{"id":4616,"date":"2013-06-29T06:56:37","date_gmt":"2013-06-29T06:56:37","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4616"},"modified":"2026-03-25T11:05:54","modified_gmt":"2026-03-25T16:05:54","slug":"boyles-law","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/boyles-law\/","title":{"rendered":"What is Boyle\u2019s Law?"},"content":{"rendered":"<p>\n\t\t\t<div id=\"mmDeferVideoEncompass_wJ8q7pXSWD0\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_wJ8q7pXSWD0\" data-source-videoID=\"wJ8q7pXSWD0\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"What is Boyle\u2019s Law? 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We\u2019ll go over a few practical examples along the way, so you\u2019ll be able to understand its application in the context of many disciplines.<\/p>\n<p>Let\u2019s get started!<\/p>\n<h2><span id=\"Inverse_Relationship_of_Pressure_and_Volume\" class=\"m-toc-anchor\"><\/span>Inverse Relationship of Pressure and Volume<\/h2>\n<p>\nBoyle\u2019s Law defines the inverse relationship between pressure and volume for a collection of particles in the gas phase.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/relationship-between-pressure-and-volume-65806e369bc24.webp\" alt=\"relationship between pressure and volume\" width=\"486.3\" height=\"186.3\" class=\"aligncenter size-full wp-image-209198\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/relationship-between-pressure-and-volume-65806e369bc24.webp 1621w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/relationship-between-pressure-and-volume-65806e369bc24-300x115.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/relationship-between-pressure-and-volume-65806e369bc24-1024x392.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/relationship-between-pressure-and-volume-65806e369bc24-768x294.webp 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/relationship-between-pressure-and-volume-65806e369bc24-1536x588.webp 1536w\" sizes=\"(max-width: 1621px) 100vw, 1621px\" \/><\/p>\n<p>As you can see from this graph, as the volume increases, the pressure decreases.<\/p>\n<p>Boyle stated that this relationship holds for a constant temperature. However, we want to put this in a simplified, more concrete context by saying that the number of gas particles (or moles of gas particles) should also be held constant.<\/p>\n<h3><span id=\"Defining_Volume_and_Pressure\" class=\"m-toc-anchor\"><\/span>Defining Volume and Pressure<\/h3>\n<p>\nTo better interpret the mathematical nature of this statement, let\u2019s remind ourselves what volume and pressure are.<\/p>\n<p>Simply put, volume is a measurement of distance in three dimensions. A rectangular box, for example, has the dimensions of length (L), width (W), and height (H). The volume of the box is \\(V = L\u00d7W\u00d7H\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/length-width-height-of-box-65806e36ba751.webp\" alt=\"Length, width, and height of a box\" width=\"423.92\" height=\"306.88\" class=\"aligncenter size-full wp-image-209201\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/length-width-height-of-box-65806e36ba751.webp 757w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/length-width-height-of-box-65806e36ba751-300x217.webp 300w\" sizes=\"(max-width: 757px) 100vw, 757px\" \/><\/p>\n<p>Now let\u2019s look at pressure. Pressure is defined as the force per unit area.<\/p>\n<p>Imagine we have a flat surface with a certain area. This would be \\(A = x^2\\).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/Area-of-a-square-01.svg\" alt=\"\" width=\"278\" height=\"303\" class=\"aligncenter size-full wp-image-209204\"  role=\"img\" \/><\/p>\n<p>Let\u2019s now say that this flat area has a force pressing against it. We could use our equation \\(P = F\/A\\) to measure the pressure.<\/p>\n<p>Now, imagine that area was part of the box from earlier whose volume is \\(L\u00d7W\u00d7H\\). <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/smaller-square-on-the-larger-box-658071dcafa4e.webp\" alt=\"Smaller square on a large box\" width=\"440.1\" height=\"256.2\" class=\"aligncenter size-full wp-image-209207\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/smaller-square-on-the-larger-box-658071dcafa4e.webp 1467w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/smaller-square-on-the-larger-box-658071dcafa4e-300x175.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/smaller-square-on-the-larger-box-658071dcafa4e-1024x596.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/smaller-square-on-the-larger-box-658071dcafa4e-768x447.webp 768w\" sizes=\"(max-width: 1467px) 100vw, 1467px\" \/><\/p>\n<p>This vessel has some gas molecules in it. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/A-vessel-with-molecules-in-it.webp\" alt=\"A vessel with molecules in it\" width=\"480\" height=\"283.8\" class=\"aligncenter size-full wp-image-209210\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/A-vessel-with-molecules-in-it.webp 800w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/A-vessel-with-molecules-in-it-300x177.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/A-vessel-with-molecules-in-it-768x454.webp 768w\" sizes=\"(max-width: 800px) 100vw, 800px\" \/><\/p>\n<p>Each time one of those particles collides with the wall of the vessel, it exerts a force. The amount of force per unit area is a function of the number of particles contained in that volume. <\/p>\n<p>You might be able to tell by looking at this that time is also a factor, because the number of particles hitting a certain area also depends on how fast the particles are moving. As we\u2019ll see a bit later, pressure is also a function of temperature. Boyle did his experiments with a constant temperature, so we\u2019re going to imagine that the velocity of the particles is constant.<\/p>\n<p>Now, let\u2019s say that we could move the top of our box downward, like how a piston in a car engine works. This would reduce the volume of the vessel. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/01\/Boyles-law-smaller-container.webp\" alt=\"A smaller vessel with molecules\" width=\"480\" height=\"306\" class=\"aligncenter size-full wp-image-211945\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/01\/Boyles-law-smaller-container.webp 800w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/01\/Boyles-law-smaller-container-300x191.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/01\/Boyles-law-smaller-container-768x490.webp 768w\" sizes=\"auto, (max-width: 480px) 100vw, 480px\" \/><\/p>\n<p>As you can see, though the particles are moving at the same speed, more particles are hitting that area more frequently, and, therefore, the pressure inside the vessel has increased. A smaller volume leads to larger pressure at constant temperature and a constant number of particles.<\/p>\n<h2><span id=\"The_Ideal_Gas_Law\" class=\"m-toc-anchor\"><\/span>The Ideal Gas Law<\/h2>\n<p>\nAs we talk about Boyle\u2019s Law, it\u2019s important to have a qualitative understanding of the inverse nature that this law indicates. To do this, let\u2019s put this law in the context of what is known as the ideal gas law.<\/p>\n<p>In this equation, n is the number of moles, T is the temperature, and R is the ideal gas constant.<\/p>\n<p>The meaning of \u201cideal\u201d in the phrase \u201cideal gas law\u201d implies the molecules in an ideal gas have no intermolecular interactions due to electrostatic charges or charge distributions surrounding the atoms or molecules.<\/p>\n<p>The general term for intermolecular interactions is \u201cVan der Waals forces.\u201d Here\u2019s a quick look at the different forces included here. There are dipole-dipole interactions, hydrogen bonding, ion-dipole interactions, and London dispersion forces:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9.webp\" alt=\"Different forces\" width=\"638.25\" height=\"358.25\" class=\"aligncenter size-full wp-image-209222\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9.webp 2553w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9-300x168.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9-1024x575.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9-768x431.webp 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9-1536x862.webp 1536w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/different-forces-658079dbabcf9-2048x1150.webp 2048w\" sizes=\"(max-width: 2553px) 100vw, 2553px\" \/><\/p>\n<p>When none of these intermolecular forces are present in a collection of gas molecules, the ideal gas law accurately represents the dynamics of a collection of \u201cn\u201d moles of a gas at a given temperature, pressure, and volume.<\/p>\n<p>Let\u2019s consider a system under a constant temperature and look at the relationship between the pressure and the volume. We can note that assuming a constant number of particles, the equation\u2019s right side is a constant. <\/p>\n<p>So, if we look at a system with an initial P<sub>1<\/sub> and V<sub>1<\/sub> and then change the volume and note the corresponding pressure change, then we can set P<sub>1<\/sub>V<sub>1<\/sub> = P<sub>2<\/sub>V<sub>2<\/sub>.<\/p>\n<p>This is an expression of Boyle\u2019s Law for a system with a constant temperature and number of moles of gas.<\/p>\n<h2><span id=\"Examples\" class=\"m-toc-anchor\"><\/span>Examples<\/h2>\n<p>\nLet\u2019s look at a couple of examples illustrating how Boyle\u2019s Law is applied and do some simple calculations to understand the nature of the inverse relationship.<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nFor our first example, let\u2019s look at a piston. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/pistons.webp\" alt=\"Pistons\" width=\"470.34\" height=\"360.99\" class=\"aligncenter size-full wp-image-209225\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/pistons.webp 1742w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/pistons-300x230.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/pistons-1024x786.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/pistons-768x589.webp 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/pistons-1536x1179.webp 1536w\" sizes=\"(max-width: 1742px) 100vw, 1742px\" \/><\/p>\n<p>If we move the piston down inside the vessel, decreasing the volume, the pressure of the gas particles will increase. Since we are assuming the number of particles in the cylinder is constant and the temperature is constant, we can use Boyle\u2019s Law to calculate the pressure as the volume changes, using the ideal gas law.<\/p>\n<p>Let\u2019s say that the original pressure in the cylinder was 2 atmospheres, the original volume was 3 liters, and the piston compressed the gas into a volume of 0.2 liters. <\/p>\n<p>We can solve the equation for P<sub>2<\/sub> = P<sub>1<\/sub> \u00d7 (V<sub>1<\/sub>\/V<sub>2<\/sub>) by plugging in our numbers:<\/p>\n<div class=\"examplesentence\">\\(P_2 = 2 \\text{ atm} \u00d7 (3 \\text{ L}\/0.2 \\text{ L})\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis gives us our numerical answer, which is 30 atmospheres.<\/p>\n<div class=\"examplesentence\">\\(P_2 = 2 \\text{ atm} \u00d7 (3 \\text{ L}\/0.2 \\text{ L}) = 30 \\text{ atm}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNote that because the volume went down, the pressure went up, confirming the inverse nature that Boyle identified.<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nAs a second example, let\u2019s look at a zeppelin, which is a lighter-than-air vehicle. <\/p>\n<p>Imagine that the zeppelin is on the ground, where the pressure is 1 atmosphere, and that we fill the zeppelin with a fixed number of moles of helium. We\u2019ll also say that the volume of our zeppelin is 2.40\u00d7105 L.<\/p>\n<p>Now, let\u2019s say that we want to send this zeppelin into the sky to a height of 4 kilometers. At that height, the air pressure is significantly lower, measuring approximately 0.61 atmospheres.<\/p>\n<p>Furthermore, if the temperature is constant, we know that the zeppelin will expand with a decrease in pressure. If the zeppelin can expand to a maximum volume of 3\u00d7105 L before bursting, will the zeppelin explode when it reaches that height?<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin.webp\" alt=\"zeppelin\" width=\"613.68\" height=\"345.6\" class=\"aligncenter size-full wp-image-209228\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin.webp 2557w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin-300x169.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin-1024x577.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin-768x433.webp 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin-1536x865.webp 1536w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/12\/zeppelin-2048x1153.webp 2048w\" sizes=\"(max-width: 2557px) 100vw, 2557px\" \/><\/p>\n<p>Of course, with the given conditions, we can use Boyle\u2019s Law to figure this out.<\/p>\n<p>This time, we need to solve for V<sub>2<\/sub> = V<sub>1<\/sub> (P<sub>1<\/sub>\/P<sub>2<\/sub>).<\/p>\n<p>Then, plug in the numbers.<\/p>\n<div class=\"examplesentence\">\\(V_2 = (P_1\/P_2)V_1 = 2.40\u00d7105 \\text{ L} (1 \\text{ atm}\/0.61 \\text{ atm}) = 4.92\u00d7105 \\text{ L}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe value obtained from Boyle\u2019s Law exceeds the maximum volume allowed, which means that the zeppelin would indeed explode.<\/p>\n<p>Again, note the inverse relationship: as the pressure decreased, the volume increased.<\/p>\n<hr>\n<h2><span id=\"Review\" class=\"m-toc-anchor\"><\/span>Review<\/h2>\n<p>\nNow that we\u2019ve looked at a couple of examples, let\u2019s finish up with a few review questions:<\/p>\n<p>1. Which of the following is Boyle\u2019s Law related to?<\/p>\n<ol style=\"list-style: upper-alpha;\">\n<li>The ideal gas law<\/li>\n<li>The first law of thermodynamics<\/li>\n<li>The boiling point and its relation to vapor pressure<\/li>\n<li>The Van der Waals equation<\/li>\n<\/ol>\n<div style=\"text-align: center; margin-bottom: 20px;\">\n   <button class=\"buttontranscript\" onClick=\"toggle('Answer1')\">Show Answer<\/button>\n<\/div>\n<div id=\"Answer1\" class=\"showanswer\">\n<strong>The correct answer is A.<\/strong><\/p>\n<p style=\"text-align: left;\">Boyle\u2019s Law can be considered an expression of the ideal gas law for constant temperature (and a fixed number of particles). Choice B is otherwise known as the law of conservation of energy. Choice C makes sense in the sense that there is an actual relationship between the two but is only loosely related at best. Choice D is mentioned as a type of intermolecular force, and though we used the lack of intermolecular forces to discuss the ideal gas law, it is not a precondition for Boyle\u2019s Law.<\/p>\n<\/div>\n<p>\n&nbsp;<br \/>\n2. Which of the following are held constant in the ideal gas law PV=nRT to show Boyle\u2019s Law?<\/p>\n<ol style=\"list-style: upper-alpha;\">\n<li>The number of moles of a gas and the pressure<\/li>\n<li>The pressure and the temperature<\/li>\n<li>The pressure and the number of moles<\/li>\n<li>The temperature and the number of moles<\/li>\n<\/ol>\n<div style=\"text-align: center; margin-bottom: 20px;\">\n   <button class=\"buttontranscript\" onClick=\"toggle('Answer2')\">Show Answer<\/button>\n<\/div>\n<div id=\"Answer2\" class=\"showanswershort\">\n   <strong>The correct answer is D.<\/strong>\n<\/div>\n<p>\n&nbsp;<br \/>\nThat\u2019s all for this review! 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