High School Chemistry : Using Boyle's Law

Study concepts, example questions & explanations for High School Chemistry

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Example Questions

Example Question #11 : Gases And Gas Laws

An ideal gas exerts a pressure of \(\displaystyle \small 3atm\) in a \(\displaystyle \small 3L\) container. The container is at a temperature of \(\displaystyle \small 298K\).

What will be the final pressure if the volume of the container changes to \(\displaystyle \small 2L\)?

Possible Answers:

\(\displaystyle 2 atm\)

\(\displaystyle 4.5atm\)

\(\displaystyle 4atm\)

\(\displaystyle 3.5atm\)

Correct answer:

\(\displaystyle 4.5atm\)

Explanation:

Since the volume of the gas is the only variable that has changed, we can use Boyle's law in order to find the final pressure. Since pressure and volume are on the same side of the ideal gas law, they are inversely proportional to one another. In other words, as one increases, the other will decrease, and vice versa.

Boyle's law can be written as follows:

\(\displaystyle P_{1}V_{1} = P_{2}V_{2}\)

Use the given volumes and the initial pressure to solve for the final pressure.

\(\displaystyle (3atm)(3L) = (2L)P_{2}\)

\(\displaystyle P_{2} = 4.5atm\)

Example Question #2 : Using Boyle's Law

What law is the following formula?

\(\displaystyle P_{1}V_{1}= P_{2}V_{2}\)

Possible Answers:

Charles's law

Gay-Lussac's law

Combined gas law

Boyle's law

Ideal gas law

Correct answer:

Boyle's law

Explanation:

Boyle's law relates the pressure and volume of a system, which are inversely proportional to one another. When the parameters of a system change, Boyle's law helps us anticipate the effect the changes have on pressure and volume.

\(\displaystyle P_1V_1=P_2V_2\)

Charles's law relates temperature and volume: \(\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}\)

Gay-Lussac's law relates temperature and pressure: \(\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}\)

The combined gas law takes Boyle's, Charles's, and Gay-Lussac's law and combines it into one law: \(\displaystyle \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

The ideal gas law relates temperature, pressure, volume, and moles in coordination with the ideal gas constant: \(\displaystyle PV=nRT\)

Example Question #12 : Gases And Gas Laws

A helium balloon has a volume of \(\displaystyle 735mL\) when it is at ground level. The balloon is transported to an elevation of \(\displaystyle 5km\), where the pressure is only \(\displaystyle 0.8atm\). At this altitude the gas occupies a volume of \(\displaystyle 1286mL\). Assuming the temperature has remained the same, what was the ground level pressure?

Possible Answers:

\(\displaystyle 1.2atm\)

\(\displaystyle 1.0atm\)

\(\displaystyle 0.7atm\)

\(\displaystyle 1.4atm\)

Correct answer:

\(\displaystyle 1.4atm\)

Explanation:

To solve this question we will need to use Boyle's law:

\(\displaystyle P_1V_2=P_2V_2\)

We are given the final pressure and volume, along with the initial volume. Using these values, we can calculate the initial pressure.

\(\displaystyle P_1(735mL)=(0.8atm)(1286mL)\)

\(\displaystyle P_1=\frac{(0.8atm)(1286mL)}{735mL}\)

\(\displaystyle P_1=1.4atm\)

Note that the pressure at sea level is equal to \(\displaystyle 1.0atm\). A pressure greater than \(\displaystyle 1.0atm\) simply indicates that the ground level is below sea level at this point.

Example Question #1 : Using Boyle's Law

The graph depicted here represents which of the gas laws?

Boyle_s_law_graph

Possible Answers:

Gay-Lussac's law

Hund's law

Boyle's law

Charles's law

Correct answer:

Boyle's law

Explanation:

The graph shows that there is an inverse relationship between the volume and pressure of a gas, when kept at a constant temperature. This was described by Robert Boyle and can be represented mathematically as Boyle's law:

\(\displaystyle P_1V_1=P_2V_2\)

Gay-Lussac's law shows the relationship between pressure and temperature. Charles's law shows the relationship between volume and temperature. Hund's rule (Hund's law) is not related to gases, and states that electron orbitals of an element will be filled with single electrons before any electrons will form pairs within a single orbital.

Example Question #2 : Using Boyle's Law

A sample of oxygen gas has a volume of \(\displaystyle 225mL\) when its pressure is \(\displaystyle 1.12atm\). What will the volume of the gas be at a pressure of \(\displaystyle 0.98atm\) if the temperature remains constant?

Possible Answers:

\(\displaystyle 257mL\)

\(\displaystyle 250mL\)

\(\displaystyle 316mL\)

\(\displaystyle 220mL\)

Correct answer:

\(\displaystyle 257mL\)

Explanation:

To solve this question we will need to use Boyle's law:

\(\displaystyle P_1V_2=P_2V_2\)

We are given the initial pressure and volume, along with the final pressure. Using these values, we can calculate the final volume.

\(\displaystyle (1.12atm)(225mL)=(0.98atm)V_2\)

\(\displaystyle V_2=\frac{(1.12atm)(225mL)}{0.98atm}\)

\(\displaystyle V_2=257mL\)

Example Question #6 : Using Boyle's Law

A gas is initially in a 5L piston with a pressure of 1atm.

If pressure changes to 3.5atm by moving the piston down, what is new volume?

Possible Answers:

\(\displaystyle 2.7 L\)

\(\displaystyle 1. 31 L\)

\(\displaystyle 1.75 L\)

\(\displaystyle 2.03 L\)

\(\displaystyle 1.43 L\)

Correct answer:

\(\displaystyle 1.43 L\)

Explanation:

Use Boyle's Law:

\(\displaystyle P_1*V_1 = P_2*V_2\)

Plug in known values and solve for final volume.

\(\displaystyle 5L+1atm=3.5atm*V_2\)

\(\displaystyle V_2=1.43L\)

Example Question #7 : Using Boyle's Law

A balloon of volume \(\displaystyle .666L\) at \(\displaystyle 1.03atm\) is placed in a pressure chamber where the pressure becomes \(\displaystyle 5.68atm\), determine the new volume.

Possible Answers:

\(\displaystyle .033L\)

\(\displaystyle .467L\)

\(\displaystyle 1.27L\)

\(\displaystyle .121L\)

\(\displaystyle .236L\)

Correct answer:

\(\displaystyle .121L\)

Explanation:

Use Boyle's law and plug in appropriate parameters:

\(\displaystyle P_1V_1=P_2V_2\)

\(\displaystyle .666L*1.03atm=5.68atm*V_2\)

\(\displaystyle \frac{.666L*1.03atm}{5.68L}=V_2\)

\(\displaystyle .121L=V_2\)

Example Question #8 : Using Boyle's Law

A gas in a \(\displaystyle 30.0\, mL\) container is at \(\displaystyle 1.05\, atm\) is compressed to a volume of \(\displaystyle 15.0\, mL\). What is the new pressure of the container?

Possible Answers:

\(\displaystyle P_{2}=1.05\, atm\)

\(\displaystyle P_{2}=0.525\, atm\)

\(\displaystyle P_{2}=472.5\, atm\)

\(\displaystyle P_{2}=2.10\, atm\)

Correct answer:

\(\displaystyle P_{2}=2.10\, atm\)

Explanation:

Boyle's Law is:

\(\displaystyle P_{1} V_{1}=P_{2} V_{2}\)

The initial volume (\(\displaystyle V_{1}\)) and pressure (\(\displaystyle P_{1}\)) of the gas is given. The volume changes to a new volume (\(\displaystyle V_{2}\)). Our goal is to find the new pressure (\(\displaystyle P_{2}\)). Solving for the new pressure gives:

\(\displaystyle \frac{P_{1} V_{1}}{V_{2}}=\frac{P_{2} V_{2}}{V_{2}}\rightarrow P_{2}=\frac{P_{1} V_{1}}{V_{2}}\)

\(\displaystyle P_{2}=\frac{(1.05\, atm)(30\, mL)}{15\, mL}=2.10\, atm\)

Notice the answer has 3 significant figures.

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