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A mixture containing \(0.765 \mathrm{~mol} \mathrm{He}(g), 0.330 \mathrm{~mol} \mathrm{Ne}(g),\) and \(0.110 \mathrm{~mol} \mathrm{Ar}(g)\) is confined in a \(10.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). Calculate the partial pressure of each of the gases in the mixture. (b) Calculate the total pressure of the mixture.

Short Answer

Expert verified
The partial pressures of Helium, Neon, and Argon in the mixture are 1.8559 \(atm\), 0.8077 \(atm\), and 0.2677 \(atm\) respectively, and the total pressure of the gas mixture is 2.9313 \(atm\).

Step by step solution

01

Convert temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin. We do this by adding 273.15 to the Celsius temperature. \(T(K) = T(°C) + 273.15\) \(T(K) = 25 + 273.15\) \(T(K) = 298.15\)
02

Calculate the partial pressures

Next, we'll find the partial pressures of the three gases (Helium, Neon, and Argon) using the Ideal Gas Law: \(P_i = \frac{n_i R T}{V}\), where \(P_i\) is the partial pressure of the gas, \(n_i\) is the number of moles of the gas, R is the ideal gas constant (0.0821 \(L\cdot atm/mol\cdot K\)), T is the temperature in Kelvin, and V is the volume of the vessel. For Helium: \(P_{He} = \frac{n_{He} R T}{V}\) \(P_{He} = \frac{0.765 \mathrm{~mol} \cdot 0.0821 \frac{L \cdot atm}{ mol \cdot K} \cdot 298.15 K}{10.00 L}\) \(P_{He} = 1.8559 \, atm\) For Neon: \(P_{Ne} = \frac{n_{Ne} R T}{V}\) \(P_{Ne} = \frac{0.330 \mathrm{~mol} \cdot 0.0821 \frac{L \cdot atm}{ mol \cdot K} \cdot 298.15 K}{10.00 L}\) \(P_{Ne} = 0.8077 \, atm\) For Argon: \(P_{Ar} = \frac{n_{Ar} R T}{V}\) \(P_{Ar} = \frac{0.110 \mathrm{~mol} \cdot 0.0821 \frac{L \cdot atm}{ mol \cdot K} \cdot 298.15 K}{10.00 L}\) \(P_{Ar} = 0.2677 \, atm\)
03

Calculate the total pressure

Now that we have the partial pressures of each gas, we can find the total pressure of the gas mixture using Dalton's Law of partial pressures: \(P_{total} = P_{He} + P_{Ne} + P_{Ar}\) \(P_{total} = 1.8559 \, atm + 0.8077 \, atm + 0.2677 \, atm\) \(P_{total} = 2.9313 \, atm\) So, the partial pressures of Helium, Neon, and Argon in the mixture are 1.8559 \(atm\), 0.8077 \(atm\), and 0.2677 \(atm\), respectively, and the total pressure of the gas mixture is 2.9313 \(atm\).

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Key Concepts

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

Ideal Gas Law
Understanding the Ideal Gas Law goes beyond plugging values into an equation— it’s essential for predicting how gases will behave under different conditions. The law is usually stated as \( PV = nRT \), where P is pressure, V is volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

When it comes to practical applications, such as calculating the partial pressure of a gas in a mixture, the Ideal Gas Law lets us find out how much pressure one single gas contributes to the overall mixture. Remember, since gases in a mixture share the same volume and temperature, the only variable affecting each gas’s partial pressure is the amount of substance, or moles, of that specific gas.

To apply the law effectively, always ensure you convert the temperature to Kelvin, and use the value for R that matches the units of pressure and volume you need, as was demonstrated in the step-by-step solution for the exercise.
Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a crucial concept when dealing with gas mixtures. It states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Each gas in a mixture creates pressure as if it were alone in the container.

Mathematically, Dalton’s Law is expressed as \( P_{total} = P_1 + P_2 + P_3 + ... + P_n \), where each \( P \) represents the partial pressure of a gas in the mixture. The ease and beauty of this law shine when you realize it allows us to break down complex systems into simple, solvable parts. Find each gas's partial pressure — as if working with multiple miniature gas containers — and then add them up for the total pressure, just like in the solution for the given exercise.

This concept is particularly useful in a variety of scientific fields such as chemistry, environmental science, and respiratory physiology, where understanding the behavior of gas mixtures is critical.
Gas mixtures
Gas mixtures, like the ones often found in our atmosphere or industrial processes, consist of different gases occupying the same space. Unlike liquid or solid mixtures, which might show heterogeneity, gases mix uniformly due to their high kinetic energy and the space between gas particles.

In a mixture, each gas follows its own set of physical behaviors and properties, yet they don't interact with each other chemically if they are non-reactive. When we calculate the properties of gas mixtures, such as density or pressure, it's key to consider each particle type independently before summarising the mixture's overall behavior. The calculation of partial pressures is just one example where we see the individual contributions of each gas amalgamating to reveal the big picture of the gas mixture's behavior, reinforcing concepts like Dalton’s Law and Ideal Gas Law in the context of gas mixtures.

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Most popular questions from this chapter

In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for Xe, calculate the radius of a Xe atom and compare it to the value found in Figure 7.6, \(1.30 \AA\) A. Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s) $$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of 1.50 atm at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

The temperature of a 5.00-L container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean-square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls ner second.

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5} \mathrm{~mol}\) of argon is placed in a \(600-\mathrm{cm}^{3}\) lightbulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

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