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A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) pressure in a 1.00 - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

Short Answer

Expert verified
The total pressure in the new container is given by Dalton's Law of Partial Pressures, which is the sum of the final pressures for N₂ and O₂ gases. Calculate the moles of each gas using the Ideal Gas Law, then compute their respective final pressures based on the new volume and temperature. Finally, add these partial pressures to obtain the total pressure: \(P_{total} = P_3 + P_4\).

Step by step solution

01

Ideal Gas Law for initial conditions

Using the Ideal Gas Law, \(PV = nRT\), we can calculate the number of moles (n) for each gas in their respective initial conditions. The given values are \(R = 8.314\ \mathrm{J \cdot K^{-1} \cdot mol^{-1}}\), and temperature must be in Kelvin. For N₂ gas: Initial pressure (P₁): \(531.96\ \mathrm{kPa}\) Initial volume (V₁): \(1.00\ \mathrm{L}\) Initial temperature (T₁): \(26^{\circ} \mathrm{C} = 299\ \mathrm{K}\) For O₂ gas: Initial pressure (P₂): \(531.96\ \mathrm{kPa}\) Initial volume (V₂): \(5.00\ \mathrm{L}\) Initial temperature (T₂): \(26^{\circ} \mathrm{C} = 299\ \mathrm{K}\)
02

Calculate moles (n) for N₂ and O₂

Solve the Ideal Gas Law for n: n = \(\frac{PV}{RT}\) For N₂ gas: n₁ = \(\frac{(531.96\ \mathrm{kPa}) * (1.00\ \mathrm{L})}{(8.314\ \mathrm{J \cdot K^{-1} \cdot mol^{-1}}) * (299\ \mathrm{K})}\) For O₂ gas: n₂ = \(\frac{(531.96\ \mathrm{kPa}) * (5.00\ \mathrm{L})}{(8.314\ \mathrm{J \cdot K^{-1} \cdot mol^{-1}}) * (299\ \mathrm{K})}\)
03

Use the Ideal Gas Law for final conditions

Using the calculated moles of both gases and the new final conditions, we can calculate the final pressures for both gases. The final conditions are: Final volume (V₃): \(12.5\ \mathrm{L}\) Final temperature (T₃): \(20^{\circ} \mathrm{C} = 293\ \mathrm{K}\)
04

Calculate final pressure (P) for N₂ and O₂

Using the Ideal Gas Law and the final conditions, we can obtain the final pressures for N₂ and O₂: For N₂ gas: P₃ = \(\frac{n₁ * R * T₃}{V₃}\) For O₂ gas: P₄ = \(\frac{n₂ * R * T₃}{V₃}\)
05

Use Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures states that the total pressure is the sum of the partial pressures of the individual gases: Total pressure (P_total) = P₃ + P₄ Calculate the total pressure using the values obtained in Step 4.

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Key Concepts

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

Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is essential when dealing with mixtures of gases. This law tells us that, in a mixture of non-reactive gases, the total pressure exerted is equal to the sum of the partial pressures of each individual gas. A partial pressure is effectively the pressure that a gas would exert if it occupied the entire volume by itself. For example, if you have a container with nitrogen and oxygen gases, and each gas is given an initial pressure, you can find the total pressure exerted by adding the pressures of the nitrogen and the oxygen. The formula is simple: - Total Pressure, \( P_{\text{total}} = P_{\text{N}_2} + P_{\text{O}_2} \).This becomes particularly handy when gases expand or are combined in different containers. Understanding this law helps ensure accurate predictions of behavior in gaseous systems, like when solving for total pressures as shown in our exercise.
Gas Laws
Gas laws, like the Ideal Gas Law, describe relationships between measurable gas properties like pressure, volume, and temperature. They are founded on principles of behavior under different conditions. The Ideal Gas Law, represented by the equation \( PV = nRT \), combines several simpler gas laws such as Boyle's Law, Charles's Law, and Avogadro's Law. - **Boyle's Law** explores how pressure increases as volume decreases for a constant temperature.- **Charles's Law** shows how volume increases as temperature rises at constant pressure.- **Avogadro's Law** reveals that equal volumes of gas at the same temperature and pressure contain an equal number of molecules.To use the Ideal Gas Law in practical scenarios, it's crucial to have temperature in Kelvin, as the zero point for this scale reflects absolute zero where molecular movement stops. Using the law accurately helps predict changes in state, as seen in determining initial moles and pressures of gases moving between containers.
Molar Calculations
Molar calculations are vital in understanding the quantities of gases involved in chemical reactions or other transformations. To apply these, our key tool is the Ideal Gas Law, which allows us to solve for the number of moles () in a given condition.For instance, rearranging the Ideal Gas Law formula: \( n = \frac{PV}{RT} \) lets you calculate the moles from known values of pressure, volume, and temperature. In scenarios like the exercise, calculating moles helps understand the quantity of gas transferred to a new container, thereby aiding subsequent calculations of pressure in changed conditions.Moreover, molar calculations are foundational in stoichiometry, which measures reactants' and products' relative amounts. This principle supports predicting how gases will behave under altered conditions, allowing scientists and engineers to model systems accurately in fields ranging from chemistry to environmental science.

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Most popular questions from this chapter

Arsenic(III) sulfide sublimes readily, even below its melting point of \(320^{\circ} \mathrm{C}\). The molecules of the vapor phase are found to effuse through a tiny hole at 0.52 times the rate of effusion of Xe atoms under the same conditions of temperature and pressure. What is the molecular formula of arsenic(III) sulfide in the gas phase?

Radon (Rn) is the heaviest (and only radioactive) member of the noble gases. How much slower is the root-mean-square speed of Rn than He at \(300 \mathrm{K?}\)

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(210.0 \mathrm{~L}\) that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal \(15.2 \mathrm{MPa} ?\) (d) What would be the pressure of the gas, in \(\mathrm{kPa}\), if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

Imagine that the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,\) (b) the volume increases by \(33 \%\), (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\mathbf{e})\) the volume decreases by \(50 \%\). [Sections 10.3 and 10.4\(]\)

Which of the following statements is false? (a) Gases are far less dense than liquids. (b) Gases are far more compressible than liquids. (c) Because liquid water and liquid carbon tetrachloride do not mix, neither do their vapors. (d) The volume occupied by a gas is determined by the volume of its container.

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