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Two glass bulbs are connected by a valve. One bulb has a volume of \(650.0\) milliliters and is occupied by \(\mathrm{N}_{2}(g)\) at 825 Torr. The other has a volume of \(500.0\) milliliters and is occupied by \(\mathrm{O}_{2}(g)\) at 732 Torr. The valve is opened and the two gases mix. Calculate the total pressure and the partial pressures of \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) in the resulting mixture.

Short Answer

Expert verified
The total pressure is 784.565 Torr; partial pressures are 466.304 Torr for Nâ‚‚ and 318.261 Torr for Oâ‚‚.

Step by step solution

01

Determine Total Volume

Calculate the total volume of the system by adding the volumes of both bulbs. The total volume is the sum of the initial volume of the bulb containing nitrogen and the bulb containing oxygen: \[ V_{\text{total}} = V_{\text{N}_2} + V_{\text{O}_2} = 650.0 \text{ mL} + 500.0 \text{ mL} = 1150.0 \text{ mL} \]
02

Apply Boyle's Law for Nâ‚‚

Using Boyle's Law, calculate the partial pressure of nitrogen gas in the combined volume. We assume temperature remains constant. Boyle's Law is given by \( P_1V_1 = P_2V_2 \) :\[ P_{\text{N}_2\text{ final}} \times V_{\text{total}} = P_{\text{N}_2\text{ initial}} \times V_{\text{N}_2} \]Substitute the known values:\[ P_{\text{N}_2\text{ final}} \times 1150.0 = 825 \times 650.0 \]\[ P_{\text{N}_2\text{ final}} = \frac{825 \times 650.0}{1150.0} = 466.304 \text{ Torr} \]
03

Apply Boyle's Law for Oâ‚‚

Similarly, calculate the partial pressure of oxygen gas in the combined volume using Boyle's Law:\[ P_{\text{O}_2\text{ final}} \times V_{\text{total}} = P_{\text{O}_2\text{ initial}} \times V_{\text{O}_2} \]Substitute the known values:\[ P_{\text{O}_2\text{ final}} \times 1150.0 = 732 \times 500.0 \]\[ P_{\text{O}_2\text{ final}} = \frac{732 \times 500.0}{1150.0} = 318.261 \text{ Torr} \]
04

Calculate Total Pressure

Add the partial pressures of nitrogen and oxygen to calculate the total pressure in the system.\[ P_{\text{total}} = P_{\text{N}_2\text{ final}} + P_{\text{O}_2\text{ final}} \]\[ P_{\text{total}} = 466.304 + 318.261 = 784.565 \text{ Torr} \]

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Key Concepts

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

Boyle's Law
Boyle's Law is a fundamental principle in physics and chemistry that describes how the pressure of a gas tends to increase as the volume of the container decreases, and vice versa, provided the temperature remains constant. This relationship can be expressed mathematically as: \( P_1V_1 = P_2V_2 \) Where:
  • \( P_1 \) and \( P_2 \) are the initial and final pressures of the gas respectively.
  • \( V_1 \) and \( V_2 \) are the initial and final volumes respectively.
In the given exercise, Boyle's Law is utilized to find the new partial pressures of nitrogen and oxygen gases after mixing them in a larger volume. For example, starting with nitrogen gas at 825 Torr in 650 mL, we calculate its new pressure in the combined 1150 mL volume by rearranging the formula to find \( P_2 \). This step demonstrates the inverse relationship between pressure and volume, key to Boyle's Law. The same approach is taken for oxygen gas with its initial conditions. Boyle's Law is quite handy in scenarios involving gas mixtures and understanding gas behavior under varying volumes.
partial pressure
In a mixture of gases, each gas exerts a pressure independently, called its partial pressure. It is the pressure that the gas would exert if it were the only gas present in the entire volume. This concept is crucial to solving problems involving gas mixtures like in the exercise. To calculate partial pressures using Boyle's Law, we're essentially seeing how each individual gas contributes to the total pressure when volumes are combined. For nitrogen, the calculation involves adjusting its initial pressure from 825 Torr, found in its initial volume of 650 mL, to fit into the total volume of 1150 mL when the valve is opened. This yields a partial pressure of 466.304 Torr for nitrogen after mixing. Similarly, for oxygen gas starting at 732 Torr in 500 mL, Boyle's Law helps us find it's new pressure, 318.261 Torr, in the combined volume. These partial pressures play a significant role in predicting how gases behave in a mixed system, making it easier to calculate total pressure.
total pressure
The total pressure of a gas mixture is the sum of the partial pressures of all the gases present. This idea comes from Dalton's Law of Partial Pressures, which states that each gas in a mixture exerts pressure independent of the other gases. In our exercise, once we have found the partial pressures for nitrogen and oxygen, which are 466.304 Torr and 318.261 Torr respectively, we simply add them to get the total pressure. \[ P_{\text{total}} = P_{\text{N}_2\text{ final}} + P_{\text{O}_2\text{ final}} \] \[ P_{\text{total}} = 466.304 + 318.261 = 784.565 \text{ Torr} \] This calculation shows how the pressures combine to give the overall pressure in the container after the gases have mixed. This total pressure is crucial in understanding how gases will behave in the system, and it is often used to predict outcomes in chemical reactions involving gases. Understanding both partial and total pressures helps clarify how gases interact when mixed.

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Most popular questions from this chapter

A 0.622-gram mixture of gases contained in a vessel at 1450 Torr is found to be \(22.2 \%\) argon, \(68.5 \%\) helium, and \(9.3 \%\) fluorine by mass. What is the partial pressure of each of the gases in the mixture?

The Dumas method was one of the first reliable techniques for determining the molar mass of a volatile liquid. In this method a liquid is boiled inside an open glass bulb so that its vapor completely fills the bulb, displacing any air present. The vapor inside the bulb is then cooled and the condensate weighed. From this mass, the boiling point of the liquid, and the atmospheric pressure, one can determine the molar mass of the liquid. A chemist using this technique finds that a pure volatile liquid with a boiling point of \(80.1^{\circ} \mathrm{C}\) fills a \(500.0-\mathrm{cm}^{3}\) bulb at an atmospheric pressure of \(755.3\) Torr. If the weighed condensate has a mass of \(1.335\) grams, what is the molecular mass of the liquid? If the empirical formula of the unknown liquid is \(\mathrm{CH}\), what is the molecular formula?

Nitrogen is often injected into cans and bottles to keep the contents from oxidizing. If the nitrogen injected into a particular bottle occupies a volume of \(10.0\) milliliters at a pressure of \(1.2\) bar, what volume will it occupy at a pressure of \(0.95\) bar at the same temperature?

Nitrous oxide, \(\mathrm{N}_{2} \mathrm{O}(g)\), can be prepared by gently heating equimolar quantities of potassium nitrate and ammonium chloride: $$ \mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{KNO}_{3}(s) \rightarrow \mathrm{KCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{N}_{2} \mathrm{O}(g) $$ How many grams of \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) and \(\mathrm{KNO}_{3}(s)\) are required to produce 825 milliliters of \(\mathrm{N}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C}\) and 748 Torr?

A gaseous mixture consisting of \(0.513\) grams of \(\mathrm{H}_{2}(g)\) and \(16.1\) grams of \(\mathrm{N}_{2}(g)\) occupies \(10.0\) liters at \(20.0^{\circ} \mathrm{C}\). Calculate the partial pressures of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) in the mixture in units of atmospheres.

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