Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process $\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)$ a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and $\Delta S_{\text { univ }}$ for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and $\Delta S_{\text { univ }}\( for the process in vessel \)2 .$ (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

Step by step solution

01

Understand the process in Vessel 1

Vessel 1 contains an ice cube and water. As the ice cube receives energy, it goes through a phase change and becomes liquid water. We want to determine what happens to the entropy during this transformation.

02

Calculate the sign of ∆S for Vessel 1

The process of melting (H2O(s) → H2O(l)) involves a change from a more ordered state (solid) to a less ordered state (liquid). As a result, the entropy change, ∆S, for this process will be positive.

03

Determine the sign of ∆S_sum for Vessel 1

Both the system and surroundings are perfectly insulated, meaning no exchange of energy with the outside environment. Thus, the overall entropy change (∆S_sum) is the same as the entropy change of the system (∆S). In this case, ∆S_sum is positive as well.

04

Determine the sign of ∆S_univ for Vessel 1

Under these conditions, the universe consists only of the system (Vessel 1) and not the surroundings. Therefore, the entropy change of the universe, ∆S_univ, is equal to the entropy change of the system, ∆S. For this situation in Vessel 1, ∆S_univ is also positive. b. Vessel 2: Ice and Saltwater mixture

05

Understand the process in Vessel 2

Vessel 2 contains an ice cube and saltwater solution. Due to the salt, the freezing point of water is lowered. We want to determine what happens to the entropy during the ice melting process in this case.

06

Calculate the sign of ∆S for Vessel 2

Similar to Vessel 1, in the process of ice melting (H2O(s) → H2O(l)), the ice cube turns into a liquid. This results in a more disordered state. Thus, the entropy change, ∆S, for this process in Vessel 2 will also be positive.

07

Determine the sign of ∆S_sum for Vessel 2

Like in Vessel 1, the system and surroundings are perfectly insulated, meaning no exchange of energy with the environment. Therefore, the overall entropy change (∆S_sum) is the same as the entropy change of the system (∆S). In this case, ∆S_sum is also positive.

08

Determine the sign of ∆S_univ for Vessel 2

Again, considering that the universe only consists of the system (Vessel 2), the entropy change of the universe, ∆S_univ, is equal to the entropy change of the system, ∆S. For this situation in Vessel 2, ∆S_univ is also positive.

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