{"id":470,"date":"2026-07-12T01:43:06","date_gmt":"2026-07-12T01:43:06","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=470"},"modified":"2026-07-12T01:43:07","modified_gmt":"2026-07-12T01:43:07","slug":"boyles-law","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/boyles-law\/","title":{"rendered":"Boyle&#8217;s Law: Pressure, Volume and the P\u2081V\u2081 = P\u2082V\u2082 Rule"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nBoyle&#8217;s Law states that for a fixed mass of gas at constant temperature, absolute pressure is inversely proportional to volume: pressure times volume stays constant, so P\u2081V\u2081 = P\u2082V\u2082. Halve the volume and the pressure doubles. It is exact for an ideal gas and accurate for real gases at ordinary pressures.\n\n<\/p><\/div>\n\n<p>Take a breath. Right now, a sheet of muscle under your lungs has pulled downward, your chest cavity has grown by roughly half a litre, and the pressure inside it has dropped about one per cent below the pressure of the room. Air rushed in to close the gap.<\/p>\n\n<p>You did not pull the air in. You made space, and the atmosphere did the rest. That trade \u2014 more room, less pressure \u2014 is Boyle&#8217;s Law, and you have been running it about twenty thousand times a day since the moment you were born.<\/p>\n\n<h2>What Is Boyle&#8217;s Law?<\/h2>\n\n<p>Boyle&#8217;s Law says that if you seal a fixed amount of gas in a container and keep its temperature steady, squeezing the gas into a smaller space raises its pressure by exactly the same factor that the volume shrinks. Squeeze it to one-third the volume, and the pressure triples.<\/p>\n\n<p>Stated formally: <strong>at constant temperature, the absolute pressure of a fixed mass of gas is inversely proportional to its volume.<\/strong> &#8220;Inversely proportional&#8221; is the whole idea in two words. As one goes up, the other goes down, and their product refuses to move.<\/p>\n\n<p>Three conditions are doing quiet work in that sentence, and every one of them matters:<\/p>\n\n<ul>\n<li><strong>Fixed amount of gas<\/strong> \u2014 nothing leaks in or out. The number of molecules is locked.<\/li>\n<li><strong>Constant temperature<\/strong> \u2014 the process is <em>isothermal<\/em>. The gas has time to shed or absorb heat and settle back to its surroundings&#8217; temperature.<\/li>\n<li><strong>Absolute pressure<\/strong> \u2014 measured from a perfect vacuum, not from atmospheric pressure. This is where most mistakes begin.<\/li>\n<\/ul>\n\n<h3>Who Was Robert Boyle?<\/h3>\n\n<p>In 1662 the Anglo-Irish natural philosopher Robert Boyle published the result that carries his name. He poured mercury into a sealed J-shaped glass tube, trapping air in the short arm, and recorded how the trapped column shortened as he added more mercury.<\/p>\n\n<p>Boyle was building on a suggestion from Richard Towneley and Henry Power, and the apparatus was largely the work of his assistant Robert Hooke \u2014 the same Hooke of spring fame. The French physicist Edme Mariotte reported the relationship independently in the 1670s, which is why continental textbooks often call it the <em>Boyle\u2013Mariotte law<\/em>. NASA&#8217;s Glenn Research Center still teaches it to aeronautics students in <a href=\"https:\/\/www1.grc.nasa.gov\/beginners-guide-to-aeronautics\/boyles-law\/\" target=\"_blank\" rel=\"noopener\">exactly the form Boyle left it<\/a>.<\/p>\n\n<figure style=\"margin:32px auto;max-width:600px;text-align:center;\">\n\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/images-1-1.jpeg\"\n\n       alt=\"Portrait of Robert Boyle, who published Boyle's Law in 1662\"\n\n       loading=\"lazy\"\n\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Robert Boyle published the pressure\u2013volume relationship in 1662, using a J-shaped mercury tube.<\/figcaption>\n\n<\/figure>\n\n<h2>The Boyle&#8217;s Law Formula<\/h2>\n\n<p>The law is usually written as a statement about a constant:<\/p>\n\n<div class=\"pf-formula\">PV = constant   (fixed amount of gas, constant temperature)<\/div>\n\n<p>Because the product cannot change, the initial state and the final state must give the same answer. That gives the form you will actually use:<\/p>\n\n<div class=\"pf-formula\">P\u2081V\u2081 = P\u2082V\u2082<\/div>\n\n<p>Rearranged for whichever quantity is missing:<\/p>\n\n<div class=\"pf-formula\">P\u2082 = P\u2081V\u2081 \/ V\u2082        V\u2082 = P\u2081V\u2081 \/ P\u2082<\/div>\n\n<p>Every symbol, with its SI unit:<\/p>\n\n<ul>\n<li><strong>P\u2081<\/strong> \u2014 initial <em>absolute<\/em> pressure of the gas. SI unit: pascal (Pa). 1 Pa = 1 N\/m\u00b2.<\/li>\n<li><strong>V\u2081<\/strong> \u2014 initial volume of the gas. SI unit: cubic metre (m\u00b3).<\/li>\n<li><strong>P\u2082<\/strong> \u2014 final absolute pressure. SI unit: pascal (Pa).<\/li>\n<li><strong>V\u2082<\/strong> \u2014 final volume. SI unit: cubic metre (m\u00b3).<\/li>\n<\/ul>\n\n<p>Here is the practical mercy of this equation: <strong>the units cancel.<\/strong> Because pressure appears on both sides and volume appears on both sides, you may work in kilopascals and litres, or atmospheres and millilitres, provided you are consistent within each pair.<\/p>\n\n<p>What you may <em>not<\/em> do is mix. Litres on the left and cubic metres on the right will hand you an answer that is wrong by a factor of a thousand \u2014 and it will look perfectly reasonable.<\/p>\n\n<p>Prefer to skip the algebra? Drop your three known values into our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/boyles-law\">Boyle&#8217;s Law Calculator<\/a> and it returns the missing pressure or volume, in whatever units you choose.<\/p>\n\n<svg role=\"img\" aria-label=\"Boyle's Law diagram: a piston halves the volume of a gas at constant temperature and the pressure doubles\" viewBox=\"0 0 720 410\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:720px;display:block;margin:28px auto;\">\n<rect x=\"0\" y=\"0\" width=\"720\" height=\"410\" fill=\"#F5F2EA\"><\/rect>\n<rect x=\"90\" y=\"70\" width=\"160\" height=\"250\" fill=\"#FAF6EE\" stroke=\"#0A1628\" stroke-width=\"4\"><\/rect>\n<rect x=\"94\" y=\"150\" width=\"152\" height=\"166\" fill=\"#C8932A\" opacity=\"0.20\"><\/rect>\n<rect x=\"94\" y=\"138\" width=\"152\" height=\"12\" fill=\"#0A1628\"><\/rect>\n<rect x=\"164\" y=\"98\" width=\"12\" height=\"40\" fill=\"#0A1628\"><\/rect>\n<rect x=\"150\" y=\"78\" width=\"40\" height=\"20\" fill=\"#7A1F2B\"><\/rect>\n<circle cx=\"115\" cy=\"175\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"160\" cy=\"165\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"205\" cy=\"190\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"130\" cy=\"215\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"180\" cy=\"230\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"225\" cy=\"220\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"105\" cy=\"255\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"150\" cy=\"265\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"200\" cy=\"270\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"235\" cy=\"290\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"120\" cy=\"300\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"170\" cy=\"300\" r=\"5\" fill=\"#7A1F2B\"><\/circle>\n<rect x=\"440\" y=\"70\" width=\"160\" height=\"250\" fill=\"#FAF6EE\" stroke=\"#0A1628\" stroke-width=\"4\"><\/rect>\n<rect x=\"444\" y=\"233\" width=\"152\" height=\"83\" fill=\"#C8932A\" opacity=\"0.42\"><\/rect>\n<rect x=\"444\" y=\"221\" width=\"152\" height=\"12\" fill=\"#0A1628\"><\/rect>\n<rect x=\"514\" y=\"181\" width=\"12\" height=\"40\" fill=\"#0A1628\"><\/rect>\n<rect x=\"500\" y=\"161\" width=\"40\" height=\"20\" fill=\"#7A1F2B\"><\/rect>\n<rect x=\"500\" y=\"141\" width=\"40\" height=\"20\" fill=\"#7A1F2B\"><\/rect>\n<circle cx=\"465\" cy=\"248\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"505\" cy=\"242\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"545\" cy=\"255\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"580\" cy=\"248\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"455\" cy=\"275\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"495\" cy=\"268\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"535\" cy=\"278\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"575\" cy=\"272\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"470\" cy=\"300\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"510\" cy=\"295\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"550\" cy=\"302\" r=\"5\" fill=\"#7A1F2B\"><\/circle><circle cx=\"585\" cy=\"296\" r=\"5\" fill=\"#7A1F2B\"><\/circle>\n<defs><marker id=\"pfArrowGold\" markerWidth=\"10\" markerHeight=\"8\" refX=\"9\" refY=\"4\" orient=\"auto\"><path d=\"M0,0 L10,4 L0,8 z\" fill=\"#C8932A\"><\/path><\/marker><\/defs>\n<line x1=\"275\" y1=\"205\" x2=\"415\" y2=\"205\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#pfArrowGold)\"><\/line>\n<text x=\"345\" y=\"192\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\">compress slowly<\/text>\n<text x=\"345\" y=\"228\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#7A1F2B\">T constant, n constant<\/text>\n<text x=\"170\" y=\"352\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" font-weight=\"700\" fill=\"#0A1628\">State 1: pressure P\u2081, volume V\u2081<\/text>\n<text x=\"520\" y=\"352\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" font-weight=\"700\" fill=\"#0A1628\">State 2: pressure 2P\u2081, volume \u00bdV\u2081<\/text>\n<text x=\"170\" y=\"373\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#1F2E47\">12 molecules, roomy<\/text>\n<text x=\"520\" y=\"373\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#1F2E47\">same 12 molecules, half the room<\/text>\n<text x=\"360\" y=\"398\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"18\" font-weight=\"700\" fill=\"#7A1F2B\">P\u2081V\u2081 = P\u2082V\u2082<\/text>\n<\/svg>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Halving the volume of a sealed gas at constant temperature doubles its absolute pressure. The molecules never speed up \u2014 they simply hit the walls twice as often.<\/p>\n\n<h2>How Boyle&#8217;s Law Works: Inside the Gas<\/h2>\n\n<p>Pressure is not a substance sitting inside a container. It is a drumroll. Billions of molecules are hammering the walls every second, and pressure is nothing more than the average force of that hammering, divided by the wall area.<\/p>\n\n<p>So ask the obvious question: what happens to the drumroll when the room shrinks?<\/p>\n\n<p>Two things, and they reinforce each other. First, each molecule has less distance to cross before it slams into a wall, so it returns more often. Second, the same crowd is now packed into less space, so more of them are near any given patch of wall at any moment.<\/p>\n\n<p>Halve the volume and you double the collision rate on every square metre. Double the collision rate and you double the pressure. Nothing about the individual molecule has changed \u2014 it is still travelling at the same speed, delivering the same punch. It is just punching more often.<\/p>\n\n<h3>Why Temperature Must Stay Constant<\/h3>\n\n<p>That last point is the hinge of the whole law. The <em>speed<\/em> of the molecules is set by temperature: absolute temperature is a direct measure of the average <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a> of the molecules.<\/p>\n\n<p>Hold the temperature fixed and you hold the molecular speeds fixed. Only the geometry changes, and the pressure obeys a clean inverse rule. Let the temperature climb, and the molecules start hitting harder as well as more often \u2014 a second effect, on top of Boyle&#8217;s, which the equation P\u2081V\u2081 = P\u2082V\u2082 knows nothing about. If the distinction between <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/heat-vs-temperature\/\">heat and temperature<\/a> feels slippery, this is the sentence to anchor it to.<\/p>\n\n<h3>The Shape of the Graph<\/h3>\n\n<p>Plot pressure against volume for a gas at fixed temperature and you get a curve that falls steeply, then flattens, never quite touching either axis. That curve is a rectangular hyperbola, and physicists call it an <strong>isotherm<\/strong>.<\/p>\n\n<p>Plot pressure against <em>1\/V<\/em> instead, and the same data snaps into a straight line through the origin, with gradient equal to the constant PV. That is the plot examiners want when they ask you to &#8220;verify Boyle&#8217;s law graphically&#8221; \u2014 a hyperbola is hard to judge by eye, a straight line is not.<\/p>\n\n<svg role=\"img\" aria-label=\"Two graphs illustrating Boyle's Law: pressure against volume gives a hyperbola, pressure against one over volume gives a straight line\" viewBox=\"0 0 760 400\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:760px;display:block;margin:28px auto;\">\n<rect x=\"0\" y=\"0\" width=\"760\" height=\"400\" fill=\"#F5F2EA\"><\/rect>\n<text x=\"235\" y=\"34\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" font-weight=\"700\" fill=\"#7A1F2B\">P against V \u2014 a hyperbola<\/text>\n<text x=\"605\" y=\"34\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" font-weight=\"700\" fill=\"#7A1F2B\">P against 1\/V \u2014 a straight line<\/text>\n<line x1=\"70\" y1=\"330\" x2=\"405\" y2=\"330\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<line x1=\"70\" y1=\"330\" x2=\"70\" y2=\"62\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<polyline points=\"70,98.9 81.4,145.1 92.9,175.9 115.7,214.4 138.6,237.6 161.4,253.0 207.1,272.2 252.9,283.8 298.6,291.5 344.3,297.0 390,301.1\" fill=\"none\" stroke=\"#C8932A\" stroke-width=\"3.5\"><\/polyline>\n<line x1=\"115.7\" y1=\"330\" x2=\"115.7\" y2=\"214.4\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"4,4\"><\/line>\n<line x1=\"70\" y1=\"214.4\" x2=\"115.7\" y2=\"214.4\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"4,4\"><\/line>\n<line x1=\"207.1\" y1=\"330\" x2=\"207.1\" y2=\"272.2\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"4,4\"><\/line>\n<line x1=\"70\" y1=\"272.2\" x2=\"207.1\" y2=\"272.2\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"4,4\"><\/line>\n<circle cx=\"115.7\" cy=\"214.4\" r=\"6\" fill=\"#7A1F2B\"><\/circle>\n<circle cx=\"207.1\" cy=\"272.2\" r=\"6\" fill=\"#7A1F2B\"><\/circle>\n<text x=\"126\" y=\"208\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">A<\/text>\n<text x=\"217\" y=\"266\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">B<\/text>\n<text x=\"235\" y=\"358\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\">Volume V<\/text>\n<text x=\"34\" y=\"200\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\" transform=\"rotate(-90 34 200)\">Pressure P<\/text>\n<text x=\"235\" y=\"382\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" fill=\"#1F2E47\">B has twice the volume of A, so half the pressure<\/text>\n<line x1=\"480\" y1=\"330\" x2=\"740\" y2=\"330\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<line x1=\"480\" y1=\"330\" x2=\"480\" y2=\"62\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<line x1=\"480\" y1=\"330\" x2=\"720\" y2=\"98.9\" stroke=\"#C8932A\" stroke-width=\"3.5\"><\/line>\n<line x1=\"600\" y1=\"330\" x2=\"600\" y2=\"214.4\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"4,4\"><\/line>\n<line x1=\"480\" y1=\"214.4\" x2=\"600\" y2=\"214.4\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"4,4\"><\/line>\n<circle cx=\"600\" cy=\"214.4\" r=\"6\" fill=\"#7A1F2B\"><\/circle>\n<text x=\"610\" y=\"208\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">A<\/text>\n<text x=\"605\" y=\"358\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\">1 \/ Volume<\/text>\n<text x=\"444\" y=\"200\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\" transform=\"rotate(-90 444 200)\">Pressure P<\/text>\n<text x=\"605\" y=\"382\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" fill=\"#1F2E47\">gradient of this line = the constant PV<\/text>\n<\/svg>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">The same Boyle&#8217;s Law data, plotted two ways. The hyperbola is the physics; the straight line is the proof.<\/p>\n\n<h2>See Boyle&#8217;s Law in Action<\/h2>\n\n<p>Drag the piston and watch the pressure gauge climb while the product PV sits stubbornly still. Push the volume down to a tenth and the needle goes tenfold \u2014 the hyperbola on the graph traces itself as you go.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Boyle&#039;s Law Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/boyles-law.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>Boyle&#8217;s Law vs the Other Gas Laws<\/h2>\n\n<p>Boyle&#8217;s Law is one of four experimental laws that were eventually stitched into a single equation. Each one freezes two variables and watches the other two dance. Learn which variable is being <em>held<\/em> and you will never confuse them again.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr style=\"background:#0A1628;color:#FAF6EE;\">\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Law<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Held constant<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Relationship<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Equation<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Graph shape<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Boyle&#8217;s<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Temperature, amount<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P \u221d 1\/V<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P\u2081V\u2081 = P\u2082V\u2082<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Hyperbola on a P\u2013V plot<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Charles&#8217;s<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Pressure, amount<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V \u221d T<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V\u2081\/T\u2081 = V\u2082\/T\u2082<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Straight line aimed at 0 K<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Gay-Lussac&#8217;s<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Volume, amount<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P \u221d T<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P\u2081\/T\u2081 = P\u2082\/T\u2082<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Straight line aimed at 0 K<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Avogadro&#8217;s<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Pressure, temperature<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V \u221d n<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V\u2081\/n\u2081 = V\u2082\/n\u2082<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Straight line through origin<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Combined gas law<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Amount only<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">PV\/T fixed<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P\u2081V\u2081\/T\u2081 = P\u2082V\u2082\/T\u2082<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Family of isotherms<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Ideal gas law<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Nothing<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">All four linked<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">PV = nRT<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">A surface in P\u2013V\u2013T space<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p><strong>Every temperature in that table is an absolute temperature, in kelvin.<\/strong> Celsius will destroy Charles&#8217;s law and Gay-Lussac&#8217;s law instantly. Boyle&#8217;s law is the one law in the family where temperature never appears \u2014 which is precisely why it is the one students reach for by reflex, sometimes when they shouldn&#8217;t.<\/p>\n\n<h2>Real-World Examples of Boyle&#8217;s Law<\/h2>\n\n<h3>1. Breathing<\/h3>\n\n<p>Your diaphragm contracts and flattens; the chest cavity expands. By Boyle&#8217;s Law the pressure inside your lungs drops roughly 1\u20132 kPa below atmospheric, and air floods in. Exhaling reverses it. You are a soft-walled piston, running at about twelve strokes a minute.<\/p>\n\n<h3>2. Scuba Diving and the Growing Bubble<\/h3>\n\n<p>Sea water adds about one atmosphere of pressure for every 10 metres of depth. At 10 m a diver breathes air at roughly twice atmospheric pressure; at 30 m, four times.<\/p>\n\n<p>Release a bubble at 30 m and it swells to about four times its volume by the time it reaches the surface. So does the air in a diver&#8217;s lungs. This is why the first rule of scuba is never to hold your breath while ascending \u2014 a full breath taken at 30 m would expand to four lungfuls at the surface, and lungs do not stretch.<\/p>\n\n<h3>3. The Bicycle Pump and the Syringe<\/h3>\n\n<p>Block the outlet of a bicycle pump and push. The barrel volume falls, the pressure climbs, and the handle fights back harder with every centimetre. A syringe with its nozzle capped does the same thing with your thumb.<\/p>\n\n<h3>4. Weather Balloons<\/h3>\n\n<p>A meteorological balloon launched limp and floppy at ground level swells as it climbs, because the outside pressure falls away. By around 30 km it may be many times its launch volume \u2014 and eventually it bursts, which is exactly what the design expects.<\/p>\n\n<h3>5. Aerosols, Vacuum Packing and the Marshmallow Trick<\/h3>\n\n<p>Put a marshmallow under a bell jar and pump the air out. The tiny bubbles trapped in the sugar foam find themselves at a fraction of their original external pressure, so they expand, and the marshmallow inflates like a balloon. Let the air back in and it collapses into a sad, dense lump.<\/p>\n\n<h2>Common Misconceptions About Boyle&#8217;s Law<\/h2>\n\n<h3>Trap 1: Using the Pressure Your Gauge Shows<\/h3>\n\n<p>This is the single most expensive error in gas-law questions. A tyre gauge, a pressure sensor on a lab bench, most everyday instruments \u2014 they read <strong>gauge pressure<\/strong>, the excess above atmospheric. Boyle&#8217;s Law demands <strong>absolute pressure<\/strong>.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr style=\"background:#0A1628;color:#FAF6EE;\">\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Reading<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Measured from<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Safe to put into P\u2081V\u2081 = P\u2082V\u2082?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Gauge pressure<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Local atmospheric pressure<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>No<\/strong> \u2014 convert it first<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Absolute pressure<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">A perfect vacuum<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Yes<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Conversion<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P<sub>abs<\/sub> = P<sub>gauge<\/sub> + P<sub>atm<\/sub><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P<sub>atm<\/sub> \u2248 101 kPa at sea level<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>In practice: a flat tyre reads zero on a gauge, but it is emphatically not a vacuum. It still holds air at 101 kPa absolute.<\/p>\n\n<h3>Trap 2: Assuming Any Squeeze Is a Boyle&#8217;s Law Squeeze<\/h3>\n\n<p>Pump a bicycle tyre vigorously and grip the barrel. It is hot. That heat is real work being converted into internal energy, and it means the temperature did <em>not<\/em> stay constant \u2014 so Boyle&#8217;s Law does not describe what just happened.<\/p>\n\n<p>A fast compression is approximately <em>adiabatic<\/em>, following PV<sup>\u03b3<\/sup> = constant, where \u03b3 \u2248 1.4 for air. The pressure rises faster than Boyle predicts. Boyle&#8217;s Law needs a slow squeeze, or a container that leaks heat away quickly enough for the gas to keep pace with its surroundings. How fast that happens depends on the material&#8217;s <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/specific-heat-capacity\/\">specific heat capacity<\/a>.<\/p>\n\n<h3>Trap 3: Expecting a Straight Line on a P\u2013V Graph<\/h3>\n\n<p>&#8220;Inversely proportional&#8221; is not &#8220;decreasing&#8221;. A student who sketches a downward-sloping straight line has drawn P = a \u2212 bV, which is a different physical claim entirely. The Boyle isotherm curves, hugging both axes without ever meeting them. Only the P against 1\/V plot is straight.<\/p>\n\n<h3>Trap 4: Believing It Works Everywhere<\/h3>\n\n<p>Boyle&#8217;s Law is exact for an ideal gas \u2014 one whose molecules have no volume and no attraction for one another. Real molecules have both.<\/p>\n\n<p>Squeeze hard enough and the molecules&#8217; own volume stops being negligible, so the gas resists more than Boyle predicts. Cool it enough and intermolecular attraction pulls molecules inward, so the pressure comes in <em>lower<\/em> than predicted. Near the point where a gas is about to liquefy, PV stops being constant altogether. The van der Waals equation exists to patch exactly these two effects.<\/p>\n\n<h2>How Boyle&#8217;s Law Relates to the Ideal Gas Law and Thermodynamics<\/h2>\n\n<p>Boyle&#8217;s Law is not a separate truth. It is a special case, and it falls out of the ideal gas law in one line. Start with:<\/p>\n\n<div class=\"pf-formula\">PV = nRT<\/div>\n\n<p>where <strong>n<\/strong> is the amount of gas in moles, <strong>R<\/strong> is the universal gas constant (8.314 J mol\u207b\u00b9 K\u207b\u00b9) and <strong>T<\/strong> is the absolute temperature in kelvin. Hold n and T fixed. The entire right-hand side is now a number that cannot change \u2014 so PV must be constant, and P\u2081V\u2081 = P\u2082V\u2082. Georgia State&#8217;s HyperPhysics sets out <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/Kinetic\/idegas.html\" target=\"_blank\" rel=\"noopener\">the same one-line reduction<\/a>, alongside the Charles&#8217;s law case.<\/p>\n\n<p>Freeze pressure instead and you recover Charles&#8217;s law. Freeze volume and you get Gay-Lussac&#8217;s. Three famous laws, one parent equation.<\/p>\n\n<h3>Boyle&#8217;s Law and the First Law of Thermodynamics<\/h3>\n\n<p>Compress a gas isothermally and you do work on it. Where does that <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-energy-in-physics\/\">energy<\/a> go? Not into internal energy \u2014 internal energy depends only on temperature for an ideal gas, and temperature has not moved.<\/p>\n\n<p>It leaves as heat. Every joule you put in through the piston flows straight out through the walls into the surroundings. That bookkeeping is the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/laws-of-thermodynamics\/\">first law of thermodynamics<\/a> in its cleanest possible costume, and the work done comes out as W = nRT ln(V\u2081\/V\u2082). Problem 7 below puts numbers on it.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A sealed syringe holds 2.00 L of air at an absolute pressure of 100 kPa. The plunger is pushed in slowly until the volume is 0.500 L. The temperature does not change. What is the new absolute pressure?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Temperature and amount of gas are constant, so Boyle&#8217;s Law applies: P\u2081V\u2081 = P\u2082V\u2082\n\nStep 2: Rearrange for the unknown: P\u2082 = P\u2081V\u2081 \/ V\u2082\n\nStep 3: Substitute with units: P\u2082 = (100 kPa \u00d7 2.00 L) \/ 0.500 L = 200 kPa\u00b7L \/ 0.500 L\n\nStep 4: Solve: P\u2082 = 400 kPa\n\nSanity check: the volume fell to one quarter, so the pressure rose four-fold. It did.\n\n<strong>Answer: P\u2082 = 400 kPa (absolute)<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A gas occupies 3.60 L at an absolute pressure of 1.50 atm. At constant temperature, to what volume must it be compressed to raise the pressure to 4.00 atm?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Apply Boyle&#8217;s Law: P\u2081V\u2081 = P\u2082V\u2082\n\nStep 2: Rearrange for volume: V\u2082 = P\u2081V\u2081 \/ P\u2082\n\nStep 3: Substitute with units: V\u2082 = (1.50 atm \u00d7 3.60 L) \/ 4.00 atm = 5.40 atm\u00b7L \/ 4.00 atm\n\nStep 4: Solve: V\u2082 = 1.35 L\n\nThe atmospheres cancel, leaving litres. Mixed but consistent units are fine.\n\n<strong>Answer: V\u2082 = 1.35 L<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">The volume of a sealed sample of gas is reduced by 20 per cent at constant temperature. By what percentage does its absolute pressure increase?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: A 20 per cent reduction means V\u2082 = 0.800 V\u2081\n\nStep 2: From Boyle&#8217;s Law: P\u2082 = P\u2081V\u2081 \/ V\u2082 = P\u2081V\u2081 \/ (0.800 V\u2081)\n\nStep 3: The V\u2081 terms cancel: P\u2082 = P\u2081 \/ 0.800 = 1.25 P\u2081\n\nStep 4: The pressure is 1.25 times its original value, an increase of 0.25 P\u2081\n\nA common student slip is to answer &#8220;20 per cent&#8221;. Inverse proportionality is not symmetric \u2014 a 20 per cent cut in volume gives a 25 per cent rise in pressure.\n\n<strong>Answer: the absolute pressure increases by 25 per cent<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A bicycle pump barrel contains 500 cm\u00b3 of air at atmospheric pressure, with the outlet blocked. The gauge reads 0 kPa. The plunger is pushed in slowly until the trapped air occupies 125 cm\u00b3. Take atmospheric pressure as 101 kPa. What will the gauge now read?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Convert to absolute pressure before touching Boyle&#8217;s Law: P\u2081 = P_gauge + P_atm = 0 + 101 = 101 kPa\n\nStep 2: Apply P\u2081V\u2081 = P\u2082V\u2082, so P\u2082 = P\u2081V\u2081 \/ V\u2082\n\nStep 3: Substitute with units: P\u2082 = (101 kPa \u00d7 500 cm\u00b3) \/ 125 cm\u00b3 = 50 500 kPa\u00b7cm\u00b3 \/ 125 cm\u00b3\n\nStep 4: Solve for absolute pressure: P\u2082 = 404 kPa\n\nStep 5: Convert back to gauge pressure: P_gauge = 404 \u2212 101 = 303 kPa\n\nSkipping Step 1 gives 0 \u00d7 500 \/ 125 = 0 kPa \u2014 an &#8220;answer&#8221; that says compressing air does nothing.\n\n<strong>Answer: the gauge reads 303 kPa (404 kPa absolute)<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A bubble of volume 25.0 mL is released at a depth of 30.0 m in sea water of density 1025 kg\/m\u00b3. Take g = 9.81 m\/s\u00b2 and atmospheric pressure as 101 kPa. Assuming the water temperature is uniform, what is the bubble&#039;s volume just before it reaches the surface?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Absolute pressure at depth is atmospheric plus the water column: P\u2081 = P_atm + \u03c1gh\n\nStep 2: Substitute with units: \u03c1gh = 1025 kg\/m\u00b3 \u00d7 9.81 m\/s\u00b2 \u00d7 30.0 m = 301 658 Pa = 301.7 kPa\n\nSo P\u2081 = 101 + 301.7 = 402.7 kPa\n\nStep 3: At the surface, P\u2082 = 101 kPa\n\nStep 4: Apply Boyle&#8217;s Law: V\u2082 = P\u2081V\u2081 \/ P\u2082 = (402.7 kPa \u00d7 25.0 mL) \/ 101 kPa\n\nStep 5: Solve: V\u2082 = 99.7 mL\n\nSanity check: 30 m of sea water is roughly three extra atmospheres, so the total is about four. The bubble should grow about four-fold, and 99.7 \/ 25.0 = 3.99.\n\n<strong>Answer: V\u2082 \u2248 99.7 mL, about four times its original volume<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A rigid 2.00 L cylinder holds gas at an absolute pressure of 500 kPa. It is connected by a valve to a rigid, fully evacuated 3.00 L cylinder. The valve is opened and the gas is allowed to settle back to its original temperature. What is the final pressure?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: The amount of gas is unchanged and the final temperature equals the initial temperature, so Boyle&#8217;s Law applies between the initial and final equilibrium states.\n\nStep 2: The gas now fills both cylinders: V\u2082 = 2.00 L + 3.00 L = 5.00 L\n\nStep 3: Apply P\u2081V\u2081 = P\u2082V\u2082, so P\u2082 = P\u2081V\u2081 \/ V\u2082\n\nStep 4: Substitute with units: P\u2082 = (500 kPa \u00d7 2.00 L) \/ 5.00 L = 1000 kPa\u00b7L \/ 5.00 L\n\nStep 5: Solve: P\u2082 = 200 kPa\n\nNote the evacuated cylinder contributes volume but no gas, so its initial pressure of 0 kPa never enters the arithmetic.\n\n<strong>Answer: P\u2082 = 200 kPa (absolute)<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">One mole of an ideal gas at 300 K is compressed isothermally and reversibly from 20.0 L to 10.0 L. Take R = 8.314 J\/(mol\u00b7K). Find the initial and final pressures, verify Boyle&#039;s Law, and calculate the work done on the gas.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Find the initial pressure from PV = nRT, so P\u2081 = nRT \/ V\u2081\n\nnRT = 1.00 mol \u00d7 8.314 J\/(mol\u00b7K) \u00d7 300 K = 2494.2 J\n\nV\u2081 = 20.0 L = 0.0200 m\u00b3\n\nP\u2081 = 2494.2 J \/ 0.0200 m\u00b3 = 124 710 Pa = 125 kPa (3 s.f.)\n\nStep 2: Find the final pressure the same way, with V\u2082 = 0.0100 m\u00b3\n\nP\u2082 = 2494.2 J \/ 0.0100 m\u00b3 = 249 420 Pa = 249 kPa (3 s.f.)\n\nStep 3: Verify Boyle&#8217;s Law: P\u2081V\u2081 = 124 710 \u00d7 0.0200 = 2494.2 J and P\u2082V\u2082 = 249 420 \u00d7 0.0100 = 2494.2 J. The products match, and the pressure has doubled as the volume halved.\n\nStep 4: The work done on the gas during a reversible isothermal compression is W = nRT ln(V\u2081 \/ V\u2082)\n\nStep 5: Substitute with units: W = 2494.2 J \u00d7 ln(20.0 \/ 10.0) = 2494.2 J \u00d7 ln 2 = 2494.2 \u00d7 0.6931\n\nStep 6: Solve: W = 1729 J\n\nBecause the temperature never changed, the internal energy did not change either \u2014 all 1729 J left the gas as heat.\n\n<strong>Answer: P\u2081 = 125 kPa, P\u2082 = 249 kPa, and the work done on the gas is W \u2248 1.73 kJ, all of it released as heat<\/strong>\n\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is Boyle&#039;s law in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nBoyle&#8217;s law says that squeezing a fixed amount of gas into a smaller space raises its pressure by the same factor, as long as the temperature stays the same. Halve the volume and the absolute pressure doubles. The product of pressure and volume never changes, which is written as P\u2081V\u2081 = P\u2082V\u2082.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for Boyle&#039;s law?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe formula for Boyle&#8217;s law is P\u2081V\u2081 = P\u2082V\u2082, where P\u2081 and V\u2081 are the initial absolute pressure and volume, and P\u2082 and V\u2082 are the final absolute pressure and volume. Rearranged, P\u2082 = P\u2081V\u2081\/V\u2082 and V\u2082 = P\u2081V\u2081\/P\u2082. Units cancel, so any consistent pressure and volume units work.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does Boyle&#039;s law use gauge pressure or absolute pressure?<\/summary><div class=\"pf-faq-item-answer\">\n\nBoyle&#8217;s law requires absolute pressure, measured from a perfect vacuum. Most instruments report gauge pressure, which is the excess above atmospheric pressure. Convert first using P_absolute = P_gauge + P_atmospheric, taking atmospheric pressure as roughly 101 kPa at sea level. Using gauge pressure directly is the most common source of wrong answers.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why does Boyle&#039;s law require constant temperature?<\/summary><div class=\"pf-faq-item-answer\">\n\nTemperature sets the average speed of the gas molecules. If the temperature is constant, the molecules hit the walls with the same force as before, and only the collision rate changes when the volume changes. Raise the temperature and the molecules hit harder as well as more often, which breaks the simple inverse relationship.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is Boyle&#039;s law the same as the ideal gas law?<\/summary><div class=\"pf-faq-item-answer\">\n\nNo \u2014 Boyle&#8217;s law is a special case of the ideal gas law. Starting from PV = nRT, hold the amount of gas n and the absolute temperature T constant. The right-hand side becomes a fixed number, so PV must also be fixed, giving P\u2081V\u2081 = P\u2082V\u2082. The ideal gas law is more general.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What are real-life examples of Boyle&#039;s law?<\/summary><div class=\"pf-faq-item-answer\">\n\nBreathing is the everyday example: your diaphragm enlarges the chest cavity, the pressure inside drops, and air flows in. Others include a bubble expanding as it rises from a diver, a weather balloon swelling as it climbs, a syringe or bicycle pump building pressure, and a marshmallow inflating inside a vacuum jar.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>When does Boyle&#039;s law break down?<\/summary><div class=\"pf-faq-item-answer\">\n\nBoyle&#8217;s law breaks down at very high pressures and very low temperatures, where a real gas stops behaving ideally. At high pressure the molecules&#8217; own volume becomes significant, so the gas resists compression more than predicted. Near liquefaction, intermolecular attraction lowers the pressure below prediction. The van der Waals equation corrects for both effects.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Boyle&#8217;s Law states that the absolute pressure of a fixed mass of gas is inversely proportional to its volume at constant temperature, so P\u2081V\u2081 = P\u2082V\u2082. This guide covers the formula, the graph, five real-world examples, four common traps and seven worked problems.<\/p>\n","protected":false},"author":1,"featured_media":472,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[291,283,282,281,163,28],"class_list":["post-470","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-thermodynamics","tag-boyles-law","tag-gas-laws","tag-ideal-gas-law","tag-kinetic-theory","tag-pressure","tag-thermodynamics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/470","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=470"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/470\/revisions"}],"predecessor-version":[{"id":473,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/470\/revisions\/473"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/472"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=470"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=470"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=470"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}