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An isolated system has two phases, denoted by \(\mathrm{A}\) and B, each of which consists of the same two substances, denoted by 1 and \(2 .\) Show that necessary conditions for equilibrium are 1\. the temperature of each phase is the same, \(T_{\mathrm{A}}=T_{\mathrm{B}}\). 2\. the pressure of each phase is the same, \(p_{\mathrm{A}}=p_{\mathrm{B}}\). 3\. the chemical potential of each component has the same value in each phase, \(\mu_{1}^{\mathrm{A}}=\mu_{1}^{\mathrm{B}}, \mu_{2}^{\mathrm{A}}=\mu_{2}^{\mathrm{B}}\).

Short Answer

Expert verified
For equilibrium: \[ T_{\text{A}} = T_{\text{B}}, p_{\text{A}} = p_{\text{B}}, \mu_{1}^{\text{A}} = \mu_{1}^{\text{B}}, \mu_{2}^{\text{A}} = \mu_{2}^{\text{B}} \]

Step by step solution

01

Understanding the Isolated System

An isolated system means that there is no exchange of matter or energy with the surroundings. Therefore, the total energy, mass, and composition of the system remain constant.
02

Defining Thermodynamic Equilibrium

Thermodynamic equilibrium in a closed system with multiple phases requires that properties like temperature, pressure, and chemical potential are uniform across all phases.
03

Temperature Equilibrium

For thermal equilibrium (no heat flow between phases), the temperature must be the same in both phases: \[ T_{\text{A}} = T_{\text{B}} \]
04

Pressure Equilibrium

For mechanical equilibrium (no work can be done between phases), the pressure must be identical in both phases: \[ p_{\text{A}} = p_{\text{B}} \]
05

Chemical Potential Equilibrium

For chemical equilibrium (no net transfer of components between phases), the chemical potential of each substance must be the same in both phases: \[ \ \mu_{1}^{\text{A}} = \mu_{1}^{\text{B}} \mu_{2}^{\text{A}} = \mu_{2}^{\text{B}} \]

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Key Concepts

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

Isolated System
An isolated system is a specific type of thermodynamic system that doesn't interact with its surroundings. Imagine a perfectly sealed container that neither gains nor loses matter and energy. It's like a thermos bottle that remarkably keeps your coffee hot or cold for a long time. In this imaginary container, aside from what’s already inside, nothing else can enter or exit. This means:
  • no heat exchange
  • no exchange of matter
  • total energy and mass remain constant
Such a system is crucial when studying fundamental thermodynamic principles, as it eliminates external factors, providing a clearer view of the internal processes.
Thermal Equilibrium
Thermal equilibrium is when there is no net flow of heat between two or more phases in a system. Think of it as each phase being at the same temperature. It’s like two rooms connected by an open door. If they’re both at the same temperature, no heat flows between them. Mathematically, for phases A and B:
\[ T_{\text{A}} = T_{\text{B}} \]
This condition ensures that both phases are comfortable, making sure no energy is wasted in the form of heat trying to balance out temperature differences.
Mechanical Equilibrium
Mechanical equilibrium is characterized by no net force acting within the system, which typically translates to no pressure difference between the phases. Imagine two adjacent inflatables with a connecting valve. If both are at the same pressure, the valve stays shut, and no air moves between the inflatables. In thermodynamic terms, this is expressed as:
\[ p_{\text{A}} = p_{\text{B}} \]
This condition ensures that no physical work is done within the system, thereby maintaining a balanced and steady state.
Chemical Equilibrium
Chemical equilibrium ensures that the chemical potential of each substance is the same in all phases. It’s similar to having the same concentration of sugar dissolved equally in two cups of water connected by a thin tube. No sugar will move between the cups if the concentration (chemical potential) is the same. Mathematically, this can be written for components 1 and 2 in phases A and B:
\[ \mu_{1}^{\text{A}} = \mu_{1}^{\text{B}} \]
\[ \mu_{2}^{\text{A}} = \mu_{2}^{\text{B}} \]
This means that no component travels between the phases, maintaining a chemical balance. It’s crucial in systems where multiple chemical reactions occur and are interconnected.

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Most popular questions from this chapter

If the ionization-equilibrium constants for \(\mathrm{Cs} \rightleftarrows \mathrm{Cs}^{+}+\mathrm{e}^{-}\) at 1600 and \(2000 \mathrm{~K}\) are \(K=0.78\) and \(K=15.63\), respectively, estimate the enthalpy of ionization, in \(\mathrm{kJ} / \mathrm{kmol}\), at \(1800 \mathrm{~K}\) using the van't Hoff equation.

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Estimate the equilibrium constant at \(2800 \mathrm{~K}\) for \(\mathrm{CO}_{2} \rightleftarrows \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2}\) using the equilibrium constant at \(2000 \mathrm{~K}\) from Table A-27, together with the van't Hoff equation and enthalpy data. Compare with the value for the equilibrium constant obtained from Table A-27.

The following exercises involve oxides of nitrogen: (a) One \(\mathrm{kmol}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) to form an equilibrium ideal gas mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) in which the amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) present is \(0.8154 \mathrm{kmol}\). Determine the amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) that would be present in an equilibrium mixture at \(25^{\circ} \mathrm{C}, 0.5 \mathrm{~atm}\). (b) A gaseous mixture consisting of \(1 \mathrm{kmol}\) of \(\mathrm{NO}, 10 \mathrm{kmol}\) of \(\mathrm{O}_{2}\), and \(40 \mathrm{kmol}\) of \(\mathrm{N}_{2}\) reacts to form an equilibrium ideal gas mixture of \(\mathrm{NO}_{2}, \mathrm{NO}\), and \(\mathrm{O}_{2}\) at \(500 \mathrm{~K}, 0.1 \mathrm{~atm}\). Determine the composition of the equilibrium mixture. For \(\mathrm{NO}+\frac{1}{2} \mathrm{O}_{2} \rightleftarrows \mathrm{NO}_{2}, K=120\) at \(500 \mathrm{~K}\) (c) An equimolar mixture of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) reacts to form an equilibrium ideal gas mixture of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and NO. Plot the mole fraction of \(\mathrm{NO}\) in the equilibrium mixture versus equilibrium temperature ranging from 1200 to \(2000 \mathrm{~K}\). Why are oxides of nitrogen of concern?

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