Boyle's Law Calculator
- Boyle's law definition
- Boyle's law formula
- Boyle's law examples
- Where is Boyle's law applied?
Other thermodynamic processes
This Boyle's law calculator is a great tool when you need to estimate the parameters of a gas in an isothermal process . You will find the answer to "What is Boyle's law?" in the text, so read on to find out about the Boyle's law formula , see some practical examples of Boyle's law exercises and learn how to recognize when a process satisfies Boyle's law on a graph.
In case you need to work out the results for an isobaric process, check our Charles' law calculator .
Boyle's law definition
Boyle's law (also known as Boyle-Mariotte law) tells us about the relationship between the pressure of a gas and its volume at a constant temperature and mass of gas . It states that the absolute pressure is inversely proportional to the volume .
Boyle's law definition can also be phrased in the following way: the product of the pressure and the volume of a gas in a closed system is constant as long as the temperature is unchanged.
Boyle's law describes the behavior of an ideal gas. We can characterize this gas by the ideal gas equation, which you can read more about in our ideal gas law calculator . Boyle's law tells us about an isothermal process , which means that the temperature of the gas remains constant during the transition, as does the internal energy of the gas.
Boyle's law formula
We can write the Boyle's law equation in the following way:
p 1 × V 1 = p 2 × V 2 ,
where p 1 and V 1 are initial pressure and volume, respectively. Similarly, p 2 and V 2 are the final values of these gas parameters.
We can write Boyle's law formula in various ways depending on which parameter we want to estimate. Let's say we change the volume of a gas under isothermal conditions, and we want to find the resulting pressure. Then, the equation of Boyle's law states that:
p 2 = p 1 × V 1 / V 2 or p 2 / p 1 = V 1 / V 2 .
As we can see, the ratio of the final and initial pressure is the inverse of the ratio for volumes . This Boyle's law calculator works in any direction you like. Just insert any three parameters, and the fourth one will be calculated immediately!
We can visualize the whole process on Boyle's law graph. The most commonly used type is where the pressure is a volume function. For this process, the curve is a hyperbola. The transition can progress in both ways, so both compression and gas expansion satisfy Boyle's law.
🔎 If a transition is an isochoric process (the constant volume), you'll find Omni's Gay-Lussac's law calculator helpful.
Boyle's law examples
We can use Boyle's law in several ways, so let's take a look at some examples:
Imagine that we have an elastic container that holds a gas. The initial pressure is 100 kPa (or 10 5 Pa if we use scientific notation), and the volume of the container equals 2 m 3 . We decide to compress the box down to 1 m 3 , but we don't change the overall temperature. The question is: "How does the pressure of the gas change?". We can use Boyle's law formula:
p 2 = p 1 × V 1 / V 2 = 100 kPa × 2 m 3 / 1 m 3 = 200 kPa .
After halving the volume, the internal pressure is doubled. This is a consequence of the fact that the product of the pressure and the volume must be constant during this process.
The next Boyle's law example concerns a gas under 2.5 atm pressure while occupying 6 liters of space. It is then decompressed isothermally to the pressure of 0.2 atm . Let's find out its final volume. We have to rewrite Boyle's law equation:
V 2 = p 1 × V 1 / p 2 = 2.5 atm × 6 l / 0.2 atm = 75 l .
You can always use our Boyle's law calculator to check if your evaluations are correct!
Where is Boyle's law applied?
Boyle's law describes all processes for which temperature remains constant . In thermodynamics, temperature measures the average kinetic energy that atoms or molecules have. In other words, we can say that the average velocity of gas particles doesn't change during that transition. Boyle's law formula is valid for a wide range of temperatures.
In advanced mode , you can choose any temperature you like, and we will calculate the number of molecules contained in the gas. You only have to ensure that the substance remains in the gas form (e.g., neither condensates nor crystallizes) at this temperature.
There are a few areas where Boyle's law is applicable:
Carnot heat engine – Consists of four thermodynamic processes, two of which are isothermal ones, satisfying Boyle's law. This model can tell us what the maximal efficiency of a heat engine is.
Breathing also can be described by Boyle's law. Whenever you take a breath, your diaphragm and intercostal muscles increase the volume of your lungs, which decreases gas pressure. As air flows from an area of higher pressure to a place of lower pressure, air enters the lungs and allows us to take in oxygen from the environment. During exhalation, the volume of the lungs decreases, so the pressure inside is higher than outside, so the air flows in the opposite direction.
Syringe – Whenever you have an injection, a doctor or a nurse draws a liquid from the small vial first. To do so, they use a syringe. By pulling the plunger, the accessible volume increases, which decreases pressure and, according to Boyle's law formula, causes the suction of the fluid.
Boyle's law, along with Charles's and Gay-Lussac's law, are among the fundamental laws describing the vast majority of thermodynamic processes.
Other than working out the values of specific parameters like pressure or volume, it is also possible to discover something about the heat transfer and the work done by the gas during these transitions, as well as the internal energy change . We have gathered them all in our combined gas law calculator , where you can choose whichever process you like and evaluate the outcomes for real gases.
Why is Boyle's law also called isotherm?
Boyle's law is one of the three fundamental thermodynamic processes . In each of them, we study a variation of two out of three quantities :
- The pressure ;
- The temperature ; and
- The volume .
The third quantity remains constant during the process. In the case of Boyle's law, we don't change the temperature, thus we call the process isothermal .
How much will a balloon with initial volume 1000 cm³ expand at cruising altitude?
Follow these steps:
- Find the initial pressure. We will take atmospheric pressure at sea level: P i = 1 atm = 101,325 Pa .
- Find the final pressure. In a cruising plane, the cabin is usually pressurized at about P f = 0.8 atm = 81,060 Pa .
- Calculate the final volume with Boyle's law: V f = P i · V i /P f = (101,325 Pa · 0.001 m 3 )/81,060 Pa = 0.00125 m 3 .
- Find the expansion by subtracting the final and initial volumes : ΔV = V f - V i = (0.00125 - 0.001) m 3 = 0.00025 m 3 = 250 cm 3 .
How do I calculate Boyle's law?
To calculate Boyle's law, we need to perform a few simple steps, which depend on the initial data we know. To calculate the final pressure given the initial volume and pressure and the final volume:
- Compute the product of the initial volume and pressure : V i · P i .
- Divide the result by the final volume . The final pressure is P f = (V i · P i )/V f .
- You can invert final and initial values freely (reversible transformation).
- To find the final volume, switch it with the final pressure on the right-hand side of the formula.
What is the final pressure if the volume reduces by half with intial pressure 1 atm?
The final pressure in a process where the volume reduces by half, starting from P i = 1 atm , is 2 . To find this result:
- Write Boyle's law for the final pressure: P f = (V i · P i )/V f .
- In this formula, identify the ratio of the volumes . Since we know that the volume reduces by half, we can write V i = 2 · V f , hence V i /V f = 2 .
- The final pressure is, then: P f = 2 · P i = 2 atm .
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Boyle’s Law Core Concepts
In this tutorial, you will learn about the Boyle’s law formula, and how this equation can be used to determine relationships between changes in pressure and changes in volume in a given closed system.
Robert Boyle, sometimes called the “Father of Modern Chemistry” worked in the fields of physics and chemistry. In 1660, Boyle published The Spring and Weight of the Air in which he described different experiments he created using a vacuum pump which he designed.
In 1662, Boyle published version two of The Spring and Weight of the Air . This is were he described the inverse relationship between pressure and volume, now known as Boyle’s Law. Boyle’s Law states that pressure and volume are inversely proportional to each other. As the pressure increases the volume decreases, and as the pressure decreases the volume increases. He made these observations by using mercury in a J-tube, and then made measurements of the volume of the gas at pressures both lower and higher than standard atmospheric pressure.
Topics Covered in Other Articles
- What is Pressure
- Dalton’s Law of Partial Pressure
- Charles’ Law
- Avogadro’s Law
- The Ideal Gas Law
- Van der Waal’s Equation of State
- Gay Lussac’s Law
- Combined Gas Law
- Henry’s Law
- Graham’s Law of Effusion
Important Things to Consider
The gas law described in this article only applies to ideal gases, which you can read about on our article, The Ideal Gas Law . Under nonideal conditions, such as high pressures and low temperatures, repulsive or attractive forces between gas particles results in pressure and volume not having as close of a relationship. Specifically, too much attractive force results in a gas occupying a smaller volume at a given pressure than Boyle’s Law would suggest. Conversely, too much repulsive force results in a larger volume.
Relationship Between Pressure and Volume
We can make a graph of this relationship as follows:
Change in Pressure and Volume
This proportionality can enable us to solve specific problems relating to the changes in pressure and volume in a closed system.
Consider, for example, a piston full of oxygen. From this proportionality, we know that if the piston is compressed, the pressure of the gas will increase.
Boyle’s Law Equation – Example Problem
Here is an example of how you can solve a Boyle’s Law problem.
You can read about how a student used Boyle’s original data to verify Boyle’s Law.
Boyle’s Law Practice Problems
You have a sample of gas within Container A. Firstly, you transfer the gas completely into Container B and notice that the gas’s pressure decreases. Secondly, you transfer the gas completely into Container C and notice the pressure is higher than when it was in Container A. Thirdly, you transfer half of the gas into Container D and notice the pressure is equal to its pressure in Container C. Rank each container from largest to smallest volume.
Boyle’s Law Practice Problem Solutions
2: B > A > C > D
Ideal Gas vs Non-Ideal Gas Example Problem
Van Der Waals Equation Example Problem
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This example problem demonstrates how to calculate the pressure of a gas system using the ideal gas law and the van der Waal's equation. It also demonstrates the difference between an ideal gas and a non-ideal gas.
Van der Waals Equation Problem
Calculate the pressure exerted by 0.3000 mol of helium in a 0.2000 L container at -25 °C using a. ideal gas law b. van der Waals equation What is the difference between the non-ideal and ideal gases? Given: a He = 0.0341 atm·L 2 /mol 2 b He = 0.0237 L·mol
How to Solve the Problem
Part 1: Ideal Gas Law The ideal gas law is expressed by the formula: PV = nRT where P = pressure V = volume n = number of moles of gas R = ideal gas constant = 0.08206 L·atm/mol·K T = absolute temperature Find absolute temperature T = °C + 273.15 T = -25 + 273.15 T = 248.15 K Find the pressure PV = nRT P = nRT/V P = (0.3000 mol)(0.08206 L·atm/mol·K)(248.15)/0.2000 L P ideal = 30.55 atm Part 2: Van der Waals Equation Van der Waals equation is expressed by the formula P + a(n/V) 2 = nRT/(V-nb) where P = pressure V = volume n = number of moles of gas a = attraction between individual gas particles b = average volume of individual gas particles R = ideal gas constant = 0.08206 L·atm/mol·K T = absolute temperature Solve for pressure P = nRT/(V-nb) - a(n/V) 2 To make the math easier to follow, the equation will be broken into two parts where P = X - Y where X = nRT/(V-nb) Y = a(n/V) 2 X = P = nRT/(V-nb) X = (0.3000 mol)(0.08206 L·atm/mol·K)(248.15)/[0.2000 L - (0.3000 mol)(0.0237 L/mol)] X = 6.109 L·atm/(0.2000 L - .007 L) X = 6.109 L·atm/0.19 L X = 32.152 atm Y = a(n/V) 2 Y = 0.0341 atm·L 2 /mol 2 x [0.3000 mol/0.2000 L] 2 Y = 0.0341 atm·L 2 /mol 2 x (1.5 mol/L) 2 Y = 0.0341 atm·L 2 /mol 2 x 2.25 mol 2 /L 2 Y = 0.077 atm Recombine to find pressure P = X - Y P = 32.152 atm - 0.077 atm P non-ideal = 32.075 atm Part 3 - Find the difference between ideal and non-ideal conditions P non-ideal - P ideal = 32.152 atm - 30.55 atm P non-ideal - P ideal = 1.602 atm Answer: The pressure for the ideal gas is 30.55 atm and the pressure for van der Waals equation of the non-ideal gas was 32.152 atm. The non-ideal gas had a greater pressure by 1.602 atm.
Ideal vs Non-Ideal Gases
An ideal gas is one in which the molecules don't interact with each other and don't take up any space. In an ideal world, collisions between gas molecules are completely elastic. All gases in the real world have molecules with diameters and which interact with each other, so there's always a bit of error involved in using any form of the Ideal Gas Law and van der Waals equation.
However, noble gases act much like ideal gases because they don't participate in chemical reactions with other gases. Helium, in particular, acts like an ideal gas because each atom is so tiny.
Other gases behave much like ideal gases when they are at low pressures and temperatures. Low pressure means few interactions between gas molecules occur. Low temperature means the gas molecules have less kinetic energy, so they don't move around as much to interact with each other or their container.
- Gases Study Guide
- 3 Ways To Increase the Pressure of a Gas
- What Is the Most Ideal Gas?
- Gas Definition and Examples in Chemistry
- Topics Typically Covered in Grade 11 Chemistry
- Kinetic Molecular Theory of Gases
- AP Chemistry Course and Exam Topics
- How to Calculate Osmotic Pressure
- The Isochoric Process
- The Combined Gas Law in Chemistry
- What Is the Ideal Gas Law?
- Gases - General Properties of Gases
- How to Calculate Density of a Gas
- Ideal Gas Law Example Problem
- Graham's Law Example: Gas Diffusion-Effusion
- How to Calculate the Density of a Gas
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In these lessons, we will learn and apply:
Charles' law, gay-lussac’s law, combined gas law, ideal gas law.
Related Pages Solving Gas Law Problems High School Chemistry Chemistry Lessons
Boyle’s Law states that volume of a given amount of gas held at a constant temperature varies inversely the with pressure. The relationship between pressure and volume of Boyle’s Law is expressed in mathematical terms as P 1 V 1 = P 2 V 2 .
An Introduction To The Relationship Between Pressure And Volume
A lesson on how to solve gas problems with Boyle’s Law.
Example: At 1.70 atm, a sample of gas takes up 4.25L. If the pressure in the gas is increased to 2.40 atm, what will the new volume be?
Understanding And Applying Boyle’s Law
Example: A sample of Ne gas occupies 0.220L at 0.86 atm. What will be its volume at 29.4kPa?
Charles' Law states that the volume of a given mass of a gas is directly proportional to its Kelvin temperature at constant pressure. In mathematical terms, the relationship between temperature and volume is expressed as V 1 /T 1 =V 2 /T 2 .
What Is The Relationship Between Volume And Temperature Of A Gas
A lesson on how to solve problems using Charles' Law.
Example: A balloon takes up 625L at 0°C. If it is heated to 80°C, what will its new volume be?
Understanding And Applying Charles' Law
Example: A gas at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the new volume be if the pressure is constant?
Gay-Lussac’s Law states that the pressure of a given mass of gas varies directly with the Kelvin temperature when the volume remains constant. Gay-Lussac’s Law is expressed in a formula form as P 1 /T 1 = P 2 /T 2 . When dealing with Gay-Lussac’s Law, the unit of the temperature should always be in Kelvin.
Using Gay-Lussac’s Law To Understand The Relationship Between A Gas' Pressure And Temperature
Example: If the pressure in a car tire is 1.88 atm at 25°C, what will be the pressure if the temperature warms to 37°C?
How To Solve Word Problems That Show How To Use Gay-Lussac’s Law?
- The pressure in a sealed can of gas is 235kPA when it sits at room temperature (20°). If the can is warmed to 48°C, what will the new pressure inside the can be?
- A car tire has a pressure of 2.38 atm at 15.2°C. If the pressure inside reached 4.08 atm, the tire will explode. How hot would the tire have to get for this to happen? Report the temperature in degrees Celsius.
Practice Problem to show how to use Gay-Lussac’s Law
Example: In the morning, a paintball pressure tank is at 306 atm. The weather heats up over the course of the day, and by 3 PM, the outside temperature is roasting at 38.5°C, and the pressure inside the tank is 324 atm. What was the temperature (in degree Celsius) in the morning?
The Combined Gas Law combines Charles' Law, Boyle’s Law and Gay Lussac’s Law. The Combined Gas Law states that a gas' (pressure × volume)/temperature = constant.
Example: A gas at 110kPa at 30.0°C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80,0°C and the pressure increases to 440Kpa, what is the new volume?
How To Solve Problems With The Combined Gas Equation?
Example: A 40.0L balloon is filled with air at sea level (1.00 atm, 25.0°C). It is tied to a rock and thrown in a cold body of water, and it sinks to the point where the temperature is 4.0°C and the pressure is 11.0 atm. What will its new volume be?
The Ideal Gas Law mathematically relates the pressure, volume, amount and temperature of a gas with the equation: pressure × volume = moles × ideal gas constant × temperature; PV = nRT.
The Ideal Gas Law is ideal because it ignores interactions between the gas particles in order to simplify the equation. There is also a Real Gas Law which is much more complicated and produces a result which, under most circumstances, is almost identical to that predicted by the Ideal Gas Law.
Understanding And Applying The Ideal Gas Law
Example: What is the pressure in atm of a 0.108 mol sample of the gas at a temperature of 20.0°C if its volume is 0.505L?
Sample Problems For Using The Ideal Gas Law, PV = nRT
- 2.3 moles of Helium gas are at a pressure of 1.70 atm, and the temperature is 41°C. What is the volume of the gas?
- At a certain temperature, 3.24 moles of CO 2 gas at 2.15 atm take up a colume of 35.28L. What is this temperature (in Celsius)?
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How do you find volume with pressure and work?
Explanation: pressure-volume work done occurs when the volume of the gas or system changes. work done by a system is the energy transferred from it to its surroundings. express that work in joules and calories, p e x t e r n a l × △ v . vloume can be found from this equation. if the gas is heated, the gas molecules are energized. the increase in the average kinetic energy of the gas molecules can be observed by measuring how the temperature of the gas particles increased. the gas molecules collide with the piston more frequently, by the fast movement of the gas particles. these frequent collisions transfer energy to the piston and allow it to move against an external pressure thus, increasing the net volume of the gas. to calculate how much work is done (or has been done to it) by the gas against a constant external pressure, we have to use an equation: w o r k ( w ) = p e x t e r n a l × △ v where external pressure is denoted by p e x t e r n a l . the external pressure opposes the pressure of the gas in the system. △ v is a change in the volume of the gas, which is calculated as v f i n a l − v in i t i a l . sign convention of work: when a system does work on the surroundings, negative work is said to have occurred. when the gas does work, the volume of the gas particles decreases ( △ v < 0 ) . the work done is positive. when the gas does work, the volume of the gas particles increases ( △ v > 0 ) . the work done is negative. units conversion: work has its units in joules..
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do you think air pressure has to do bsomething with the working of vacuum cleaner?
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All About Physics
All about physics will help you solve complicated physics problems with just a few clicks. it can solve for many things such as acceleration, depth, force, friction, heat, impulse, kinetic energy, mirror, momentum, power, pressure, projectile height, projectile range, projectile time, refractive essay, speed of light, velocity and work. it also features a unit converter that can convert energy, force, length, mass, power, pressure, speed, temperature, time and volume., 11/12/2012 8:20:25 am.
4 ways to solve pressure vessel problems
US FUSION / PALA Group
November 1, 2023
When an ASME Code stamped pressure vessel is built, it is fabricated to the specifications of the ASME Boiler and Pressure Vessel Code.
Repairs, however, are covered by the National Board Inspection Code, commonly referred to as the NBIC. Some of the most common types of repairs to pressure vessels covered by the NBIC include:
Anywhere there’s a joint, there’s a potential weakness in a pressure vessel. Especially in pressure differentials, welds can fail under stress. There are many causes of weld failure, some occurring in production, such as porosity in the weld or the inclusion of slag. Others — namely corrosion — occur over long stretches of time as the weld is exposed to the elements and gradually corrodes, eventually causing cracks and other issues.
Nozzle replacement and installation
Nozzles on pressure vessels see a lot of use, and as such, they comprise a common type of pressure vessel repair. Nozzles can wear out over time, and their components can become degraded, sparking a need for replacement. Additionally, installing new nozzles when the need arises is common and falls under the NBIC. Technology can change over the lifespan of a pressure vessel, which can be over 20 years, and replacing older nozzle designs with new, more efficient ones can be a repair well worth making.
Mounting clips for ladders
Working on and around pressure vessels carries some inherent risk. For jobs where ladders must be installed on pressure vessels for access to gauges or making repairs, mounting clips are to be installed according to the code. This will prevent damage to the vessel itself and ensure adequate, ergonomic access to necessary components of the equipment for technicians.
Any repairs or additions that require welding non-pressurized components to pressurized components or boundary components of the vessel fall under the NBIC. Such cases require special care to guarantee the soundness of the weld and avoid compromising the vessel’s ability to maintain the pressurized environment.
Inspecting for necessary repairs
Because pressure vessels are pressurized and contain sensitive materials, keeping up with maintenance and repairs is critical. To know when repairs are needed, reduce downtime and keep the pressure vessel operational, one should have a regular inspection program.
Pressure vessels should be inspected at least every five years to check for these commonly needed repairs, if not more frequently, depending on the code to which the vessel needs to adhere.
Who performs pressure vessel repair services?
The NBIC clearly defines who is and who is not authorized to conduct repairs that fall under the code, like fortifying corroded welds. To perform repairs governed by the code, repair shops must be certified by the board to use the official R-stamp. This certification ensures the qualification of the servicer to meet the necessary design, materials and quality standards of the code.
PALA, an employee-owned, turnkey industrial construction company, offers NBIC-compliant pressure vessel repairs and boasts a wide range of unique capabilities in their 70,000 SF shop facility. In 2017, the company further diversified when it acquired specialty construction company US FUSION. The joint teams from US FUSION and PALA Group are certified to use the NBIC R-stamp and can perform all repairs governed by Part 3, Section 3 of the NBIC. Within their expansive facility, they offer specialized services, including a separate shop dedicated to alloy fabrication, on-site blasting, an on-site rubber shop for linings and repairs, subcontracted shop coatings and paints and a 20’ × 40’ HYDRO test pad with a 1-million-pound capacity.
PALA Group excels in fabricating and servicing a variety of pressure vessels, including ASME Div 1 and 2 code vessels, API 650 and API 620 tanks and Section VIII pressure parts. Not only do they hold the NBIC R-stamp accreditation, but also accreditation for U-, U2- and S-stamps, demonstrating their commitment to versatility in the industry.
For more information, visit palagroup.com or usfusion.com .
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