For the process H2O(l) - H2O(g) at T = 100°C and 1 atmosphere pressure, the correct choice is
(A) Δ

Q: Why is the boiling of water considered a reversible process here?

Because the process is occurring exactly at the normal boiling point ($100^\circ \text{C}$ at $1 \text{ atm}$). At this specific temperature and pressure, the liquid and vapor phases are in dynamic equilibrium, making the phase transition thermodynamically reversible.

Q: Doesn't the entropy of the universe always increase?

The entropy of the universe always increases for spontaneous, irreversible processes. However, for a perfectly reversible process (like a phase change at equilibrium), the entropy of the universe remains constant, meaning $\Delta S_{\text{universe}} = 0$.

Q: Why does the entropy of the system increase?

The system is converting from a liquid to a gas. Gas molecules have much more freedom of movement and are far more disordered than liquid molecules. This increase in randomness means $\Delta S_{\text{system}}$ is positive.

Q: Why does the entropy of the surroundings decrease?

To boil the water, the surroundings must supply heat to the system. Because the surroundings are losing heat energy ($q_{\text{surr}}$ is negative), their thermal disorder decreases, leading to a negative $\Delta S_{\text{surroundings}}$.

Q: What if the water was boiling at $110^\circ \text{C}$ at $1 \text{ atm}$?

At $110^\circ \text{C}$ and $1 \text{ atm}$, liquid water is unstable and the vaporization would be a spontaneous, irreversible process. In that case, $\Delta S_{\text{universe}} > 0$, and you could not simply say $\Delta S_{\text{surroundings}} = -\Delta S_{\text{system}}$.

Question

For the process $H_2O(l) ightarrow H_2O(g)$ at $T = 100^\circ ext{C}$ and $1$ atmosphere pressure, the correct choice is Options: (A) $\Delta S_{ ext{system}} > 0$ and $\Delta S_{ ext{surroundings}} > 0$ (B) $\Delta S_{ ext{system}} > 0$ and $\Delta S_{ ext{surroundings}} < 0$ (C) $\Delta S_{ ext{system}} < 0$ and $\Delta S_{ ext{surroundings}} > 0$ (D) $\Delta S_{ ext{system}} < 0$ and $\Delta S_{ ext{surroundings}} < 0$

Sigma Percentile · Accepted Answer

### The Thermodynamics of Boiling Water: A Tale of Two Entropies

Imagine a beaker filled with water, placed on a hot stove. The water is bubbling and boiling vigorously. This everyday process of liquid water converting into steam is happening at exactly $100^\circ 	ext{C}$ and $1 	ext{ atm}$ of pressure. Now, if you recall your basic chemistry, these are the exact conditions for the normal boiling point of water. This physical setup is the key to unlocking the thermodynamic mystery of this problem.

### The Reversibility Catch

There is a crucial catch here that many students miss. When a phase transition, like boiling or melting, occurs exactly at its normal transition temperature and pressure, it is not just any ordinary process. It is a perfectly **reversible process**.

At $100^\circ 	ext{C}$, liquid water and water vapor are in a state of dynamic equilibrium. The system (our water) and the surroundings are perfectly balanced thermally. If you were to drop the temperature by an infinitesimally small amount, the steam would condense. If you raised it infinitesimally, more water would boil. This delicate balance is the hallmark of reversibility.

### The Universe's Balance Sheet

Now, let us bring in the heavy machinery of thermodynamics. The Second Law tells us about the entropy of the universe. For any spontaneous, irreversible process, the entropy of the universe increases. But for a strictly reversible process, the total entropy change of the universe is exactly zero.

The universe is simply the sum of our system and the surroundings. Therefore, we can write our master equation:

$$\Delta S_{	ext{universe}} = \Delta S_{	ext{system}} + \Delta S_{	ext{surroundings}} = 0$$

### The System's Chaos

Let us zoom into the system. We are going from a liquid state to a gaseous state. In a liquid, molecules are somewhat restricted by intermolecular forces. But in a gas, they are free to move chaotically anywhere they want!

The randomness, the disorder, is massively increasing as the liquid vaporizes. Therefore, the entropy change of the system must be strictly positive:

$$\Delta S_{	ext{system}} > 0$$

### The Surroundings' Sacrifice

Let us look back at our universe equation. The sum of the two entropies is zero. If the system's entropy is a positive number, the only mathematical way for their sum to be zero is if the surroundings' entropy is a negative number of the exact same magnitude.

$$\Delta S_{	ext{surroundings}} = - \Delta S_{	ext{system}}$$

Physically, what is happening? The surroundings (the stove, the air) are losing heat to boil the water. Because they are losing thermal energy ($q_{	ext{surr}} < 0$), their entropy drops. Therefore:

$$\Delta S_{	ext{surroundings}} < 0$$

We have our two pieces of the puzzle. The entropy of the system increases, and the entropy of the surroundings decreases. This perfectly matches Option (B). Always remember to check if the phase change is at its normal transition point before applying the reversible universe rule!

Animated Solution for Chemistry - Chemical Thermodynamics: For the process $H_{2} O (l) \to H_{2} O (g)$ at $T = 10 0^{\circ} C$ and $1$ atmosphere pressure, the correct choice is

Select Answer:

Visualized Solution

$H_{2} O (l) \to H_{2} O (g)$

$Reversibility of Phase Change$

$Δ S_{universe} = 0$

$Δ S_{system} > 0$

$Δ S_{surroundings} < 0$

$Conclusion$

The Sigma Insight: Entropy and Free Energy

The Thermodynamics of Boiling Water

The Reversibility Catch

The Universe's Balance Sheet

The System's Chaos

The Surroundings' Sacrifice

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Animated Solution for Chemistry - Chemical Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the process H2​O(l)→H2​O(g) at T=100∘C and 1 atmosphere pressure, the correct choice is