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Chapter 5: Problem 5

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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Most popular questions from this chapter

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