boyle's law problem solving examples

(40.0 mmHg) (12.3 liters) = (60.0 mmHg) (x) x = 8.20 L Note three significant figures.

(1.00 atm) ( 3.60 liters) = (2.50 atm) (x) x = 1.44 L

(400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot) x = 133 atm

(1.56 L) (1.00 atm) = (3.00 atm) (x) 0.520 L

(11.2 liters) (0.860 atm) = (x) (15.0 L) x = 0.642 atm

(745.0 mmHg) (500.0 mL) = (760.0 mmHg) (x) x = 490.1 mL

(740.0 mmHg) (350.0 mL) = (760.0 mmHg) (x)

(63.0 atm) (338 L) = (1.00 atm) (x)

(166.0 kPa) (273.15 mL) = (101.325 kPa) (x)

(18.0 mmHg) (77.0 L) = (760.0 mmHg) (x)

Volume will decrease.

It will double in size.

(0.755 atm) (4.31 liters) = (1.25 atm) (x)

(8.00 atm) (600.0 mL) = (2.00 atm) (x)

(800.0 torr) (400.0 mL) = (1000.0 torr) (x)

100 °C is to 101.3 kPa as 88 °C is to x x = 89.144 kPa

P 1 V 1 = P 2 V 2 (101.3) (2.0) = (88.144) (x) x = 2.27 L The balloon will not burst.

(1.00 atm) (2.00 L) = (x) (5.00 L) x = 0.400 atm (1.50 atm) (3.00 L) = (y) (5.00 L) y = 0.900 atm

0.400 atm + 0.900 atm = 1.30 atm

(1.00) (2.00) = n 1 RT in the first bulb moles gas = n 1 = 2.00/RT (1.50) (3.00) = n 2 RT in the second bulb moles gas = n 2 = 4.50/RT

total volume = 2.00 + 3.00 = 5.00 (P 3 ) (5.00) = (n 1 + n 2 )RT (P 3 ) (5.00) = (2.00/RT + 4.50/RT)RT (P 3 ) (5.00) = 6.50 P 3 = 6.50 / 5.00 = 1.30 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.

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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

Problems and solutions.

Boyle’s law is an experimental gas law that explains the relationship between pressure and volume. According to Boyle’s law, the pressure and volume are inversely proportional, provided the temperature and mass remain unchanged [1-4] .

The law was named after Anglo-Irish physicist and chemist Robert Boyle, who published the law in 1662.

Suppose P is the pressure and V is the volume of the gas. Mathematically, Boyle’s law is given by [1-6]

Or, P = k/V

P : Pressure

k : Proportionality constant

This equation states that the product of pressure and volume is a constant for a given mass confined to a container as long as the temperature remains unchanged. A graphical representation of the above equation is shown below.

Boyle’s law can be used to establish a relationship between two states of a gas. Suppose the gas with pressure P 1 and volume V 1 expands or shrinks to pressure P 2 and volume V 2 . Then, using Boyle’s law equation,

P 1 V 1 = k and P 2 V 2 = k

From the above two equations

P 1 V 1 = P 2 V 2

This equation shows that as the pressure increases, the volume decreases and vice versa. For example, when the pressure doubles, the volume is decreased by half. Also, the units of pressure and volume must be consistent. P 1 and P 2 must be expressed in Pa or atm. V 1 and V 2 must be expressed in m 3 or L.

Here are some examples of Boyle’s law in real life [3] .

Respiration and Breathing : When we inhale, our lungs expand. As a result, the volume increases, and pressure decreases. The air pressure inside the lungs is less than the environmental pressure. Hence, oxygen from the environment fills up the lungs. The reverse happens during exhalation. When we exhale, the lungs shrink. Its volume decreases, and pressure increases. The air pressure inside the lungs is higher than the environmental pressure. Hence, the air is expelled out of the lungs.

Syringe : A syringe draws liquid from a small vial and injects it into a body. It consists of a barrel, needle, and plunger. When the plunger is pulled, the volume inside the barrel increases, resulting in a decrease in pressure. Any fluid flows from high pressure to a low-pressure region. Since the pressure outside the syringe, near the needle, is higher than the pressure inside it, fluid will flow into the syringe. The reverse also holds. When the plunger is pushed, it creates a low volume and high pressure inside the syringe. As the pressure inside is higher than outside, the fluid will flow out of the syringe.

Balloon : When a balloon filled with air is squeezed, its volume decreases, and pressure increases. If the balloon is squeezed further, the pressure inside the balloon is so high that it causes the balloon to burst.

Problem 1 : A gas confined to a volume of 2 L at a pressure of 10 atm. It can flow into a 10 L container by opening the valve that connects the two containers. What is the final pressure of the gas?

P 1 = 10 atm

From Boyle’s law,

Or, P 2 = P 1 V 1 / V 2

Or, P 2 = 10 atm x 2 L/ 10 L

Or, P 2 = 2 atm

Problem 2 : A gas exerts a pressure of 5 kPa on the walls of a container. When the container is emptied into a 12 L container, the pressure exerted by the gas increases to 8 kPa. Find the volume of container 1.

P 1 = 5 kPa

P 2 = 8 kPa

Or, V 1 = P 2 V 2 /P 1

Or, V 1 = 8 kPa x 12 L/5 kPa

Or, V 1 = 19.2 L

Ans. The variables involved in Boyle’s law are pressure and volume. On the other hand, temperature and the number of moles of gas are the constants.

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Boyle-Mariotte law: formula, examples, solved exercises

The Boyle-Mariotte Law (or simply Boyle's Law) is a law that establishes the relationship between pressure and volume in the thermodynamic processes of a gas.

Boyle's Law is one of the fundamental laws of gas physics and is widely applied in areas such as thermodynamics, physics, chemistry , and engineering. It is especially useful in the study of the behavior of gases and has practical applications in gas compression, ventilation systems, engine design, refrigeration systems, and many other gas-related fields.

The Boyle-Mariotte law was first formulated in the 17th century by the English scientist Robert Boyle, who performed a series of experiments to investigate the properties of gases. However, it was rediscovered soon after by the French scientist Edme Mariotte.

What does Boyle's Law state?

Boyle's Law states that, at constant temperature, the volume of a fixed amount of gas is inversely proportional to the pressure applied to it.

In other words, when the temperature is held constant, if the pressure of a gas increases, its volume decreases, and if the pressure decreases, the volume increases.

This law is based on the idea that, at the microscopic level, gases consist of constantly moving particles. When the pressure on a gas is increased, the particles are compressed more tightly, reducing the space they occupy and decreasing the total volume of the gas.

Conversely, if the pressure is reduced, the particles have more space to move and the gas expands, occupying a larger volume.

Boyle-Mariotte law formula

Mathematically, Boyle's Law is expressed by the formula:

P₁ * V₁ = P₂ * V₂

P₁ and V₁ represent the initial pressure and volume of the gas.

P₂ and V₂ represent the final pressure and volume of the gas.

Examples of Boyle's Law

Here are some examples of Boyle's Law that illustrate the relationship between pressure and volume of a gas at constant temperature:

Inflatable balloon: If you have an inflatable balloon and you compress it by applying pressure to it, its volume will decrease. As the pressure increases, the gas inside the balloon is compressed and takes up less space, causing the balloon to shrink. On the contrary, by releasing the pressure, the volume of the gas increases and the balloon inflates again.

Syringe: Pulling back on the plunger of a medical syringe reduces the volume inside the syringe. This creates a lower pressure in the syringe space, making it possible to draw in liquids or medications. Pushing the plunger forward increases the pressure and expels fluid from the syringe.

Car tires: Car tires contain compressed air. If you need to increase tire pressure, you use an air pump to add more air to the inside of the tire. As the pressure increases, the air volume decreases and the tire inflates. On the other hand, if you decrease the pressure, the tire goes flat and the air volume increases.

Diving: When a diver descends to greater depths under water, the pressure increases considerably. This causes a decrease in the volume of air in the lungs. If the diver does not gradually exhale the compressed air upon ascent, he or she may experience pulmonary barotrauma due to the sudden expansion of air in the lungs as the pressure decreases.

Importance of Boyle's Law

The Boyle-Mariotte law is especially useful in fields such as thermodynamics, physics, and chemistry. For example, in industry, this law is applied to the compression of gases for storage and transportation.

In addition, it is essential in the design and operation of internal combustion engines, refrigeration and air conditioning systems, and in the manufacture of chemical products.

Another important application of this law is found in medicine. For example, in the field of pulmonary ventilation, the Boyle-Mariotte law is essential to understanding how the lungs expand and contract during respiration. The increase in pressure in the lungs during inspiration causes a decrease in their volume, allowing air to enter.

Solved exercises on the Boyle-Mariotte Law

A gas occupies a volume of 4 liters at a pressure of 3 atmospheres. If the pressure is reduced to 2 atmospheres, what will be the new volume of the gas?

Using Boyle's Law formula (P₁ * V₁ = P₂ * V₂), we can solve the problem.

Applying the formula:

(3 atm) * (4 L) = (2 atm) * V₂

12 L atm = 2 atm * V₂

Dividing both sides of the equation by 2 atm:

(12 L atm) / (2 atm) = V₂

Therefore, the new volume of the gas will be 6 liters when the pressure is reduced to 2 atmospheres.

A cylinder contains a gas at a pressure of 2 atmospheres and a volume of 10 liters. If the pressure is increased to 4 atmospheres, what will be the new volume of the gas?

(2 atm) * (10 L) = (4 atm) * V₂

20 L atm = 4 atm * V₂

Dividing both sides of the equation by 4 atm:

(20 L atm) / (4 atm) = V₂

Therefore, the new volume of the gas will be 5 liters when the pressure increases to 4 atmospheres.

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Solving Numerical problems using Boyle’s law

Last updated on October 23rd, 2021 at 10:23 am

Here we will present a set of solved numerical problems that use Boyle’s law formula to solve the numerical problems on gas.

The primary formula that will be used to solve these numerical problems is p 1V1 = p 2V2 = constant . where p1 and V1 are initial pressure and initial volume of the gas respectively. Similarly, p2 and V2 are the final pressure and final volume of the gas respectively.

Numerical problems based on Boyle’s law – with solution

A gas occupies 200 mL at a pressure of 0.820 bar at 20°C. How much volume will it occupy when it is subjected to external pressure of 1.025 bar at the same temperature ?

p 1 = 0.820 bar

p 2 = 1.025 bar

V1 = 200 ml

Since temperature is constant, therefore, by applying Boyle’s law,

p 1V1 = p 2V2

v2 = P1V1/P2 = 160 ml

A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure ?

Solution: Since temperature and amount of gas remain constant, therefore, Boyle’s law is applicable.

p 1 = 1.2 bar

V 1 = 120 mL

V 2 = 180 mL

∴ 1.2 × 120 = p 2 × 180

p2 =0.8 bar

A gas occupies a volume of 250 mL at 745 mm Hg and 25°C. What additional pressure is required to reduce the gas volume to 200 mL at the same temperature ?

p 1 = 745 mm Hg

V1 = 250 mL

V2 = 200 mL

Since temperature remains constant, therefore, by applying Boyle’s law,

P2 = P1 V1 / V2

p2 = 931.25 mm Hg

The additional pressure required =P2 – P1= 931.25 – 745

= 186.25 mm.

A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, up to what volume can the balloon be expanded ?

Solution: According to Boyle’s law, at constant temperature,

p 1 V1 = p 2 V2

If p 1 = 1 bar then V1 = 2.27 L

If p 2 = 0.2 bar, then

V2 = p1v1/p2= 1 bar x 2.27 L/0.2 bar = 11.35 L

Since the balloon bursts at 0.2 bar pressure, the volume of the balloon should be less than 11.35 L.

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What is an example of a Boyle's law practice problem?

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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?

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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 #12 : Gases And Gas Laws

What law is the following formula?

Ideal gas law

Gay-Lussac's law

Combined gas law

Boyle's law

Charles'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.

Example Question #13 : Gases And Gas Laws

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.

Example Question #14 : Gases And Gas Laws

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 #1 : Using Boyle's Law

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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Chemistry Steps

General Chemistry

In this post, we will discuss one of the gas laws using an illustration with a pump filled with some gas and a freely moving plunger and by changing the gas parameters determine their correlation.

Boyle’s Law

Boyle’s Law shows the correlation between the pressure and the volume of a gas. To see how these two parameters change based on one another, suppose we need to alter them. Let’s start with the volume. If we push down the plunger, it’s evident that the volume of the gas is decreasing .  Now, as push down, the volume is decreasing and because the gas molecules have less space, the pressure is increasing :

The observation is that the volume of a fixed quantity of gas at constant temperature is inversely proportional to its pressure.

This relationship can be written as:

\[{\rm{V}} \sim \,\frac{{\rm{1}}}{{\rm{P}}}\]

To bring in the equal sign, we introduce a constant:

                                                                         PV = constant              

This can be explained using the example of a car dealership income. The income depends on the number of sales which we can represent as:

Income  ∼ number of cars

However, we cannot say income = number of cars sold, so to switch an equal sign, we need to introduce a constant. This can be the price of the car transforming the equation to:

Income  = price x number of cars

So, for the gas pressure and volume, we are not interested too much in the constant, but rather in its linkage of pressure and volume at positions 1 and 2. Because the P x V product is constant, we can write that:

P 1 V 1 = constant = P 2 V 2

This is the practical implication of the Boyle’s law that is used for solving gas problems.

For example ,

The pressure of a gas is 2.30 atm in a 1.80 L container. Calculate the final pressure of the gas if the volume is decreased to 1.20 liters.

First , write down what you have and what needs to be determined. If nothing is mentioned about any parameter, for example, the moles and the temperature in this case, it is assumed that they are constant, so you don’t need to worry about them.

P 1 = 2.30 atm

V 1 = 1.80 L

V 2 = 1.20 L

\[{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\; = \;{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\]

Now, rearrange to calculate P 2 :

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}\]

\[{{\rm{P}}_{\rm{2}}}\; = \;\frac{{{\rm{2}}{\rm{.30}}\;{\rm{atm}}\;{\rm{ \times 1}}{\rm{.80}}\;\cancel{{\rm{L}}}}}{{{\rm{1}}{\rm{.20}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.45}}\;{\rm{atm}}\]

• The Ideal Gas Law

There are other gas laws such as the Charles’ law, Avogadro’s law etc., and they are all together summarized in the ideal gas law equation. Unlike the individual gas laws, the ideal gas law links all the variables into one equation:

The R is called the ideal gas constant. Although it has different values and units, you will mostly be using this:

\[R\;{\rm{ = }}\;{\rm{0}}{\rm{.08206}}\;\frac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}}\]

Remember, to change the pressure to atm when the ideal gas law equation is used! This is because the units of R contain atm when the 0.08206 value is used. And this is what most problems in this chapter use.

How do I know which gas law to use?

Once you cover all the gas laws, it sometimes becomes challenging to choose the appropriate one when solving problems. For this, there is what is called the combined gas law and as long as you remember it, you do not need to remember all the gas laws to solve a problem.

Let’s keep it for another article because there is quite a lot of information in this one.

• Charle’s Law
• Gay-Lussac’s Law
• Avogadro’s Law
• Celsius or Kelvin
• Ideal-Gas Laws
• Combined Gas Law Equation
• How to Know Which Gas Law Equation to Use
• Molar Mass and Density of Gases
• Graham’s Law of Effusion and Diffusion
• Graham’s Law of Effusion Practice Problems
• Dalton’s Law of Partial Pressures
• Mole Fraction and Partial Pressure of the Gas
• Gases in Chemical Reactions
• Gases-Practice Problems

2 thoughts on “<strong>Boyle’s Law</strong>”

Im pretty sure I’m going to subscribe to your 1 month deal. But I was wondering if you recommend any other books that might help me get through this general chemistry? Or do you have a book that I could buy?

I would say the first should be what your instructor recommends for the class, but aside from that, there are plenty of great books: Tro – Chemistry a Molecular approach, Chang – Chemistry, Brown – Chemistry The Central Science, Silberberg – Chemistry, Fay – Chemistry, Zumdahl – Chemistry. I do not have a book of mine.

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11.4: Boyle’s Law: Pressure and Volume

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Learning Objectives

• Learn what is meant by the term gas laws .
• Learn and apply Boyle’s Law.

When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure ( P ) and volume ( V ), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [ n ]), if the temperature ( T ) of the gas was kept constant, pressure and volume were related: as one increases, the other decreases. As one decreases, the other increases. This means that pressure and volume are inversely related .

There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:

\[P × V = \text{ constant at constant n and T} \nonumber \]

If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled \(P_1\) and \(V_1\) and the new conditions are labeled \(P_2\) and \(V_2\), we have

\[P_1V_1 = \text{constant} = P_2V_2 \nonumber \]

where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply

\[P_1V_1 = P_2V_2 \text{ at constant n and T} \nonumber \]

This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's Law , after the English scientist Robert Boyle, who first announced it in 1662. Figure \(\PageIndex{1}\) shows two representations of how Boyle’s Law works.

Boyle’s Law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle’s Law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won’t matter what the unit is, but the unit must be the same on both sides of the equation.

Example \(\PageIndex{1}\)

A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?

Exercise \(\PageIndex{1}\)

If P 1 = 334 torr, V 1 = 37.8 mL, and P 2 = 102 torr, what is V 2 ?

As mentioned, you can use any units for pressure and volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.

Example \(\PageIndex{2}\):

A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?

Exercise \(\PageIndex{2}\)

If V 1 = 456 mL, P 1 = 308 torr, and P 2 = 1.55 atm, what is V 2 ?

• The behavior of gases can be modeled with gas laws.
• Boyle’s Law relates the pressure and volume of a gas at constant temperature and amount.