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Show that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same.

Short Answer

Expert verified
If so, explain why. Answer: Yes, when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same. This can be shown by analyzing the conditions for phase equilibrium (temperature, pressure, and chemical potential being the same in all phases) and relating the chemical potential to the specific Gibbs function for each phase. Since the molar mass of a pure substance is the same in each phase, the specific Gibbs functions of all three phases equate to each other, proving that they are the same under equilibrium conditions.

Step by step solution

01

Understand the conditions for phase equilibrium

In order for the three phases of a substance to be in equilibrium, the temperature, pressure, and chemical potential must be the same in all phases. Mathematically, this is given as: \(T_{1} = T_{2} = T_{3}\) (Temperature Equilibrium) \(P_{1} = P_{2} = P_{3}\) (Pressure Equilibrium) \(\mu_{1} = \mu_{2} = \mu_{3}\) (Chemical Potential Equilibrium) Here, the subscripts 1, 2, and 3 denote the different phases of the substance.
02

Understand the definition of specific Gibbs function

The specific Gibbs function (g) is defined as the Gibbs energy (G) per unit mass (m) of a substance. It is related to the specific enthalpy (h), specific entropy (s), temperature (T), and pressure (P) as follows: \( g = \frac{G}{m} \) \( g = h - Ts \) Note that we will analyze the specific Gibbs function of each phase, so we'll add a subscript that denotes the phase (1, 2 or 3).
03

Derive the equation for chemical potential in terms of specific Gibbs function

The chemical potential of a substance is the change in the Gibbs energy per change in the amount of substance: \( \mu = \frac{dG}{dn} \) We know that the specific Gibbs function is equal to the Gibbs energy divided by mass (G = mg), and we can relate mass (m) to the number of particles (n) through the molar mass (M): \( m = \frac{n}{M} \) Using these relationships, we can rewrite the chemical potential in terms of specific Gibbs function (g): \( \mu = \frac{d(G(Mm))}{d(n)} \) \( \mu = M \frac{d(g \frac{n}{M})}{d(n)} = Mg \) Now the chemical potential equation becomes: \( \mu_{1} = M_{1}g_{1} = \mu_{2} = M_{2}g_{2} = \mu_{3} = M_{3}g_{3} \)
04

Show that the specific Gibbs function of each phase is the same

Since the substance is pure, the molar mass of each phase is the same: \( M_{1} = M_{2} = M_{3} \) Substituting this into the chemical potential equation: \( M_{1}g_{1} = M_{2}g_{2} = M_{3}g_{3} \) Now, we can divide the equation by a common molar mass (M): \( g_{1} = g_{2} = g_{3} \) This shows that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same.

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Key Concepts

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

Phase Equilibrium
In the captivating world of thermodynamics, phase equilibrium is a state where a substance's various phases, such as solid, liquid, and gas, coexist at certain conditions of temperature and pressure without changing over time. It's an astonishing dance where everything is perfectly balanced, so there's no driving force for the substance to transition from one phase to another. This balance implies that the rates of transfer between phases are equal, creating a dynamic but stable system. Understand it like an equilibrium walk on a tightrope, maintaining just the right conditions to stay perfectly level.

Let's simplify the technical talk: For phase equilibrium to happen, certain parameters must match across all phases – and the main stars here are temperature, pressure, and chemical potential. Think of them as the rhythm, beats, and harmony in music – all needing to sync for a flawless performance. These equals signs in the equations, like a symphonic conductor, keep everything in tune – making sure that the temperature (\(T\textsubscript{1} = T\textsubscript{2} = T\textsubscript{3}\) ), pressure (\(P\textsubscript{1} = P\textsubscript{2} = P\textsubscript{3}\) ), and chemical potential (\(µ\textsubscript{1} = µ\textsubscript{2} = µ\textsubscript{3}\) ) are in harmony across the solid, liquid, and gas phases.
Chemical Potential
Dive into the concept of chemical potential and you'll discover it's a cornerstone in understanding phase equilibrium. Let's demystify this term. Chemical potential, represented by the Greek letter 'µ', is essentially the 'oomph' behind any potential change in a system. In more formal terms, it's the change in Gibbs energy – the measure of a system's capacity to do non-mechanical work – when an infinitesimal amount of substance is added at a constant temperature and pressure.

Imagine you're at a party and you've got a finite amount of energy to dance. The energy you contribute to every dance move – that's akin to the chemical potential in a system. The more you groove, the more energy you spend. Similarly, for a phase equilibrium to remain motionless on that cosmic dance floor, the chemical potential must be equal across all phases (\(µ\textsubscript{1} = µ\textsubscript{2} = µ\textsubscript{3}\) ). The moment one phase gets an extra bit of energy – boom! The substance could start changing phase, like you'd move to a different dance rhythm.
Specific Gibbs Function
The specific Gibbs function, often denoted simply as 'g', brings a personal touch to the Gibbs energy by focusing on a per unit mass basis. This elegant function connects the dots between the intrinsic energy content of each phase and its tendency to maintain equilibrium or start a transformation. To put it plainly: the specific Gibbs function is an indicator of the stability of a phase. If you're wondering how to figure it out, here's a formula that can make a mathematician's heart flutter: \( g = h - Ts \) with 'h' being the specific enthalpy and 's' the specific entropy.

But why does this matter in the context of phase equilibrium? Imagine each phase of a substance holding a flag that represents its specific Gibbs function. In a parade of equilibrium, each flag needs to match. That's the only way they'll march in harmony down Equilibrium Lane. To express this mathematically, the three phases, when in equilibrium, must have the same specific Gibbs function \( g\textsubscript{1} = g\textsubscript{2} = g\textsubscript{3} \). This reveals that they're in perfect sync, and no phase has more incentive to change than the other.

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Most popular questions from this chapter

Using the enthalpy of reaction \(\bar{h}_{R}\) data and the \(K_{P}\) value at \(2200 \mathrm{K}\), estimate the \(K_{P}\) value of the dissociation process \(\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{O}\) at \(2400 \mathrm{K}\).

The value of Henry's constant for \(\mathrm{CO}_{2}\) gas dissolved in water at \(290 \mathrm{K}\) is \(12.8 \mathrm{MPa}\). Consider water exposed to atmospheric air at 100 kPa that contains 3 percent \(\mathrm{CO}_{2}\) by volume. Under phase equilibrium conditions, the mole fraction of \(\mathrm{CO}_{2}\) gas dissolved in water at \(290 \mathrm{K}\) is \((a) 2.3 \times 10^{-4}\) \((b) 3.0 \times 10^{-4}\) \((c) 0.80 \times 10^{-4}\) \((d) 2.2 \times 10^{-4}\) \((e) 5.6 \times 10^{-4}\)

An engineer suggested that high-temperature disassociation of water be used to produce a hydrogen fuel. A reactor-separator has been designed that can accommodate temperatures as high as \(4000 \mathrm{K}\) and pressures as much as 5 atm. Water enters this reactor-separator at \(25^{\circ} \mathrm{C}\). The separator separates the various constituents in the mixture into individual streams whose temperature and pressure match those of the reactor-separator. These streams are then cooled to \(25^{\circ} \mathrm{C}\) and stored in atmospheric pressure tanks with the exception of any remaining water, which is returned to the reactor to repeat the process again. Hydrogen gas from these tanks is later burned with a stoichiometric amount of air to provide heat for an electrical power plant. The parameter that characterizes this system is the ratio of the heat released by burning the hydrogen to the amount of heat used to generate the hydrogen gas. Select the operating pressure and temperature for the reactor-separator that maximizes this ratio. Can this ratio ever be bigger than unity?

A mixture of ideal gases is made up of 30 percent \(\mathrm{N}_{2}, 30\) percent \(\mathrm{O}_{2},\) and 40 percent \(\mathrm{H}_{2} \mathrm{O}\) by mole fraction. Determine the Gibbs function of the \(\mathrm{N}_{2}\) when the mixture pressure is 5 atm, and its temperature is \(600 \mathrm{K}\).

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