A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the con

Exercises · 6.1

Q. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

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Concept understanding — Phase Equilibrium

Phase Equilibrium: From Intuition to Precision

Imagine a glass of ice water on a warm day. The ice cubes are melting, but the water around them stays cold. If you wait long enough, the ice stops melting — not because the room got colder, but because the system has reached a balance. The ice and water coexist without any net change. That balance is phase equilibrium.

The Intuition: A Tug-of-War Between Two Tendencies

Every substance wants to exist in the phase (solid, liquid, or gas) that has the lowest energy under the given conditions. But there's a catch: nature also loves disorder (entropy). A solid is low-energy but highly ordered; a gas is high-energy but very disordered. At a specific temperature and pressure, these two opposing drives — minimising energy and maximising disorder — exactly cancel each other out. The result? Two phases can coexist peacefully, with molecules constantly jumping between them but no net change in the amount of each phase.

Think of it as a tug-of-war. On one side, molecules in the solid phase are tightly bound and low in energy. On the other side, molecules in the liquid phase have more freedom and higher entropy. At the melting point, the rope doesn't move — the rates of melting and freezing are equal.

The Precise Statement

Phase equilibrium is the condition in which two or more phases of a substance coexist in thermodynamic equilibrium, with no net transfer of mass between phases. For a pure substance, this occurs at a unique temperature for a given pressure (or vice versa), defined by the equality of the chemical potential (or Gibbs free energy per mole) in all phases.

In simpler terms: at equilibrium, the driving force for a molecule to move from phase A to phase B is exactly the same as the driving force to move from B to A. The system is dynamic — molecules are still crossing the boundary — but macroscopically static.

The Key Condition: Equal Chemical Potential

The rigorous condition for phase equilibrium in a pure substance is:

\mu_{\text{solid}} = \mu_{\text{liquid}} = \mu_{\text{gas}}

where $\mu$ (mu) is the chemical potential — the Gibbs free energy per mole. When chemical potentials are equal, no phase has a "preference" to grow at the expense of another.

Important

For a pure substance, phase equilibrium occurs only along specific lines on a phase diagram (the melting curve, boiling curve, and sublimation curve). At the triple point, all three phases coexist.

A Concrete Example: Water at 0°C

At 1 atm pressure and 0°C, ice and liquid water are in equilibrium. Here's what's happening at the molecular level:

Water molecules at the surface of the ice are vibrating. Some gain enough energy to break free and enter the liquid.
Simultaneously, water molecules in the liquid near the ice surface lose energy and attach to the ice crystal.
The rates of these two processes are equal. So the amount of ice and the amount of liquid water remain constant — even though individual molecules are constantly switching sides.

If you raise the temperature slightly above 0°C, the melting rate exceeds the freezing rate — the ice melts completely. Below 0°C, freezing dominates. Only at exactly 0°C (at 1 atm) do you get stable coexistence.

Why This Matters

Phase equilibrium is the foundation for understanding:

Phase diagrams — the maps that show which phase is stable at any temperature and pressure.
Boiling and condensation — the liquid-vapour equilibrium curve.
Sublimation — solid-vapour equilibrium (like dry ice "evaporating" without melting).
Colligative properties — how solutes shift equilibrium temperatures (freezing point depression, boiling point elevation).

The One Thing to Remember

Note

Phase equilibrium is not a static state where nothing happens. It is a dynamic balance where opposing processes occur at equal rates. The system appears unchanged because the forward and backward rates are identical.

When you see a phase diagram, every line on it represents a condition of phase equilibrium — a specific temperature and pressure where two phases can coexist indefinitely. That's the power of this concept: it tells you exactly where nature's tug-of-war ends in a draw.

A quick search for "Phase Equilibrium class 11 chemistry" or "NCERT chemistry syllabus phase equilibrium" will confirm what's true here: this concept is a standard, curriculum-aligned part of Class 11 Chemistry. Given how often it's tested in JEE Main, NEET and state CET Chemistry papers, it's worth revisiting this explanation until the reasoning feels automatic, not just the final formula.

When the container volume is suddenly increased at constant temperature, the vapour pressure initially drops because the same number of vapour molecules now occupy a larger space. This makes condensation slower than evaporation, so net evaporation occurs until the original equilibrium vapour pressure is restored — the final vapour pressure is the same as before.

The Core Idea: Phase Equilibrium at Fixed Temperature

A liquid and its vapour in a sealed container at a fixed temperature are in dynamic equilibrium. This means two opposing processes — evaporation and condensation — occur at equal rates. The vapour pressure at this point is called the saturated vapour pressure (SVP), and it depends only on the temperature, not on the container volume. That last point is the key to the entire problem.

Why does SVP depend only on temperature? Because at a given temperature, the molecules in the liquid have a fixed average kinetic energy. The number that can escape into the vapour (evaporation) and the number that return (condensation) balance at a specific vapour density. Since pressure is proportional to density at constant temperature (ideal gas law), the equilibrium pressure is fixed.

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Now let's apply this to the three parts of the question.

1. Initial effect on vapour pressure when volume is increased

The container is sealed — no vapour can escape. When you suddenly increase the volume, the number of vapour molecules $n$ remains the same (for an instant, before any net evaporation or condensation occurs). The temperature is fixed, so the ideal gas law applies:

P V = nRT

If $V$ increases and $n$ and $T$ are unchanged, $P$ must decrease immediately. So the vapour pressure drops below the saturated vapour pressure.

Watch out

A common mistake is to think the vapour pressure stays constant because "it's equilibrium." But equilibrium is a dynamic balance of rates — it takes time to re-establish. The instantaneous effect is purely mechanical: same number of molecules, larger volume → lower pressure.

2. How rates of evaporation and condensation change initially

Evaporation rate: This depends on the temperature of the liquid and the surface area exposed. Neither has changed (temperature is fixed, and the liquid surface area is essentially the same). So the rate of evaporation remains unchanged initially.

Condensation rate: This depends on how many vapour molecules strike the liquid surface per second. That number is proportional to the vapour density (or pressure). Since the vapour pressure has dropped, fewer molecules hit the liquid per unit time. So the condensation rate decreases immediately.

The result: evaporation rate > condensation rate. The system is no longer in equilibrium — there is a net transfer of molecules from liquid to vapour.

Tip

Think of it like a door: evaporation is people leaving a room (constant rate), condensation is people coming back in (depends on how crowded the hallway is). If you suddenly make the hallway bigger, fewer people bump into the door to come back — so more leave than enter.

3. What happens when equilibrium is restored — final vapour pressure

Because evaporation now exceeds condensation, the number of vapour molecules $n$ increases. This raises the vapour pressure. As pressure rises, the condensation rate increases (more molecules available to strike the liquid). Eventually, the condensation rate climbs back up to match the evaporation rate.

At what pressure does this happen? At the same saturated vapour pressure as before — because the temperature hasn't changed. The SVP is a function of temperature only.

So the final vapour pressure is identical to the initial equilibrium vapour pressure. The only difference is that slightly more liquid has evaporated (so the liquid level is a tiny bit lower), but the vapour pressure is unchanged.

Note

This is why, when you open a soda bottle (sudden volume increase), you see bubbles form — the pressure drops, so dissolved gas comes out of solution. For a pure liquid-vapour system, the same principle applies: the vapour "boils" momentarily until equilibrium is restored at the same pressure.

✓Final answer

a) The vapour pressure initially decreases . b) The evaporation rate stays the same, but the condensation rate decreases . c) When equilibrium is restored, the final vapour pressure is the same as the initial saturated vapour pressure at that temperature.

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The Core Idea: Phase Equilibrium at Fixed Temperature

1. Initial effect on vapour pressure when volume is increased

2. How rates of evaporation and condensation change initially

3. What happens when equilibrium is restored — final vapour pressure

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