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 vaporization of water at \(25^{\circ}\mathrm{C}\) involves a larger change in entropy compared to vaporization at \(100^{\circ}\mathrm{C}\) as calculated through the formula \(\Delta S = (\Delta H_{vap}) / T\). The entropy change in either process does not depend on whether the process is carried out reversibly or not, as the formula is based on the system's properties independent of the process path.

Step-by-step solution

Calculate entropy change for 25°C and 100°C

To calculate the entropy change for vaporizing water at both temperatures, we will use the formula: ΔS = (ΔH_vap) / T where ΔS is the change in entropy, ΔH_vap is the enthalpy of vaporization, and T is the absolute temperature (in Kelvin). Since we assume that the enthalpy of vaporization is constant for both temperatures, the difference in entropy change will be influenced by the temperatures. First, we need to convert the given temperatures to Kelvin: For 25°C: T1 = 25 + 273.15 = 298.15 K For 100°C: T2 = 100 + 273.15 = 373.15 K Now, we can calculate the entropy change for both temperatures using the same ΔH_vap: ΔS1 = (ΔH_vap) / T1 ΔS2 = (ΔH_vap) / T2

Compare the entropy change for both temperatures

To determine which process involves a larger change in entropy, we will compare ΔS1 and ΔS2: If ΔS1 > ΔS2, then vaporizing water at 25°C involves a larger change in entropy. If ΔS1 < ΔS2, then vaporizing water at 100°C involves a larger change in entropy. Since ΔH_vap is constant for both temperatures, we can see that the ratio of ΔS is inversely proportional to the ratio of T: ΔS1 / ΔS2 = T2 / T1 = 373.15 / 298.15 ≈ 1.25 Since ΔS1 / ΔS2 > 1, we can conclude that ΔS1 > ΔS2, meaning that vaporizing water at 25°C involves a larger change in entropy.

Determine the dependency of entropy change on reversibility

Entropy change depends on the heat transfer that occurs during a process. In the case of reversible processes, the system always remains in equilibrium, and the heat transfer is evenly distributed throughout the process. The entropy change, in this case, is only dependent on the system's properties and is calculated using the formula (ΔH_vap) / T. In irreversible processes, the system does not remain in equilibrium, and there might be additional sources of entropy generation, such as friction, turbulence, or non-uniform heat transfer. In these cases, the entropy change for the process would be greater than that for the equivalent reversible process. However, it is important to note that the entropy change calculated using (ΔH_vap) / T for each temperature is the same for both reversible and irreversible processes. This is because the formula is based on the system's properties and does not consider the process path. Therefore, the entropy change in either process (at 25°C or 100°C) would not depend on whether we carry out the process reversibly or not. The entropy change calculated from the formula represents the minimum entropy change, which is associated with the reversible process.

Key concepts

Enthalpy of Vaporization

When we talk about the enthalpy of vaporization, we are discussing the quantity of heat required to convert a unit mass of a liquid into a gas at constant pressure. This process is significantly important in thermodynamics because it provides us with a measure of the energy needed to overcome intermolecular forces that hold liquid molecules together.

For water, the enthalpy of vaporization is relatively high, which is why boiling water requires considerable amounts of energy. It is commonly denoted as \( \Delta H_{\text{vap}} \) and is measured in joules per mole (J/mol) or calories per gram (cal/g).

In our case, we're assuming that the enthalpy of vaporization doesn't change much between 25°C and 100°C, which is a reasonable assumption since \( \Delta H_{\text{vap}} \) typically varies slightly with temperature within a certain range. The understanding of this concept is crucial for calculating the entropy change during the vaporization process.

Reversible Processes

The concept of reversible processes is fundamental in understanding the efficiency of thermodynamic systems. A reversible process is an idealization in which the system undergoes changes in such a way that the system and the surroundings can be returned to their original states without leaving any net change.

These types of processes progress infinitely slowly to maintain equilibrium at each stage. When considering entropy change, the path of the process becomes highly relevant. In a reversible process, the entropy change of the system is precisely balanced by the entropy change of the surroundings, resulting in no net entropy change for the universe. This characteristic is critical when determining the minimum possible entropy change for any process, including vaporization.

While no real process is truly reversible, this concept serves as an ideal standard to gauge the performance of actual processes and to calculate the maximum work achievable by a thermodynamic system.

Thermodynamic Temperature

Thermodynamic temperature is one of the principal parameters of thermodynamics. It represents an absolute scale for measuring thermal energy in a system. The thermodynamic temperature provides us with an understanding of the energy at the molecular level, and it is directly related to the kinetic energy of molecues in a system.

The Kelvin scale, which is an SI base unit of temperature, is used to represent thermodynamic temperature. One key aspect of using Kelvin is that it starts at absolute zero, which is the point where no more thermal energy can be extracted from a substance.

In the context of our problem, the use of thermodynamic temperature \( T \) in Kelvin is crucial because it ensures the calculated entropy change is independent of the chosen temperature scale and solely dependent on the physical properties of the system in question. Moreover, it allows us to accurately compare the entropy change at different temperatures, as the Kelvin scale provides an absolute reference point.