For the isothermal expansion of a gas into a vacuum, $\Delta E=0,q=0,$ and $w=0.$ (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) What is the "driving force" for the expansion of the gas: enthalpy or entropy?

Step by Step Solution: Step 1. Under isothermal expansion $\Delta E=0$ and since $q=0,w=0$, this is an ideal gas process. Step 2. Spontaneous processes are those that occur on their own without any external influence. Since there is no $\Delta E$ or $q$ for heat transfer, the expansion of gas into a vacuum is not a spontaneous process. Step 3. No work is done by the system since the expansion is into a vacuum. In other words, there is no pressure against which the gas has to work to expand. Hence, $w=0$. Step 4. The driving force for the expansion of gas into a vacuum is entropy. The second law of thermodynamics states that in any process, the entropy of the universe increases or remains constant. In this case, the entropy of the gas increases as it expands into the vacuum (i.e., the number of available microstates for the gas particles increases). Therefore, the process is spontaneous in terms of an increase in entropy.