How is the Internal Energy of a Photon Gas Related to Temperature?

[Jenkz]

Homework Statement

The equation of state PV= U/3 applies to a photon gas at pressure P, volume V and internal energy U. By using the 'energy equation', show that the internal energy of a phont gas is proportional to the fourth power of temperature (i.e U $\propto$T$^{4}$). You can assume that U/V is only a function of temperature.

Homework Equations

Energy equation: (dU/dvV)$_{T}$ = T (dP/dT)$_{V}$ - P

The Attempt at a Solution

U = 3PV
(dU/dV)$_{T}$ = 3P as (dP/dV)$_{T}$ = 0 as it is only a function of T

From here.. I'm not sure what to do. Help?

 

[dynamicsolo]

Homework Statement

The equation of state PV= U/3 applies to a photon gas at pressure P, volume V and internal energy U. By using the 'energy equation', show that the internal energy of a phont gas is proportional to the fourth power of temperature (i.e U $\propto$T$^{4}$). You can assume that U/V is only a function of temperature.

Homework Equations

Energy equation: (dU/dV)$_{T}$ = T (dP/dT)$_{V}$ - P

The Attempt at a Solution

U = 3PV
(dU/dV)$_{T}$ = 3P and (dP/dV)$_{T}$ = 0 as it is only a function of T

From here.. I'm not sure what to do. Help?

(typoes corrected)

There is nothing in what you are told that allows you to state that $(\frac{dP}{dV})_{T} = 0$ . This would lead you to the result 3P = -P .

What you know is that $P = \frac{U}{3V}$ ; we also know that U must depend on temperature T , particularly since we're being asked to find a proportionality for it. You need to differentiate P implicitly with respect to T , with V held constant; this will give you a relation between $\frac{\partial P}{\partial T}$ and $\frac{\partial U}{\partial T}$ . Use this to replace $\frac{\partial P}{\partial T}$ in your energy relation, bring all the P's to one side, and use the given relation U = 3PV to replace P . You will have a separable differential equation that will lead you to the proportionality you seek.