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Derive an expression for estimating the pressure at which graphite and diamond exist in equilibrium at \(25^{\circ} \mathrm{C}\) in terms of the specific volume, specific Gibbs function, and isothermal compressibility of each phase at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Discuss.

Short Answer

Expert verified
P = P_0 + \frac{G_{\text{diamond}} - G_{\text{graphite}}}{V_{\text{graphite}} - V_{\text{diamond}}}

Step by step solution

01

Understand the Equilibrium Condition

When graphite and diamond are in equilibrium, their Gibbs free energy per mole is equal. This means that the Gibbs free energy change between the two phases is zero: \[ \text{dG} = \text{G}_{\text{diamond}} - \text{G}_{\text{graphite}} = 0 \] This implies that the chemical potential of graphite (\(\mu_{\text{graphite}}\)) is equal to the chemical potential of diamond (\(\mu_{\text{diamond}}\)).
02

Express Gibbs Free Energy in terms of Pressure and Volume

The change in Gibbs free energy with respect to pressure (\(P\)) is given by:\[ dG = V \times dP \]where \(V\) is the specific volume. Therefore, at equilibrium:\[ V_{\text{diamond}} \times (P - P_0) = G_{\text{diamond}} - G_{\text{graphite}} - V_{\text{graphite}} \times (P - P_0) \]
03

Isothermal Compressibility

Isothermal compressibility \((\beta)\) is given by:\[ \beta = - \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T \]At small changes in pressure, the change in volume can be approximated as:\[ \Delta V = -V \times \beta \times (P - P_0) \]
04

Substitute Compressibility into Volume Expression

With \(\Delta V\) from compressibility, substitute it into our equilibrium expression:\[ G_{\text{diamond}} - G_{\text{graphite}} = V_{\text{diamond}} \times (P - P_0) - V_{\text{graphite}} \times (P - P_0) \]Considering changes using isothermal compressibility, we get:\[ G_{\text{diamond}} - G_{\text{graphite}} = -\left( V_{\text{diamond}} \beta_{\text{diamond}} - V_{\text{graphite}} \beta_{\text{graphite}} \right) (P - P_0) \]
05

Final Expression

Rearranging the final equilibrium expression gives us the pressure at which graphite and diamond exist in equilibrium at 25°C:\[ P = P_0 + \frac{G_{\text{diamond}} - G_{\text{graphite}}}{V_{\text{graphite}} - V_{\text{diamond}}} \]

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Key Concepts

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

Gibbs Free Energy
To understand why graphite and diamond can exist in equilibrium, we need to understand Gibbs free energy. Gibbs free energy (\textbf{G}) is a measure of the maximum reversible work that can be done by a thermodynamic system at constant temperature and pressure. It plays a key role in predicting the direction of chemical processes and phase changes. When two phases, like graphite and diamond, are in equilibrium, their Gibbs free energies are equal. This is represented mathematically as: \[ \text{dG} = \text{G}_{\text{diamond}} - \text{G}_{\text{graphite}} = 0 \]
This implies the chemical potential (\textbf{µ}) of graphite is equal to that of diamond, meaning neither phase is energetically favored over the other under the given conditions. Understanding this equilibrium is crucial in deriving the pressure expression for the coexistence of graphite and diamond.
Isothermal Compressibility
Isothermal compressibility (\textbf{β}) is a measure of how much a material's volume changes under pressure at constant temperature. Mathematically, it is given by: \[ \beta = - \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T \]
At constant temperature, \textbf{β} helps us understand how a substance reacts to changes in pressure. For instance, for small changes in pressure, the change in volume (\textbf{\textDelta V}) can be approximated as: \[ \Delta V = -V \times \beta \times (P - P_0) \]
This relationship is essential in deriving pressure conditions for equilibrium. By substituting the compressibility into the volume expression, we can understand how the volumes of graphite and diamond change with pressure and thus find the equilibrium condition.
Specific Volume
Specific volume (\textbf{V}) is the volume occupied by a unit mass of a substance. It's essentially the reciprocal of density. In the context of our problem, it's crucial because the difference in specific volumes of graphite and diamond affects their Gibbs free energy under different pressures. When deriving the pressure at which they coexist in equilibrium, specific volumes appear in the following relationship: \[ G_{\text{diamond}} - G_{\text{graphite}} = V_{\text{diamond}} \times (P - P_0) - V_{\text{graphite}} \times (P - P_0) \]
Also, using isothermal compressibility, the equation further evolves. Hence, specific volume helps us understand how different phases will behave under pressure differences, crucial for predicting equilibrium states. Ultimately, this leads us to derive the pressure condition for equilibrium: \[ P = P_0 + \frac{G_{\text{diamond}} - G_{\text{graphite}}}{V_{\text{graphite}} - V_{\text{diamond}}} \] Understanding specific volume and its implications is key to solving such thermodynamic problems.

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Most popular questions from this chapter

Why is using the Gibbs function advantageous when studying chemical and phase equilibrium?

An isolated system has two phases, denoted by 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, \(\mu_{1}^{\mathrm{A}}=\mu_{1}^{\mathrm{B}}, \mu_{2}^{\mathrm{A}}=\mu_{2}^{\mathrm{B}}\).

Gaseous propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) enters a reactor operating at steady state and burns with \(80 \%\) of theoretical air entering separately at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}, \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) exits at \(1227^{\circ} \mathrm{C}\), \(1 \mathrm{~atm}\). Determine the heat transfer between the reactor and its surroundings, in \(\mathrm{kJ}\) per \(\mathrm{kmol}\) of propane entering. Neglect kinetic and potential energy effects.

If the ionization-equilibrium constants for \(\mathrm{Cs} \rightleftarrows\) \(\mathrm{Cs}^{+}+\mathrm{e}^{-}\)at 1800 and \(2200 \mathrm{~K}\) are \(K=8.21\) and \(K=23.06\), respectively, estimate the enthalpy of ionization, in \(\mathrm{kJ} / \mathrm{kmol}\), at \(2000 \mathrm{~K}\) using the van't Hoff equation.

Carbon monoxide at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) enters an insulated reactor operating at steady state and burns with air entering at \(258 \mathrm{C}, 1 \mathrm{~atm}\). The products exit at \(1 \mathrm{~atm}\) as an equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}, \mathrm{O}_{2}\), and \(\mathrm{N}_{2}\). Determine the temperature of the equilibrium mixture, in \(\mathrm{K}\), if the combustion occurs with (a) \(80 \%\) of theoretical air. (b) \(100 \%\) of theoretical air. Kinetic and potential energy effects are negligible.

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