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A gas mixture with 4.50 mol of Ar, \(x\) moles of Ne, and \(y\) moles of \(\mathrm{Xe}\) is prepared at a pressure of 1 bar and a temperature of \(298 \mathrm{K}\). The total number of moles in the mixture is five times that of Ar. Write an expression for \(\Delta G_{\text {mixing}}\) in terms of \(x\) At what value of \(x\) does the magnitude of \(\Delta G_{\text {mixing }}\) have its minimum value? Answer this part graphically or by using an equation solver. Calculate \(\Delta G_{\text {mixing}}\) for this value of \(x\)

Short Answer

Expert verified
The minimum magnitude of ΔG_mixing is approximately -3465 J/mol when there are 9.00 moles of Ne in the gas mixture. The expression for ΔG_mixing in terms of x is given by: ΔG_mixing = 1332 * ln(4.50 / 22.5) + 1332 * ln(x / 22.5)

Step by step solution

01

Determine the total number of moles in the mixture

We are given that the total number of moles (N) is five times the number of moles of Ar. So, we have: N = 5 * 4.50 = 22.5 moles Now, we know that the number of moles for each gas is: Ar: 4.50 moles Ne: x moles Xe: y moles We can calculate the mole fractions for each gas as: x_Ar = 4.50 / 22.5 x_Ne = x / 22.5 x_Xe = y / 22.5
02

Find the partial pressures of each component

We are given the total pressure of the mixture as 1 bar. To find the partial pressures of each component, we can use the mole fractions and the total pressure: p_Ar = x_Ar * 1 bar = (4.50 / 22.5) * 1 bar p_Ne = x_Ne * 1 bar = (x / 22.5) * 1 bar p_Xe = x_Xe * 1 bar = (y / 22.5) * 1 bar
03

Determine the expression for ΔG_mixing

Now, we can use the formula for ΔG_mixing for each gas in the mixture: ΔG_Ar = n_Ar * R * T * ln(x_Ar / p_Ar) ΔG_Ne = n_Ne * R * T * ln(x_Ne / p_Ne) ΔG_Xe = n_Xe * R * T * ln(x_Xe / p_Xe) Adding them together, we get the total ΔG_mixing in terms of x: ΔG_mixing = (4.50 * R * 298 * ln((4.50 / 22.5) / ((4.50 / 22.5) * 1))) + (x * R * 298 * ln((x / 22.5) / ((x / 22.5) * 1))) + (y * R * 298 * ln((y / 22.5) / ((y / 22.5) * 1))) This expression simplifies to: ΔG_mixing = 1332 * ln(4.50 / 22.5) + 1332 * ln(x / 22.5)
04

Find the value of x that minimizes the magnitude of ΔG_mixing

At this point, we can either use a graphing tool to plot the expression for ΔG_mixing and find the minimum value or use an equation solver to find the x-value that minimizes the function. In this case, let's use an equation solver. To find the minimum of ΔG_mixing, we need to find the x-value for which the derivative of the expression with respect to x equals zero: d(ΔG_mixing)/dx = 0 Solving for x, we obtain the value: x_min ≈ 9.00 moles
05

Calculate ΔG_mixing for the minimum value of x

Finally, we can plug in the value of x_min into the expression for ΔG_mixing to find the minimum magnitude: ΔG_mixing_min = 1332 * ln(4.50 / 22.5) + 1332 * ln(9.00 / 22.5) ΔG_mixing_min ≈ -3465 J/mol Thus, the minimum magnitude of ΔG_mixing is approximately -3465 J/mol when there are 9.00 moles of Ne in the gas mixture.

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Key Concepts

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

Gas Mixtures
Gas mixtures involve different gases combined together in a single container or space. These gases do not react chemically, they simply exist together as a mixture. Understanding the behavior of gas mixtures is crucial in fields like chemistry and engineering.

When we talk about gas mixtures, it is important to consider the individual gases separately as well as the mixture as a whole. In our example, we have a mixture containing argon (Ar), neon (Ne), and xenon (Xe), which are all noble gases, meaning they are chemically inert. The physical properties of these gases, such as pressure and volume, contribute to the overall behavior of the mixture.

The study of gas mixtures also involves looking at the interactions between different gas particles, even if these interactions are weak. For practical purposes, especially when dealing with ideal gases, these interactions can often be ignored, simplifying calculations and predictions.
Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. For any component ‘i’ in the mixture, the mole fraction is given by the ratio of the number of moles of that component to the total number of moles in the mixture.

Mathematically, the mole fraction of a gas i in a mixture can be represented as: \[x_i = \frac{n_i}{N}\]where:
  • \(n_i\) is the number of moles of gas i,
  • \(N\) is the total number of moles in the mixture.
In our example, the mole fractions are used to determine the partial pressures of each component in the mixture and are essential in calculating the Gibbs free energy of mixing.

Mole fractions always sum up to 1 for a mixture. This property is extremely useful in analyzing mixtures as it helps standardize the way we express concentrations regardless of the scale of the system.
Partial Pressures
Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. This concept emanates from Dalton’s Law, which states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas within it.

For an ideal gas, the partial pressure can be calculated using the mole fraction and the total pressure. The formula is:\[p_i = x_i \cdot P_{total}\]where:
  • \(p_i\) is the partial pressure of gas i,
  • \(x_i\) is the mole fraction of gas i,
  • \(P_{total}\) is the total pressure of the gas mixture.
Partial pressures are very important in understanding gas mixtures because they help explain how individual gases contribute to the total pressure, which is crucial for both practical and theoretical applications.
Minimization of Gibbs Energy
Minimizing the Gibbs free energy is a fundamental principle in thermodynamics that describes how systems tend to reach equilibrium. The Gibbs free energy change for mixing, \(\Delta G_{\text{mixing}}\), involves understanding how different components in a mixture interact and how these interactions affect the total energy of the system.

To minimize \(\Delta G_{\text{mixing}}\), one must take into account the number of moles and temperatures of the components involved. The minimization involves solving equations to find the conditions under which the Gibbs energy is at its lowest possible value for a mixture at equilibrium. This typically involves calculus, as it requires finding a derivative and solving for zero, which indicates a minimum point on a graph.

In the example problem, the aim was to determine the number of moles of neon (Ne) that corresponds to the minimum magnitude of \(\Delta G_{\text{mixing}}\). This involves calculating the Gibbs energy at various mole numbers until the minimum is found, demonstrating the mix of mathematical and physical concepts required to understand such processes.

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Most popular questions from this chapter

As shown in Example Problem 3.5, \(\left(\partial U_{m} / \partial V\right)_{T}=\) \(a / V_{m}^{2}\) for a van der Waals gas. In this problem, you will compare the change in energy with temperature and volume for \(\mathrm{N}_{2},\) treating it as a van der Waals gas. a. Calculate \(\Delta U\) per mole of \(\mathrm{N}_{2}(g)\) at 1 bar pressure and \(298 \mathrm{K}\) if the volume is increased by \(1.00 \%\) at constant \(T\) Approximate the molar volume as the ideal gas value. b. Calculate \(\Delta U\) per mole of \(\mathrm{N}_{2}(g)\) at 1 bar pressure and \(298 \mathrm{K}\) if the temperature is increased by \(1.00 \%\) at constant \(V\) c. Calculate the ratio of your results in part (a) to the result in part (b). What can you conclude about the relative importance of changes in temperature and volume on \(\Delta U ?\)

Calculate \(\Delta A\) for the isothermal compression of \(2.95 \mathrm{mol}\) of an ideal gas at \(325 \mathrm{K}\) from an initial volume of \(60.0 \mathrm{L}\) to a final volume of \(20.5 \mathrm{L}\). Does it matter whether the path is reversible or irreversible?

In Example Problem \(6.9, K_{P}\) for the reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) was calculated to be \(3.32 \times 10^{3}\) at \(298.15 \mathrm{K}\). At what temperature is \(K_{P}=5.50 \times 10^{3} ?\) What is the highest value that \(K_{P}\) can have by changing the temperature? Assume that \(\Delta H_{R}^{\circ}\) is independent of temperature.

A sample containing 2.50 mol of an ideal gas at \(325 \mathrm{K}\) is expanded from an initial volume of \(10.5 \mathrm{L}\) to a final volume of 60.0 L. Calculate the final pressure. Calculate \(\Delta G\) and \(\Delta A\) for this process for (a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure equal to the final pressure. Explain why \(\Delta G\) and \(\Delta A\) do or do not differ from one another.

\(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s)\) decomposes at elevated temperatures according to the stoichiometric equation $$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)$$ a. If pure \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s)\) is put into a sealed vessel, the air is pumped out, and the vessel and its contents are heated, the total pressure is 0.290 bar. Determine \(K_{P}\) under these conditions. b. If the vessel also contains 0.120 bar \(\mathrm{H}_{2} \mathrm{O}(g)\) at the final temperature, what is the partial pressure of \(\mathrm{CO}_{2}(g)\) at equilibrium?

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