boyle's law problem solving with answers

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Boyles Law Questions

Boyle’s law is also referred to as Boyle–Mariotte law or Mariotte’s law. It tells us about the behaviour of gases. Boyle’s law states that the pressure is inversely proportional to the volume of the gas at constant pressure.

P ∝ 1 / V

Boyles Law Chemistry Questions with Solutions

Q1. Suppose P, V, and T represent the gas’s pressure, volume, and temperature, then the correct representation of Boyle’s law is

• V is inversely proportional to T (at constant P)
• V inversely proportional to P (at constant T)

Answer: (b), If P, V, and T represent the gas’s pressure, volume, and temperature, then the correct representation of Boyle’s law is V inversely proportional to P (at constant T).

V ∝ 1 / P

Q2. What is the nature of Boyle’s Law’s pressure vs volume (P vs V) graph?

• Straight Line
• Rectangular Hyperbola
• None of the above

Answer: (b), The nature of Boyle’s Law’s pressure vs volume (P vs V) graph is a rectangular hyperbola.

Q3. What is the nature of Boyle’s Law’s pressure-volume vs pressure (PV vs P) graph?

• Straight-line parallel to the P axis
• Straight-line parallel to the PV axis
• Straight-line parallel to the V axis

Answer: (a), The nature of Boyle’s Law’s pressure-volume vs pressure (PV vs P) graph is a straight line parallel to the P axis.

Q4. Which of the following quantity is kept constant in Boyle’s law?

• Gas mass only
• Gas Temperature only
• Gas Mass and Gas Pressure
• Gas Mass and Gas Temperature

Answer: (d), In Boyle’s law, the mass of the gas its temperature are kept constant.

Q5. Boyle’s law is valid only for

• Ideal gases
• Non-ideal gases
• Light Gases
• Heavy Gases

Answer: (a), Boyle’s law is valid only for ideal gases.

Q6. What is Boyle’s law?

Answer: Boyle’s law depicts the relationship between the pressure, volume, and temperature of a gas. It states that the pressure of a gas is inversely proportional to its volume at a constant temperature.

Q7. How is Boyle’s law used in everyday life?

Answer: Boyle’s law can be observed in our everyday life. Filling air in the bike tire is one of the significant applications of Boyle’s law. While pumping air into the tyre, the gas molecules inside the tire are compressed and packed closer together. It increases the pressure exerted on the walls of the tyre.

Q8. What is Boyle’s temperature?

Answer: Boyle’s temperature is the temperature at which the real and non-ideal gases behave like an ideal gas over a broad spectrum of pressure. It is related to the Van der Waal’s constant a, b as TB = a / Rb

Q9. Differentiate between Boyle’s law and Charle’s law.

Q10. Match the following gas laws with the equation representing them.

Q11. A helium balloon has a volume of 735 mL at ground level. The balloon is transported to an elevation of 5 km, where the pressure is 0.8 atm. At this altitude, the gas occupies a volume of 1286 mL. Assuming that the temperature is constant, what was the ground level pressure?

Answer: Given

Initial Volume (V 1 ) = 735 mL

Final Pressure (P 2 ) = 0.8 atm

Final Volume (V 2 ) = 1286 mL

To Find: Initial Pressure (P 1 ) = ?

We can calculate the initial pressure of the gas using Boyle’s law.

P 1 V 1 = P 2 V 2

P 1 X 735 = 0.8 X 1286

P 1 = 1028.8 / 735

P 1 = 1.39 ≈ 1.4 atm

Hence the ground level pressure is 1.4 atm.

Q12. A sample of oxygen gas has a volume of 225 mL when its pressure is 1.12 atm. What will the volume of the gas be at a pressure of 0.98 atm if the temperature remains constant?

Initial Volume (V 1 ) = 225 mL

Initial Pressure (P 1 ) = 1.12 atm

Final Pressure (P 2 ) = 0.98 atm

To Find: Final Volume (V 2 ) = ?

We can calculate the final volume of the gas using Boyle’s law.

1.12 X 225 = 0.98 X V 2

252 = 0.98 X V 2

252 / 0.98 = V 2

V 2 = 257.14 mL ≈ 257mL

Hence the final volume of the gas at pressure of 0.98 atm is equivalent to 257 mL.

Q13. An ideal gas occupying a 2.0 L flask at 760 torrs is allowed to expand to a volume of 6,000 mL. Calculate the final pressure

Initial Volume (V 1 ) = 2 L

Initial Pressure (P 1 ) = 760 torrs

Final Volume (V 2 ) = 6000 mL = 6 L

To Find: Final Pressure (P 2 ) = ?

We can calculate the final pressure of the gas using Boyle’s law.

760 X 2 = P 2 X 6

1520 = P 2 X 6

P 2 = 1520 / 6

P 2 = 253.33 torrs ≈ 253 torrs

Hence the final pressure of the gas at volume of 6 L is equivalent to 253 torrs.

Q14. A gas occupies a volume of 1 L and exerts a pressure of 400 kPa on the walls of its container. What would be the pressure exerted by the gas if it is completely transferred into a new container having a volume of 3 litres (assuming that the temperature and amount of the gas remain the same.)?

Initial Volume (V 1 ) = 1 L

Initial Pressure (P 1 ) = 400 kPa

Final Volume (V 2 ) = 3 L

400 X 1 = P 2 X 3

P 2 = 400 / 3

P 2 = 133.33 ≈ 133 kPa

Hence the final pressure of the gas at of volume 3 L is equivalent to 133 kPa.

Q15. A gas exerts a pressure of 3 kPa on the walls of container 1. When container one is emptied into a 10 litre container, the pressure exerted by the gas increases to 6 kPa. Find the volume of container 1. Assume that the temperature and amount of the gas remain the same.

Initial Pressure (P 1 ) = 3 kPa

Final Volume (V 2 ) = 10 L

Final Pressure (P 2 ) = 6 kPa

To Find: Initial Volume (V 1 ) = ?

We can calculate the initial volume of the gas using Boyle’s law.

3 X V 1 = 6 X 10

3 X V 1 = 60

V 1 = 60 / 3

Hence the initial volume of the gas at pressure of 3 kPa is equivalent to 20 L.

Practise Questions on Boyle’s Law

Q1. A gas is initially in a 5 L piston with a pressure of 1 atm. What is the new volume if the pressure changes to 3.5 atm by moving the piston down?

Q2. A balloon of volume 0.666 L at 1.03atm is placed in a pressure chamber where the pressure becomes 5.68atm. Determine the new volume.

Q3. A gas in a 30.0 mL container is at a pressure of 1.05 atm and is compressed to a volume of 15.0 mL. What is the new pressure of the container?

Q4. If a gas occupies 3.60 litres at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

Q5. A gas occupies 12.3 litres at a pressure of 40.0 mmHg. What is the volume when the pressure is increased to 60.0 mmHg?

Click the PDF to check the answers for Practice Questions. Download PDF

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Boyle's Law: Worked Chemistry Problems

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If you trap a sample of air and measure its volume at different pressures (constant temperature ), then you can determine a relation between volume and pressure. If you do this experiment, you will find that as the pressure of a gas sample increases, its volume decreases. In other words, the volume of a gas sample at constant temperature is inversely proportional to its pressure. The product of the pressure multiplied by the volume is a constant:

PV = k or V = k/P or P = k/V

where P is pressure, V is volume, k is a constant, and the temperature and quantity of gas are held constant. This relationship is called Boyle's Law , after Robert Boyle , who discovered it in 1660.

Key Takeaways: Boyle's Law Chemistry Problems

• Simply put, Boyle's states that for a gas at constant temperature, pressure multiplied by volume is a constant value. The equation for this is PV = k, where k is a constant.
• At a constant temperature, if you increase the pressure of a gas, its volume decreases. If you increase its volume, the pressure decreases.
• The volume of a gas is inversely proportional to its pressure.
• Boyle's law is a form of the Ideal Gas Law. At normal temperatures and pressures, it works well for real gases. However, at high temperature or pressure, it is not a valid approximation.

Worked Example Problem

The sections on the General Properties of Gases and Ideal Gas Law Problems may also be helpful when attempting to work Boyle's Law problems .

A sample of helium gas at 25°C is compressed from 200 cm 3 to 0.240 cm 3 . Its pressure is now 3.00 cm Hg. What was the original pressure of the helium?

It's always a good idea to write down the values of all known variables, indicating whether the values are for initial or final states. Boyle's Law problems are essentially special cases of the Ideal Gas Law:

Initial: P 1 = ?; V 1 = 200 cm 3 ; n 1 = n; T 1 = T

Final: P 2 = 3.00 cm Hg; V 2 = 0.240 cm 3 ; n 2 = n; T 2 = T

P 1 V 1 = nRT ( Ideal Gas Law )

P 2 V 2 = nRT

so, P 1 V 1 = P 2 V 2

P 1 = P 2 V 2 /V 1

P 1 = 3.00 cm Hg x 0.240 cm 3 /200 cm 3

P 1 = 3.60 x 10 -3 cm Hg

Did you notice that the units for the pressure are in cm Hg? You may wish to convert this to a more common unit, such as millimeters of mercury, atmospheres, or pascals.

3.60 x 10 -3 Hg x 10mm/1 cm = 3.60 x 10 -2 mm Hg

3.60 x 10 -3 Hg x 1 atm/76.0 cm Hg = 4.74 x 10 -5 atm

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Boyle’s Law – Definition, Formula, Example

Boyle’s law or Mariotte’s law states that pressure of an ideal gas is inversely proportional to volume under conditions of constant mass and temperature. When the gas volume increases, pressure decreases. When the volume decreases, pressure increases. Boyle’s law takes its name from chemist and physicist Robert Boyle , who published the law in 1862.

Boyle’s law states that the absolute pressure of an ideal gas is inversely proportional to its volume under conditions of constant mass and temperature.

Boyle’s Law Formula

There are three common formulas for Boyle’s law:

P ∝ 1/V PV = k P 1 V 1 = P 2 V 2

P is absolute pressure, V is volume, and k is a constant.

Graphing Boyle’s Law

The graph of volume versus pressure has a characteristic downward curved shape that shows the inverse relationship between pressure and volume. Boyle used the graph of experimental data to establish the relationship between the two variables.

Richard Towneley and Henry Power described the relationship between the pressure and volume of a gas in the 17th century. Robert Boyle experimentally confirmed their results using a device constructed by his assistant, Robert Hooke. The apparatus consisted of a closed J-shaped tube. Boyle poured mercury into the tube, decreasing the air volume and increasing its pressure. He used different amounts of mercury, recording air pressure and volume measurements, and graphed the data. Boyle published his results in 1662. Sometimes the gas law is called the Boyle-Mariotte law or Mariotte’s law because French physicist Edme Mariotte independently discovered the law in 1670.

Examples of Boyle’s Law in Everyday Life

There are examples of Boyle’s law in everyday life:

• The bends : A diver ascends to the water surface slowly to avoid the bends. As a diver rises to the surface, the pressure from the water decreases, which increases the volume of gases in the blood and joints. Ascending too quickly allows these gases to form bubbles, blocking blood flow and damaging joints and even teeth.
• Air bubbles : Similarly, air bubbles expand as they rise up a column of water. If you have a tall glass, you can watch bubble expand in volume as pressure decreases. One theory about why ships disappear in the Bermuda Triangle relates to Boyle’s law. Gases released from the seafloor rise and expand so much that they essentially become a gigantic bubble by the time they reach the surface. Small boats fall into the bubbles and are engulfed by the sea.
• Deep-sea fish : Deep-sea fish die if you bring them up to the surface. As outside pressure drops, the volume of gas within their swim bladder increases. Essentially, the fish blow up or pop.
• Syringe : Depressing the plunger on a sealed syringe decreases the air volume inside it and increases its pressure. Similarly, if you have a syringe containing a small amount of water and pull back on the plunger, the volume of air increases, but it’s pressure decreases. The pressure drop is enough to boil the water within the syringe at room temperature.
• Breathing: The diaphragm expands the volume of the lungs, causing a pressure drop that allows outside air to rush into the lungs (inhalation). Relaxing the diaphragm reduces the volume of the lungs, increasing the gas pressure within them. Exhaling occurs naturally to equalize pressure.

Boyle’s Law Example Problem

For example, calculate the final volume of a balloon if it has a volume of 2.0 L and pressure of 2 atmospheres and the pressure is reduced to 1 atmosphere. Assume temperature remains constant.

P 1 V 1 = P 2 V 2 (2 atm)(2.0 L) = (1 atm)V 2 V 2 = (2 atm)(2.0 L)/(1 atm) V 2 = 4.0 L

It’s a good idea to check your work to make sure the answer makes sense. In this example, the balloon pressure decreased by a factor of two (halved). The volume increased and doubled. This is what you expect from an inverse proportion relationship.

Most of the time, homework and test questions require reasoning rather than math. For example, if volume increases by a factor of 10, what happens to pressure? You know increasing volume decreases pressure by the same amount. Pressure decreases by a factor of 10.

See another Boyle’s law example problem .

• Fullick, P. (1994).  Physics . Heinemann. ISBN 978-0-435-57078-1.
• Holton, Gerald James (2001). Physics, The Human Adventure: From Copernicus to Einstein and Beyond . Rutgers University Press. ISBN 978-0-8135-2908-0.
• Tortora, Gerald J.; Dickinson, Bryan (2006). ‘Pulmonary Ventilation’ in Principles of Anatomy and Physiology (11th ed.). Hoboken: John Wiley & Sons, Inc. pp. 863–867.
• Walsh, C.; Stride, E.; Cheema, U.; Ovenden, N. (2017). “A combined three-dimensional in vitro–in silico approach to modelling bubble dynamics in decompression sickness.” Journal of the Royal Society Interface . 14(137). doi: 10.1098/rsif.2017.0653
• Webster, Charles (1965). “The discovery of Boyle’s law, and the concept of the elasticity of air in seventeenth century”. Archive for the History of Exact Sciences . 2(6) : 441–502.

Related Posts

(6.00 atm) (4.00 L) = (2.00 atm) (x)

(790.5 mmHg) (25.3 mL) = ( 0.804 atm) (x) This is wrong!! You MUST change one of the pressures units so both are the same. I'll change the mmHg to atm: (790.5 mmHg / 760.0 mmHg/atm) (25.3 mL) = ( 0.804 atm) (x)

(1.00 atm) (12.0 L) = (2.00 atm) (x)

(4.00 atm) (30.0 mL) = (2.00 atm) (x)

(760.0 mmHg) (14.0 L) = (400.0 mmHg) (x)

(8.00 atm) (40.0 L ) = (12.0 atm) (x)

(2.00 atm) (200.0 L) = (600.0 kPa) (x) This is wrong. The pressure units must be the same. I'll change the atm to kPa. You could go the other way if you want, the answer would be the same. (2.00 atm x 101.325 kPa/atm) (200.0 L) = (600.0 kPa) (x) In fact, here's the problem with the kPa changed to atm: (2.00 atm) (200.0 L) = (600.0 kPa / 101.325 kPa/atm) (x)

Your lungs will "explode." As you go up towards the surface, the pressure on your body and lungs becomes less. The air in your lungs expands. What would happen is the alveoli and small capallaries in the lungs would rupture, causing massive bleeding in the lungs. You'd die. No, your body would not swell up and burst, like a balloon.

V 2 = (P 1 V 1 ) / P 2

(a) pressure/temperature (b) pressure/volume (c) volume/temperature (d) moles/pressure (e) none of these

(12 atm) (1.5 liter) = (2.0 atm) (x)

(645 mmHg) (300 mL) = (1.00 atm) (x) This is wrong. I will change atm to mmHg. (645 mmHg) (300 mL) = (760 mmHg) (x)

(1.00 atm) (x) = (3.00 atm) (0.50 cu. ft.)

(50.0 torr) (75.0 mL) = (8.00 torr) (x)

(1960.0 p.s.i.) (50.0 liters) = (30.00 p.s.i.) (x) Important point: 2000 psi − 40 psi = 1960 psi flowed out of tank. 2000 is not used in calculation.

[(4/3)(3.14159)(2.5 m) 3 ] (1.004 x 10 5 Pa) = [(4/3)(3.14159)(x) 3 ] (2.799 x 10 4 Pa)

(2.5 m) 3 (1.004 x 10 5 Pa) = (x) 3 (2.799 x 10 4 Pa) x 3 = [(15.625 m 3 ) (1.004 x 10 5 Pa)] / 2.799 x 10 4 Pa x 3 = 56.05 m 3 x = 3.8 m (to two significant figures)

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High School Chemistry : Using Boyle's Law

Study concepts, example questions & explanations for high school chemistry, all high school chemistry resources, example questions, example question #11 : gases and gas laws.

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:

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

Example Question #1 : Using Boyle's Law

What law is the following formula?

Charles's law

Boyle's law

Combined gas law

Ideal gas law

Gay-Lussac's law

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.

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

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

The graph depicted here represents which of the gas laws?

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:

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 #13 : Gases And Gas Laws

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

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?

Use Boyle's Law:

Plug in known values and solve for final volume.

Use Boyle's law and plug in appropriate parameters:

Boyle's Law is:

Notice the answer has 3 significant figures.

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