Thermo attributes of ideal gas in 3D harmonic potential

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SonOfOle
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Homework Statement


A classical system of N distinguishable, non-interacting particles of mass m is placed in a 3D harmonic potential,

[tex]U(r) = c \frac{x^2 + y^2 + z^2}{2 V^{2/3}}[/tex]

where V is a volume and c is a constant with units of energy.

(a) Find the partition function and the Helmholtz free energy of the system.

(b) Find the entropy, the internal energy and the total heat capacity at a constant volume for the system.

Homework Equations


[tex]Z = \Sigma \exp{-E_i / kT}[/tex]

[tex]H = U - TS[/tex]

[tex]f(E)= A \exp{-E/kT}[/tex]


The Attempt at a Solution



Unfortunately, I'm not sure where to start on this one. Anybody able to give me a tip in the right direction?
 
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Your system is a classical one, so the appropriate equation for the partition function is

[tex]Z = \int d\mathbf{r}_1 d\mathbf{r}_2 \dots d\mathbf{r}_N d\mathbf{p}_1 d\mathbf{p}_2 \dots d\mathbf{p}_N \exp\left[-\frac{E(\mathbf{r},\mathbf{p})}{k_BT}\right][/tex]

The energy is

[tex]E(\mathbf{r},\mathbf{p}) = \sum_{i=1}^{N}\left[\frac{\mathbf{p}_i^2}{2m} + U(\mathbf{r}_i)\right].[/tex]

This should get you started with that part.

For the free energy, instead of using [itex]F = U - TS[/itex], it's easy to use the equation

[tex]F = -k_BT \ln Z[/tex]

(it's more common to use F than H to denote the Helmholtz free energy, as H is typically used for the enthalpy).
 
Mute said:
Your system is a classical one, so the appropriate equation for the partition function is

[tex]Z = \int d\mathbf{r}_1 d\mathbf{r}_2 \dots d\mathbf{r}_N d\mathbf{p}_1 d\mathbf{p}_2 \dots d\mathbf{p}_N \exp\left[-\frac{E(\mathbf{r},\mathbf{p})}{k_BT}\right][/tex]

The energy is

[tex]E(\mathbf{r},\mathbf{p}) = \sum_{i=1}^{N}\left[\frac{\mathbf{p}_i^2}{2m} + U(\mathbf{r}_i)\right].[/tex]

(it's more common to use F than H to denote the Helmholtz free energy, as H is typically used for the enthalpy).

Still stuck. How would the integration over the momentums work? That is, what would the [tex]\mathbf{p_i}^2[/tex] be?
 
[tex]\mathbf{p}_i^2 = p_{i,x}^2 + p_{i,y}^2 + p_{i,z}^2[/tex]

Hence, you end up with a sum of squares of terms in the exponential that can all be split into products of exponentials.