following Einstein.

(3) Show, as thermodynamics tells us, that [itex]\delta W[/itex] in (2) agrees with the variation of [itex]\delta A[/itex] of the Helmholtz free energy [itex]A[/itex] due to the same parameter change in (2).

Homework Equations



[itex]A = -k_{B}TlogZ[/itex]

[itex]Z = \sum e^{\frac{-H}{k_{B}T}}[/itex][/B]

H = Hamiltonian
[itex]k_{B}[/itex]=Boltzmann constant

Einstein: [itex]\delta W = < \delta H >[/itex]


The Attempt at a Solution


[/B]
(1) For an ideal gas, the system hamiltonian (with the potential term added) is:

[itex]H = \sum \frac{p_{i}^2}{2m} + \lambda |r|[/itex]

so,

[itex]Z = \sum e^{\frac{-(p_{x}^2 + p_{y}^{2})}{2mk_{B}T} - \frac{\lambda |r|}{k_{B}T}}[/itex]

How is this sum computed for the canonical partition function of a 2-D ideal gas with the hamiltonian included?

Also, for (2),

[itex]\delta W = < \delta \lambda H > = \delta \lambda < |r| >[/itex]

this answer is wrong, but I really don't see what I'm missing here.

Any help would be appreciated, thanks.
 
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