PX154 - E6c - adiabatic processes

reversible

Δ S = \int_{s t a t e 1}^{s t a t e 2} \frac{d Q}{T}

however, $Δ Q = 0$ for an adiabatic process, so:

Δ S = 0

isentropic process: entropy doesn't change

irreversible

free expansion of an ideal gas [YF ex 20.8]
process is adiabatic: $Q = 0$
no work is done by the gas as there is no piston to push: $W = 0$
no change in internal energy: $Δ U = 0 ⟹ T_{2} = T_{1}$
does the entropy change?
- process is irreversible, so, come up with a reversible isothermal process that achieves the same change in state
- start and end points identical, but, the isothermal process requires:

\begin{aligned} Q = W & = n R T \ln \frac{V_{2}}{V_{1}} \\ Δ S & = n R \ln \frac{V_{2}}{V_{1}} \\ = n R \ln 2 \end{aligned}

- the entropy increases