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\(\Lambda\) cylindcr fitted with a movable piston contains liquid water in cquilibrium with water vapour at \(25^{\circ} \mathrm{C}\). Which opcration results in a decrcase in the cquilibrium vapour pressure? (1) Moving the piston downward a short distancc (2) Removing a small amount of vapour (3) Removing a small amount of the liquid (4) Dissolving the salt in water

Short Answer

Expert verified
Option (4) dissolving the salt in water decreases the equilibrium vapor pressure.

Step by step solution

01

Understanding Equilibrium Vapor Pressure

Equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its liquid at a given temperature. Any change that affects the number of liquid molecules transitioning to vapor can alter this pressure.
02

Evaluating the Options

Examine each of the given options to determine its effect on the equilibrium vapor pressure: (1) Moving the piston downward increases the pressure on the liquid, but the system re-establishes the same equilibrium vapor pressure due to the temperature remaining constant. (2) Removing a small amount of vapor initially decreases the vapor pressure, but the system compensates by vaporizing more liquid until the equilibrium vapor pressure is restored. (3) Removing a small amount of liquid decreases the surface area available for evaporation but does not change the equilibrium vapor pressure since the saturation level is governed by temperature. (4) Dissolving salt in water decreases the number of water molecules at the surface, thus lowering the vapor pressure.
03

Identifying the Correct Operation

The correct option is (4), as dissolving salt in water reduces the number of water molecules on the liquid's surface that can escape into the vapor phase, thus decreasing the equilibrium vapor pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamic Equilibrium
Thermodynamic equilibrium is a state where a system's properties do not change over time. In this state, all parts of the system are at the same temperature, and there is no net flow of energy or matter. For a system containing liquid and vapor - like liquid water and water vapor - thermodynamic equilibrium means that the rate of evaporation equals the rate of condensation. Whether it's the amount of liquid or vapor, everything remains steady if the temperature stays the same.
The equilibrium vapor pressure is a key aspect of thermodynamic equilibrium. It’s the pressure exerted by the vapor when it’s in thermodynamic equilibrium with its liquid at a given temperature. Since the system is balanced, no net change occurs in the amount of liquid and vapor.
Vapor-Liquid Equilibrium
Vapor-liquid equilibrium (VLE) occurs when a liquid and its vapor coexist at equilibrium conditions. This means that the rates of condensation and evaporation are equal. In a closed system, like the cylinder with a piston in the exercise, changes in volume or pressure will temporarily disrupt this balance. However, the system will always return to equilibrium if the temperature is kept constant.
Different factors can influence VLE. Removing vapor or liquid, changing the volume, or adding substances (like salt) can cause temporary shifts. However, the system self-corrects over time, except when the number of molecules available for transitioning between phases changes, like with salt, which can permanently alter the equilibrium vapor pressure.
Raoult's Law
Raoult’s Law explains how the presence of a solute (like salt) affects the vapor pressure of a solvent (like water). According to Raoult’s Law, the vapor pressure of a solution is proportional to the mole fraction of the solvent in the solution.
When salt dissolves in water, water molecules at the liquid surface are replaced by salt molecules, reducing the number of water molecules that can escape into the vapor phase. This results in a lower vapor pressure. The law is significant in explaining why adding solute like salt lowers the equilibrium vapor pressure of a solvent, as fewer solvent molecules are available to vaporize.
Thus, dissolving salt in water not only impacts the moles of water available but also markedly decreases the system's equilibrium vapor pressure, illustrating Raoult’s Law in action.

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Most popular questions from this chapter

At \(1000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction of the system \(2 \mathrm{II}_{2}(\mathrm{~g}) \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) is very largc. This implics that (1) \(\mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) is unstable at \(1000^{\circ} \mathrm{C}\) (2) \(\mathrm{II}_{2}(\mathrm{~g})\) is unstable at \(1000^{\circ} \mathrm{C}\) (3) \(\mathrm{II}_{2}\) and \(\mathrm{O}_{2}\) have very little tendency to combinc at \(1000^{\circ} \mathrm{C}\) (4) \(\mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) has very little tendency to decompose into \(\mathrm{II}_{2}(\mathrm{~g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\) at \(1000^{\circ} \mathrm{C}\)

The equilibrium constant for the reaction \(\mathrm{Br}_{2} \rightleftharpoons\) \(2 \mathrm{Br}\) at \(500 \mathrm{~K}\) and \(700 \mathrm{~K}\) are \(1 \times 10^{\circ}\) and \(1 \times 10^{5}\), respectively. The reaction is (1) endothermic (2) exothermic (3) fast (4) slow

For an acid \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} ; K_{1}\) and for a base \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH} ; K_{2}\) Then (1) \(K_{1} \cdot K_{2}=K_{\mathrm{w}}\) (2) \(\log K_{1}+\log K_{2}=\log K_{\mathrm{w}}\) (3) \(\mathrm{p} K_{1}+\mathrm{p} K_{2}=\mathrm{p} K_{\mathrm{w}}\) (4) all are correct

The increasing order of basic strength of \(\mathrm{Cl}, \mathrm{CO}_{3}^{2}\), \(\mathrm{CH}_{3} \mathrm{COO}, \mathrm{OH}, \mathrm{F}\) is (1) \(\mathrm{Cl}<\mathrm{F}<\mathrm{CH}_{3} \mathrm{COO}<\mathrm{CO}_{3}^{2}<\mathrm{OH}\) (2) \(\mathrm{Cl}^{-}<\mathrm{F}^{-}<\mathrm{CO}_{3}^{2-}<\mathrm{CH}_{3} \mathrm{COO}^{-}<\mathrm{OH}^{-}\) (3) \(\mathrm{CHI}_{3} \mathrm{COO}<\mathrm{Cl}<\mathrm{F}<\mathrm{CO}_{3}^{2}<\mathrm{OH}\) (4) \(\mathrm{F}<\mathrm{CO}_{3}^{2}<\mathrm{Cl}<\mathrm{CII}_{3} \mathrm{COO}<\mathrm{OII}\)

For which reaction high pressure and high temperature is helpful in obtaining a high cquilibrium yicld? (1) \(2 \mathrm{NF}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \quad 54.40 \mathrm{kcal}\) (2) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{II}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NI}_{3}(\mathrm{~g})+22.08 \mathrm{kcal}\) (3) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClO}_{3}(\mathrm{~g}) \quad 49.4 \mathrm{kcal}\) (4) \(2 \mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_{2}(\mathrm{~g}) \times 7 \mathrm{O}_{2}(\mathrm{~g})+126.8 \mathrm{kcal}\)

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