Irreversible Processes and Entropy — Question 2

Question 2

Free expansion creates entropy even though no heat crosses the boundary

One mole of ideal gas expands into vacuum inside a rigid insulated enclosure from volume $V$ to $3V$. Its internal energy depends only on temperature. To evaluate entropy, use any reversible reference path between the same equilibrium endpoints, with $dS=\delta Q_{\mathrm{rev}}/T$ and $R=8.314$ J/(mol K).

1. Find the actual heat, work and temperature change.

2. Use a reversible isothermal reference path to calculate entropy change. Identify the heat along that reference path at 300 K and explain why it is not the actual heat.

Question 2 - Solution

Free expansion creates entropy even though no heat crosses the boundary

(a) Method and calculation

$$Q_{\mathrm{actual}}=0,\quad W_{\mathrm{actual}}=\int p_{\mathrm{ext}}dV=0,\quad\Delta U=0\ \Rightarrow\Delta T=0.$$$$p_f/p_i=V_i/V_f=1/3.$$

(b) Interpretation and further calculation

$$\Delta S=\int_V^{3V}\frac{nRT\,dV/V}{T}=nR\ln3\approx9.13386\,\mathrm{J/K}.$$$$Q_{\mathrm{rev}}=nRT\ln3\approx2740.16\,\mathrm J\quad(T=300\,\mathrm K).$$The reference path is a calculation device; entropy depends on endpoints, whereas heat depends on the actual process.

Physical check

Adiabatic does not mean isentropic when the process is irreversible.