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A cylinder with a movable piston records a volume of \(12.6 \mathrm{~L}\) when \(3.0 \mathrm{~mol}\) of oxygen is added. The gas in the cylinder has a pressure of \(5.83\) atm. The cylinder develops a leak and the volume of the gas is now recorded to be \(12.1 \mathrm{~L}\) at the same pressure. How many moles of oxygen are lost?

Short Answer

Expert verified
Answer: About 0.13 mol of oxygen was lost.

Step by step solution

01

Write down the initial state of the cylinder

The initial state of the cylinder is described by a volume of \(12.6 \mathrm{~L}\), a pressure of \(5.83\,\mathrm{atm}\), and \(3.0\,\mathrm{mol}\) of oxygen.
02

Write down the final state of the cylinder

The final state of the cylinder is described by a volume of \(12.1\,\mathrm{L}\), a pressure of \(5.83\,\mathrm{atm}\), and an unknown number of moles of oxygen (which we will call \(n_2\)).
03

Set up the proportional relationship equation

Because the pressure and temperature remain constant, we can use the proportional relationships between volume and the number of moles to find the number of moles lost. The relationship can be written as: \(\frac{V_1}{n_1} = \frac{V_2}{n_2}\) Where \(V_1\) and \(V_2\) are the initial and final volumes, and \(n_1\) and \(n_2\) are the initial and final number of moles, respectively.
04

Solve the equation for \(n_2\)

Insert the given values of \(V_1\), \(V_2\), and \(n_1\) into the equation: \(\frac{12.6\,\mathrm{L}}{3.0\,\mathrm{mol}} = \frac{12.1\,\mathrm{L}}{n_2}\) Solve the equation for \(n_2\), which represents the final number of moles of oxygen: \(n_2 = \frac{12.1\,\mathrm{L} \times 3.0\,\mathrm{mol}}{12.6\,\mathrm{L}}\) \(n_2 \approx 2.87\,\mathrm{mol}\)
05

Calculate the number of moles lost

To find the number of moles of oxygen lost, simply subtract the final number of moles \(n_2\) from the initial number of moles \(n_1\): Moles lost = \(n_1 - n_2\) Moles lost = \(3.0\,\mathrm{mol} - 2.87\,\mathrm{mol}\) Moles lost = \(0.13\,\mathrm{mol}\)
06

Write down the final answer

Therefore, during the leak, about \(0.13\,\mathrm{mol}\) of oxygen was lost.

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Key Concepts

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

Ideal Gas Equation
The Ideal Gas Equation is crucial in understanding how gases behave under different conditions. It is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This equation helps us relate these different properties of a gas and predict changes. In the context of the problem, since the temperature and pressure were constant, the Ideal Gas Equation simplifies our calculations by ensuring changes in volume directly relate to changes in moles of gas.
Moles
Moles are a fundamental concept for understanding quantities in chemistry. A mole represents \( 6.022 \times 10^{23} \) entities, such as atoms or molecules, and helps chemists quantify substances at the molecular level.
For the given problem, calculating the number of moles was key to determining how much oxygen was lost from the cylinder. Initially, the cylinder had \( 3.0 \) moles of oxygen. When the volume changed due to the leak, it became necessary to calculate the new number of moles using the proportional volume-mole relationship. Ultimately, by understanding moles, you can quantify the exact amount of a substance gained or lost in a chemical setting.
Volume-Pressure Relationship
The relationship between volume and pressure is central in understanding gas behavior, particularly in systems where the temperature is held constant, following Boyle's Law. \( PV = constant \)In our given scenario, although the exercise kept the pressure constant while the volume changed, understanding this relationship helps explain why volume changes even with constant pressure.
Since the temperature and pressure were constant, the volume's change was solely due to the change in the number of moles of gas. While Boyle's Law often details the inverse relationship in varying pressure situations, here it simplifies the understanding that less gas would occupy less volume but under unchanged pressure conditions.
Chemical Calculations
Chemical Calculations in the context of gas laws often involve using proportions and relationships like the Ideal Gas Law for detailed insights. In this exercise, you applied these calculations to the proportionate relationship between volume and the number of moles, as both pressure and temperature remained constant.
  • Identified the initial and final states of the gas system.
  • Utilized the proportional equation \( \frac{V_1}{n_1} = \frac{V_2}{n_2} \) to find unknown quantities.
  • Solved for the unknown final moles \( n_2 \).
  • Calculated the moles of oxygen lost by comparing the initial and final quantities.
Understanding these calculation techniques allows for precise predictions and explanations of real-world chemical phenomena, like gas leaks, where a change in one aspect affects others predictably.

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Most popular questions from this chapter

Sketch a cylinder with ten molecules of helium (He) gas. 'The cylinder has a movable piston. Label this aketch before. Make an after abetch to represent (a) decrease in temperature at constant pressure. (b) a decrease in pressure fram \(1000 \mathrm{~mm} \mathrm{Hg}\) to \(500 \mathrm{~mm} \mathrm{Hg}\) at constant temperature. (c) five molecules of \(\mathrm{H}_{2}\) gas added at constant temperature and pressure.

Ammonium nitrate can be used as an effective explosive because it decomposes into a large number of gaseous products. At a sufficiently high temperature, ammonium nitrate decomposes into nitrogen, oxygen, and steam. (a) Write a balanced equation for the decomposition of ammonium nitrate. (b) If \(1.00 \mathrm{~kg}\) of ammonium nitrate is sealed into a \(50.0\) - \(\mathrm{L}\) steel drum and heated to \(787^{\circ} \mathrm{C}\), what is the pressure in the drum, assuming \(100 \%\) decomposition?

A mixture in which the mole ratio of hydrogen to oxygen is \(2: 1\) is used to prepare water by the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ The total pressure in the container is \(0.950 \mathrm{~atm}\) at \(25^{\circ} \mathrm{C}\) before the reaction. What is the final pressure in the container at \(125^{\circ} \mathrm{C}\) after the reaction, assuming an \(88.0 \%\) yield and no volume change?

When acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is burned in oxygen, carbon dioxide and steam are formed. A sample of acetylene with a volume of \(7.50 \mathrm{~L}\) and a pressure of \(1.00\) atm is burned in excess oxygen at \(225^{\circ} \mathrm{C}\). The products are transferred without loss to a 10.0-L flask at the same temperature. (a) Write a balanced equation for the reaction. (b) What is the total pressure of the products in the \(10.0\) - \(\mathrm{L}\) flask? (c) What is the partial pressure of each of the products in the flask?

Exhaled air contains \(74.5 \% \mathrm{~N}_{2}, 15.7 \% \mathrm{O}_{2}, 3.6 \% \mathrm{CO}_{2}\), and \(6.2 \% \mathrm{H}_{2} \mathrm{O}\) (mole percent). (a) Calculate the molar mass of exhaled air. (b) Calculate the density of exhaled air at \(37^{\circ} \mathrm{C}\) and \(757 \mathrm{~mm} \mathrm{Hg}\), and compare the value obtained with that for ordinary air (MM = \(29.0 \mathrm{~g} / \mathrm{mol}\) ).

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