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. \((\mathbf{c})\) What is the "driving force" for the expansion of the gas: enthalpy or entropy?

For a process to be spontaneous at constant temperature and pressure, the change in Gibbs free energy (∆G) must be negative. The change in Gibbs free energy is calculated using the following formula: ∆G = ∆H - T∆S However, in an isothermal process, the temperature (T) is constant. Since the expansion is happening into a vacuum, the pressure (P) is also constant. Therefore, the change in internal energy (∆E) must be equal to zero, as given in the problem statement. Thus, we have: ∆G = (∆E - P∆V) - T∆S Since ∆E = 0, we have: ∆G = -P∆V - T∆S To find the sign of ∆G, we'll need to determine the signs of both P∆V and T∆S. In this case, the change in volume is positive (∆V > 0) because the gas is expanding. However, since the expansion is happening into a vacuum, the pressure is effectively zero (P = 0), meaning P∆V = 0. Additionally, as a gas expands isothermally, its entropy increases (∆S > 0), which results in T∆S being positive. Considering these observations, we conclude that ∆G < 0, indicating that the process is a spontaneous process.

The formula to calculate work for a gas expansion is: w = -P∆V As mentioned earlier, since the expansion is happening into a vacuum, the pressure is effectively zero (P = 0). This results in: w = -0 * ∆V Thus, w = 0. In other words, no work is done by the system during this process because the pressure is zero.

The driving force behind an expansion process is determined by considering the contributions of enthalpy (∆H) and entropy change (∆S) towards the Gibbs free energy change (∆G). We previously established that the change in Gibbs free energy looks like this for our given situation: ∆G = -P∆V - T∆S Because the expansion is happening into a vacuum, the pressure is effectively zero, so -P∆V = 0. Therefore, only the entropy term contributes to the change in Gibbs free energy: ∆G = -T∆S Because ∆G is negative for a spontaneous process, and we already determined that this process is spontaneous, it means that this process is entropy-driven. The expansion of the gas is driven by an increase in entropy, which corresponds to the gas particles spreading out and occupying a larger volume. Consequently, the "driving force" for the expansion of the gas in this case is entropy, not enthalpy.