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Alternative derivation of ideal gas law

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The fundamental thermodynamic relation expresses dU in terms of dS and dV

dU = T dS - P dV (1)

If we want to consider U as a function of T and V, then all we need to do is rewrite this in terms of dT and dV. We substitute:

dS = dS/dT dT + dS/dV dV (2)

Here the derivatives are partial derivatives, the derivative w.r.t. T is at constant V and vice versa. Substituting (2) in (1) gives:

dU = T dS/dT dT + (T dS/dV - P) dV

The term T dS/dT is the heat capacity at constant volume. The term T dS is always the heat supplied to the system in case of a quasistatic change, so the T times the derivative of S w.r.t. Temperature is the heat capacity. Because we keep V constant in the derivative, it is the heat capacity at constant volume. So, we can write:

dU = Cv dT + (T dS/dV - P) dV (3)

We can simplify the term dS/dV using a Maxwell relation as follows (note that this derivative is at constant T). Starting from the fundamental relation:

dU = T dS - P dV

we switch S and T in the differentials using the Leibniz rule:

dU = d(TS) - S dT - P dV -------------->

d(U - T S) = - S dT - P dV

Then the symmetry of the second derivatives of F = U - T S implies that:

dS/dV = dP/dT

Where the derivative w.r.t. V is at constant T and the derivative w.r.t. T is at constant V. If we substitute this in (3) we get:

dU = Cv dT + (T dP/dT - P) dV (4)

If you integrate this along some arbitrary path from the point (T1, V1) to (T2, V2) you get the difference in the internal energy between these two points. The fact that ...

Solution Summary

The problem is solved in detail from first principles.