Question

The free expansion of an ideal gas is an adiabatic process and so no heat is transferred. No work is done, so the internal energy does not change. Thus,$\frac{\mathbf{Q}}{\mathbf{T}}\mathbf{=}\mathbf{0}$, yet the randomness of the system and thus its entropy have increased after the expansion. Why does Eq.$\left(20.19\right)$not apply to this situation?

Short answer

The equation 20.19 is not valid in this case.

Step-by-step solution

Step 1:Concept of ideal gas expansion

In an ideal gas, all the collisions between molecules or atoms are perfectly elastic and no intermolecular force of attraction exists in an ideal gas because of the molecules of an ideal gas move so fast, and they are so far away from each other that they do not interact at all.

Process of internal energy

The free expansion of an ideal gas is an adiabatic process and so no heat is transferred. No work is done, so the internal energy does not change.

$$\begin{gathered} ∆\mathrm{Q}=0 \\ ∆\mathrm{U}=0 \end{gathered}$$

Thus, $\frac{\mathrm{Q}}{\mathrm{T}}=0$, yet the randomness of the system and thus its entropy have increased after the expansion.

The equation 20.19 is for the reversible processes.

$∆\mathrm{S}={\int }_1^2\frac{\mathrm{dQ}}{\mathrm{T}}$

The free expansion of the gas is an irreversible.

Therefore, the equation 20.19 is not valid in this case.