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Internal Energy for Nonrelativistic and Relativistic Gasses

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(i) Show that the equation of state of an ideal gas is still PV = RT even when the gas is heated to such a high temperature that the particles are moving at relativistic speeds. (Hint: What feature of the partition function of the ideal gas determines the gas law?).

(ii) Although the equation of state does not alter when the particles in a monatomic ideal gas start to move at relativistic speeds, show that in the formula for an adiabat, PV = constant, lambda in the relativistic limit is 4/3, rather than 5/3 as in the non-relativistic case.

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Outline of solution:

The pressure is given by:

P = 1/beta dLog[Z(N)]/dV (0.1)

Here Z(N) is the partition function of the gas containing N particles. The derivation is given in the Appendix. For a dilute gas, Z(N) is given in terms of the partition function for a single particle, Z1, by:

Z(N) = Z1^N/N! (0.2)

This is derived in Section 1 below.

Z1 for an extremely relativistic gas is given by:

Z1 = V/h^3 Integral over momentum space d^3p exp[- beta |p| c] = V K

where K is the integral over momentum space which does not depend on the volume. Inserting this in (0.2) gives:

Z(N) = V^N K^N


Log[Z(N)] = N Log(V) + terms that do not depend on V.

Inserting this in (0.1) gives:

P = 1/beta dLog[Z(N)]/dV = N k T/V ---------->

P V = N k T

The internal energy of a dilute nonrelativistic gas in terms of P and V is given by

E = 3/2 P V

see section 2. The internal energy for a relativistic gas in terms of P and V is given by a different equation:

E = 3 P V

This is derived in Section 3.

To compute the relation between P and V in an adiabat we can proceed as follows. Along with an adiabat, the entropy stays the same. Recall that, by definition, during adiabatic changes, the system is assumed to be insulated so that no heat is exchanged with the environment. Also, it is assumed that the changes in the system are slow enough so that the changes do not cause irreversible changes leading to entropy increase. The Adiabatic Theorem of Quantum Mechanics:


says that in the limit of infinitely slow changes, the system will remain in the same energy level. The energy of the system will then only change because the energy of the energy level the system is in depends on the volume or other external parameters which are changed. This means that the number of microstates compatible with the macrostates will not change and therefore the entropy will remain the same. Note that in High School, students are often told that adiabatic change means a fast change of volume. This is because in practice the system may not be insulated very well. If the volume changes fast then little heat will be exchanged. The irreversible entropy increase due to the fact that the Adiabatic Theorem is not valid is in many cases, not a large effect.

So, we need to compute P as a function of V assuming that the entropy S stays constant.

The fundamental law of thermodynamics is:

d E = T dS - P d V ---------->

T dS = dE + P d V

Constant entropy means dS = 0. Therefore:

dE + P dV = 0

Inserting E = g P V in here (g = 3/2 for a classical gas and g = 3 for a relativistic gas), gives:

g d(PV) + p dV = 0 ---->

g V dP + (g+1)P dV = 0

Divide by g P V:

dP/P + (g+1)/g dV/V = 0


Log(P) + (g+1)/g Log(V) = constant ---->

P V^[(g+1)/g] = constant.

If g = 3/2 then gamma = (g+1)/g = 5/3. For a relativistic gas g = 3 and then gamma = 4/3

Detailed derivations from first principles

Table of contents:

Section 1: Partition function of a nonrelativistic gas

Section 2: Energy and Pressure of a dilute nonrelativistic ideal gas

Section 3: Energy and Pressure of a dilute relativistic ideal gas

Appendix: Derivation of P = 1/beta dLog[Z(N)]/dV


1 Partition function of a nonrelativistic gas

The partition function is in general given by:

Z = Sum over r of Exp(- beta E_r) (1.1)

Here r enumerates the energy eigenstates of the system, E_r is the energy ...

Solution Summary

This solution explains the calculations for relativistic and nonrelativistic gasses in 2683 words.