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Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C\), is added to the same vessel at the same temperature. How does the addition of gas \(C\) affect the following: (a) the partial pressure of gas A, (b) the total pressure in the vessel, (c) the mole fraction of gas B?

Short Answer

Expert verified
(a) The partial pressure of gas A will decrease, as the addition of gas C doesn't affect the moles of gas A, but decreases its mole fraction. (b) The total pressure in the vessel will increase, as the added gas C contributes to the total pressure. (c) The mole fraction of gas B will decrease, as the total number of moles in the mixture increases due to the addition of gas C.

Step by step solution

01

Understanding the ideal gas laws and partial pressure

For the given problem, we will refer to the Dalton's Law of Partial Pressures that states: The total pressure in a mixture of non-reacting gases is equal to the sum of partial pressures of individual gases. The formula for partial pressure can be expressed as: \( P_i = X_i * P_t \) where \( P_i\) is the partial pressure of gas i, \(X_i\) is the mole fraction of gas i, and \(P_t\) is the total pressure of the system.
02

Analyzing the addition of gas C

When we add gas C to the vessel, the number of moles of gas C is increased. The addition of gas C won't change the number of moles of gases A and B in the vessel, but it will influence the total pressure and the mole fractions of A and B.
03

Calculating the effect on partial pressure of gas A

The partial pressure of a gas in a mixture depends on its mole fraction. Since the addition of gas C does not change the number of moles of gas A, the mole fraction of gas A decreases as the number of moles of gas C increases. Hence, the partial pressure of gas A will decrease. (a) Effect on the partial pressure of gas A: It will decrease.
04

Calculating the effect on the total pressure in the vessel

The total pressure in the vessel is given by the sum of partial pressures of all the gases. When we add gas C to the vessel, its partial pressure will contribute to the total pressure. As a result, the total pressure will increase in the closed vessel. (b) Effect on the total pressure in the vessel: It will increase.
05

Calculating the effect on mole fraction of gas B

The mole fraction of a gas is the ratio of the number of moles of that gas to the total number of moles of all gases in the mixture. When gas C is added, the total number of moles in the mixture increases, causing the mole fractions of both gases A and B to decrease. (c) Effect on the mole fraction of gas B: It will decrease.

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Key Concepts

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

Partial Pressure
Partial pressure is a foundational concept in understanding gas behavior, particularly in mixtures. It represents the pressure that a single gas component would exert if it alone occupied the entire volume of the mixture at the same temperature. In a mixture of gases, such as in our textbook exercise, each gas exerts its own pressure independently of the others.

This concept is integral to Dalton's Law of Partial Pressures, which shows that the total pressure within a vessel is the sum of the partial pressures of each individual gas present. This can be mathematically represented by the equation: \( P_i = X_i * P_t \)where \( P_i \) is the partial pressure of gas i, \( X_i \) is the mole fraction of gas i, and \( P_t \) is the total pressure of all gases in the mixture. When a new gas is introduced to the mixture, as in the exercise, the partial pressure of the original gases, like gas A, is adjusted due to the change in total pressure and mole fraction.
Mole Fraction
Mole fraction, a unitless measure, represents the proportion of a particular gas in a mixture, relative to the total number of moles of all gases. For any individual gas, it is calculated by dividing the number of moles of that gas by the total number of moles in the mixture. In symbols, the mole fraction of gas i in a mixture is written as:\( X_i = \frac{n_i}{n_{total}} \)where \( n_i \) is the number of moles of gas i, and \( n_{total} \) is the total number of moles of all gases combined. Understanding mole fraction is crucial for predicting how the partial pressure of a gas will change if the composition of the mixture changes, as it does with the addition of a third gas, C. The more the total moles increase, the smaller each original gas's mole fraction becomes.
Ideal Gas Laws
The ideal gas laws provide a clear framework for understanding and predicting the behavior of gases under various conditions. Key formulas within this framework include the ideal gas equation \( PV = nRT \), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. This law implies that the behavior of a gas is predictable and quantifiable, assuming ideal conditions of non-interacting particles with negligible volume.

However, when dealing with real gases, certain deviations are taken into account, especially at high pressures and low temperatures. In the context of our exercise, these laws allow us to understand how adding a gas to a mixture influences the total pressure and the behavior of individual gas components.
Total Pressure in Gas Mixtures
Total pressure in gas mixtures is the cumulative effect of all individual gases exerting pressure within a container. According to Dalton's Law, it is equal to the sum of each gas's partial pressure. This relationship implies that as more gas is introduced to a container, the total pressure will rise, assuming the temperature remains constant and the volume is fixed.

For example, as a third gas C is added to our mixture of gases A and B, the total pressure will increase due to the additional partial pressure introduced by gas C. This phenomenon can be especially important to consider in chemical reactions, industrial processes, and physical systems where pressure plays a crucial role.

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Most popular questions from this chapter

The physical fitness of athletes is measured by \({ }^{~} V_{\mathrm{O}_{2}}\) max, " which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of \(45 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min, but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of \(88.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (a) Calculate the volume of oxygen, in mL, consumed in \(1 \mathrm{hr}\) by an average man who weighs \(185 \mathrm{lbs}\) and has a \(V_{\mathrm{O}_{2}}\) max reading of \(47.5 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (b) If this man lost \(20 \mathrm{lb}\), exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to \(65.0 \mathrm{~mL} \mathrm{O} / \mathrm{kg}\) body mass/min, how many \(\mathrm{mL}\) of oxygen would he consume in \(1 \mathrm{hr}\) ?

Consider the apparatus shown in the following drawing. (a) When the valve between the two containers is opened and the gases allowed to mix, how does the volume occupied by the \(\mathrm{N}_{2}\) gas change? What is the partial pressure of \(\mathrm{N}_{2}\) after mixing? (b) How does the volume of the \(\mathrm{O}_{2}\) gas change when the gases mix? What is the partial pressure of \(\mathrm{O}_{2}\) in the mixture? (c) What is the total pressure in the container after the gases mix?

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), in our bodies produces \(\mathrm{CO}_{2}\), which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and \(0.970 \mathrm{~atm}\) when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(1.00\) atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{~K}: \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{Cl}_{2}, \mathrm{HBr}\). (b) Calculate the rms speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K}\). (c) Calculate the most probable speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K}\).

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2}\), and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(0.985 \mathrm{~atm}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C}\), calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2}\) ? (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section \(3.2\) and Problem 10.57.)

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