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Van der Waals gas The thermal equation of state of van der Waals gas is given by p+ NA2 V2a ! (V −NAb) =NAkBT

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Physics Department Paderborn University

Theoretical Physics - Thermodynamics

Dr. H.T. Duc, Prof. Dr. W.G. Schmidt, WS 2008/2009 Sheet I

Preparation until Tuesday, 21.10.2008 1. dV Total

A characteristic feature of thermodynamic systems is that, in spite of many microscopic degrees of freedom, they can be described by a few variables.

a) Sum up the notations ”thermodynamic parameter” and ”minimal ensemble of thermodynamic parame- ters”.

b) A homogeneous system is defined by the thermodynamic parameterspandT. Are the following variables V and Qthermodynamic parameters as well?

δV = νR p dT −

νRT

p2 dp (1)

δQ= 5

2νRdT − νRT

p dp (2)

c) Determine, if necessary, for V and for Q respectively, an integrating factor µ(p, T) and calculate the integrals µV (µQ). Hint: the integrals are path-independent. (Why?)

2. Van der Waals gas

The thermal equation of state of van der Waals gas is given by p+ NA2

V2a

!

(V −NAb) =NAkBT. (3)

From the relation between thermal and caloric equation of state in pVT-systems, it can be shown (Section 5.1 in the Lecture) that

dU =CV(V, T)dT +

T ∂p

∂T

V −p

dV. (4)

Use this and the fact that dU is an exact differential, to prove the dependence of the internal energy U =U(V, T) on temperature and volume.

3. Black body

The radiation field in a black body with volumeV is in equilibrium with the walls which are kept in constant temperatureT. The radiation energy density is simply a function of temperature, so that the internal energy has a general form

U(V, T) =V f(T), (5)

where the function f depends only on T. The radiation pressure is given by p= 1

3bT4. (6)

How does the function f look like? Hint: use Eq. (4).

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