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A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a \(5.00-\mathrm{L}\) vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0-\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of \(2.35 \mathrm{~g} \mathrm{~N}_{2}(g)\) originally in a \(2.50-\mathrm{L}\) vessel at \(20{ }^{\circ} \mathrm{C}\) is transferred to this same 10.0 - \(\mathrm{L}\) vessel. (a) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

Short Answer

Expert verified
The partial pressure of SO鈧(g) in the larger container is \(0.139 atm\), the partial pressure of N鈧(g) is \(0.248 atm\), and the total pressure in the vessel is \(0.387 atm\).

Step by step solution

01

Find the moles of SO鈧 and N鈧 gas

To find the moles of each gas, we use the given mass of the gas and the molar mass of the gas. The molar mass of SO鈧 is 32.07 g/mol (S) + 2 脳 16 g/mol (O) = 64.07 g/mol. The molar mass of N鈧 is 2 脳 14 g/mol (N) = 28 g/mol. For SO鈧: moles = mass / molar mass = 3 g / 64.07 g/mol = 0.0468 mol For N鈧: moles = mass / molar mass = 2.35 g / 28 g/mol = 0.0839 mol
02

Calculate initial pressures of both gases

Now that we have the number of moles for each gas, we can use the ideal gas equation to find the pressure of each gas in their initial containers. The ideal gas equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.08206 Latm/molK), and T is the temperature in Kelvin. First, convert the temperatures from Celsius to Kelvin: T鈧 (SO鈧) = 21掳C + 273.15 = 294.15 K T鈧 (N鈧) = 20掳C + 273.15 = 293.15 K Then, calculate the initial pressures: P鈧 (SO鈧) = (0.0468 mol 脳 0.08206 Latm/molK 脳 294.15 K) / 5.00 L = 0.251 atm P鈧 (N鈧) = (0.0839 mol 脳 0.08206 Latm/molK 脳 293.15 K) / 2.50 L = 0.906 atm
03

Find new temperature and moles of each gas in the 10 L container

Temperature in the new 10 L container is 26掳C - convert to Kelvin. T_new = 26掳C + 273.15 = 299.15 K Now, we can use the ideal gas equation to find the new volume and partial pressures of the gases in the 10 L container using V鈧 = V鈧 脳 (T鈧/T鈧).
04

Calculate final pressures of both gases

Since both gases are in the same 10 L container and at the same temperature (T_new), we can rearrange the ideal gas equation to solve for the final pressures of each gas. P_new (SO鈧) = (0.0468 mol 脳 0.08206 Latm/molK 脳 299.15 K) / 10 L = 0.139 atm P_new (N鈧) = (0.0839 mol 脳 0.08206 Latm/molK 脳 299.15 K) / 8.5 L = 0.248 atm
05

Calculate total pressure inside the container

Now that we have the partial pressures of both gases, we can use Dalton's Law of partial pressures to find the total pressure inside the container: P_total = P_new (SO鈧) + P_new (N鈧). P_total = 0.139 atm (SO鈧) + 0.248 atm (N鈧) = 0.387 atm The partial pressure of SO鈧(g) in the larger container is \(0.139 atm\), the partial pressure of N鈧(g) is \(0.248 atm\), and the total pressure in the vessel is \(0.387 atm\).

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Key Concepts

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

Partial Pressure
Partial pressure refers to the pressure that a single gas in a mixture contributes to the total pressure. In a container with mixed gases, each type of gas exerts its own pressure, known as its partial pressure. This is an essential concept, especially when dealing with reactions and processes involving gases in a mixed state. For instance, if a vessel contains both sulfur dioxide (SO鈧) and nitrogen gas (N鈧), the total pressure inside the vessel is the sum of the individual partial pressures of these gases.

The concept is mathematically expressed using the ideal gas law. For each gas in a mixture, you use its moles and the ideal gas equation: \[ P = \frac{{nRT}}{{V}} \] where:
  • P is the partial pressure of the gas.
  • n is the number of moles.
  • R is the ideal gas constant.
  • T is the temperature in Kelvin.
  • V is the volume of the container.
This formula helps in calculating how much each gas contributes to the total pressure in the container. By understanding partial pressures, we can predict how the mixture will behave under different conditions.
Dalton's Law
Dalton's Law is a key principle in chemistry regarding gases. It states that the total pressure exerted by a mixture of non-reacting gases is the sum of their individual partial pressures. This means that each gas in a mixture behaves independently and the total pressure is simply the addition of the pressures each gas would exert if it were alone in the entire volume.

Understanding Dalton's Law is crucial for solving problems involving gas mixtures. The law is mathematically represented as:\[ P_\text{total} = P_1 + P_2 + ... + P_n \]where:
  • \( P_\text{total} \) is the total pressure of the mixture.
  • \( P_1, P_2, ..., P_n \) are the partial pressures of each individual gas.
This law is particularly useful when calculating the pressure within a container holding multiple gases, like in the exercise where both SO鈧 and N鈧 are present. Applying Dalton's Law helps determine how each gas contributes to the overall pressure, ensuring a clear understanding of the behavior of gas mixtures.
Gas Molar Mass
Gas molar mass is a fundamental concept in chemistry. It refers to the mass of one mole of a given gas and is typically measured in grams per mole (g/mol). Molar mass is crucial when converting the mass of a gas to moles, which is necessary for applying the ideal gas law and computing various properties of gases, such as pressure or volume.

To determine a gas's molar mass, you add up the atomic masses of the elements in the compound. For example:
  • Sulfur dioxide (SO鈧) has a molar mass of 64.07 g/mol, calculated as 32.07 g/mol (for S) + 2 脳 16 g/mol (for O).
  • Nitrogen gas (N鈧) has a molar mass of 28 g/mol, derived from 2 脳 14 g/mol (for N).
Understanding molar mass enables us to find the number of moles in a given sample, which is essential for using the ideal gas law to solve real-world problems. In our exercise, knowing the molar masses allows us to compute how many moles of SO鈧 and N鈧 are present, setting the stage for calculating their partial pressures.
Temperature Conversion
Temperature conversion is a pivotal step in calculations involving gas laws. Typically, the ideal gas law requires temperature input in Kelvin. Since many temperatures are given in degrees Celsius, conversion is necessary. This conversion ensures temperatures align with the absolute temperature scale required by gas equations.

To convert Celsius to Kelvin, use the following formula:\[ T(K) = T(掳C) + 273.15 \]For instance:
  • 21掳C changes to 294.15 K.
  • 26掳C converts to 299.15 K.
  • 20掳C becomes 293.15 K.
Kelvin is used because it avoids negative values, simplifying computations involving gas laws. Consistently using Kelvin in calculations ensures accuracy, especially when applying the ideal gas law to determine properties like pressure or volume in gas mixtures.

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Most popular questions from this chapter

A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of 5.12 L. (a) Calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into (a) atmospheres, (b) torr, and (c) inches of \(\mathrm{Hg}\).

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?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathrm{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 0.984 atm and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5},\) can be prepared by the reaction of solid iodine and gaseous fluorine: $$ \mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g) $$ A 5.00-L flask containing \(10.0 \mathrm{~g} \mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g} \mathrm{~F}_{2}\), and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\). (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of \(\mathrm{IF}_{5}\). (d) What is the total mass of reactants and products in the flask?

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