Question

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

Short answer

The larger entropy change occurs when vaporizing water at \(25^{\circ} \mathrm{C}\) compared to \(100{ }^{\circ} \mathrm{C}\) since the ratio of the entropy changes is greater than 1. The entropy change in either process does not depend on whether the process is carried out reversibly or not, as it is determined by enthalpy change and temperature during the process.

Step-by-step solution

Calculate the entropy change

In both cases, the process can be considered as isothermal since the temperature remains constant throughout the process (either at \(25^{\circ} \mathrm{C}\) or at \(100{ }^{\circ} \mathrm{C}\)). For an isothermal process, we can calculate the entropy change using the relation: \[ \Delta S = \frac{\Delta H}{T} \] Where \(\Delta S\) is the entropy change, \(\Delta H\) is the enthalpy change, and \(T\) is the temperature.

Calculate the entropy change at 25°C and 100°C

For process (a) at \(25^{\circ} \mathrm{C}\) (\(298 K\)), and process (b) at \(100{ }^{\circ} \mathrm{C}\)(\(373 K\)), we can calculate the entropy change using the given relation: \[ \Delta S_{25} = \frac{\Delta H}{298 K} \] \[ \Delta S_{100} = \frac{\Delta H}{373 K} \] Since the enthalpy of vaporization is assumed to be constant, we can compare the ratios of the entropy changes directly.

Compare the entropy changes

From the equations in step 2, we can observe that \[\frac{\Delta S_{25}}{\Delta S_{100}} = \frac{373}{298}\] Since the ratio is greater than 1, the entropy change \(\Delta S_{25}\) at \(25^{\circ} \mathrm{C}\) is larger than the entropy change \(\Delta S_{100}\) at \(100{ }^{\circ} \mathrm{C}\).

Effects of a reversible or irreversible process

Since the entropy change is given by the ratio of the enthalpy change to the temperature, this relationship holds in both reversible and irreversible processes, as long as the temperature and enthalpy are constant during the process. Therefore, the entropy change in either process does not depend on whether the process is carried out reversibly or not. In summary: (a) The entropy change is larger when vaporizing water at \(25^{\circ} \mathrm{C}\) compared to vaporizing water at \(100{ }^{\circ} \mathrm{C}\). (b) The entropy change in either process does not depend on whether the process is carried out reversibly or not.