{"id":464,"date":"2026-07-12T01:38:17","date_gmt":"2026-07-12T01:38:17","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=464"},"modified":"2026-07-12T01:38:19","modified_gmt":"2026-07-12T01:38:19","slug":"charles-law","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/charles-law\/","title":{"rendered":"Charles&#8217;s Law Explained"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nCharles law, properly written Charles&#8217;s law, states that the volume of a fixed mass of gas held at constant pressure is directly proportional to its absolute temperature. Double the temperature in kelvin and the volume doubles. In symbols, V\u2081 divided by T\u2081 equals V\u2082 divided by T\u2082, with T always in kelvin.\n<\/p><\/div>\n\n<p>Leave a balloon in a hot car and it strains at the knot, taut and glossy. Bring it out into a cold night and within minutes it sags, wrinkled, looking half-empty.<\/p>\n\n<p>Nothing leaked. The same air sits inside, pressed by the same atmosphere \u2014 only the temperature moved, and the volume obediently followed. That is Charles&#8217;s law, doing its work in a car park.<\/p>\n\n<h2>What Is Charles&#8217;s Law?<\/h2>\n\n<p>Ask a class what heat does to a gas and someone will say &#8220;it makes it expand.&#8221; True. But that is a description, not a law \u2014 and a law has to tell you <em>exactly how much<\/em>.<\/p>\n\n<p>Charles&#8217;s law supplies the number. Hold the pressure steady, keep the same quantity of gas, and the volume tracks the absolute temperature in strict proportion. Ten per cent hotter in kelvin means ten per cent bigger, every time.<\/p>\n\n<p>That proportionality only works if you measure temperature from the right zero. Start counting at absolute zero, in kelvin, and the relationship is a clean straight line through the origin. Start at the freezing point of water, in Celsius, and it falls apart completely.<\/p>\n\n<p>Two conditions do the heavy lifting here, and both are easy to lose sight of:<\/p>\n\n<ul>\n<li><strong>Constant pressure.<\/strong> The gas must be free to push its container outward \u2014 a balloon, a piston, a flexible bag. A sealed steel cylinder does not qualify.<\/li>\n<li><strong>Fixed amount of gas.<\/strong> No molecules added, none allowed to escape. The number of moles stays put.<\/li>\n<\/ul>\n\n<h3>Who Actually Discovered It?<\/h3>\n\n<p>Jacques Charles, a French physicist and balloonist, ran the experiments around 1787 and never wrote them up. Joseph Louis Gay-Lussac published the volume\u2013temperature relationship in 1802 and pointedly credited Charles&#8217;s unpublished work \u2014 which is why the law bears the name of the man who did not publish it.<\/p>\n\n<p>John Dalton reached much the same conclusion independently in 1801. NASA&#8217;s Glenn Research Center still teaches it as <a href=\"https:\/\/www1.grc.nasa.gov\/beginners-guide-to-aeronautics\/charles-and-gay-lussacs-law\/\" target=\"_blank\" rel=\"noopener\">Charles and Gay-Lussac&#8217;s law<\/a>, which is the fairer title, if a clumsier one.<\/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-2.jpeg\"\n\n       alt=\"Portrait of Jacques Charles, the French physicist behind Charles law\"\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;\">Jacques Charles (1746\u20131823). His gas experiments went unpublished; Gay-Lussac gave him the credit anyway.<\/figcaption>\n\n<\/figure>\n\n<h2>The Charles&#8217;s Law Formula<\/h2>\n\n<p>Charles&#8217;s law wears two faces. The proportional form tells you what the physics <em>is<\/em>. The two-state form is what you actually calculate with.<\/p>\n\n<div class=\"pf-formula\">V \u221d T   (constant pressure, fixed amount of gas)<\/div>\n\n<div class=\"pf-formula\">V\u2081 \/ T\u2081 = V\u2082 \/ T\u2082<\/div>\n\n<p>Both say the same thing: the ratio V\/T is a constant. Heat the gas, and the volume climbs by whatever factor keeps that ratio fixed.<\/p>\n\n<ul>\n<li><strong>V\u2081<\/strong> \u2014 initial volume, in cubic metres (m\u00b3). Litres are fine, provided V\u2082 uses the same unit.<\/li>\n<li><strong>T\u2081<\/strong> \u2014 initial absolute temperature, in kelvin (K). Never Celsius.<\/li>\n<li><strong>V\u2082<\/strong> \u2014 final volume, in the same unit as V\u2081.<\/li>\n<li><strong>T\u2082<\/strong> \u2014 final absolute temperature, in kelvin (K).<\/li>\n<li><strong>k<\/strong> \u2014 the constant of proportionality, k = V\/T, in cubic metres per kelvin (m\u00b3\/K). Its value depends on the pressure and on how much gas you have.<\/li>\n<\/ul>\n\n<p>Because both volumes carry the same unit, they cancel in the ratio. Temperature enjoys no such freedom \u2014 kelvin, or nothing.<\/p>\n\n<div class=\"pf-formula\">T (K) = \u03b8 (\u00b0C) + 273.15<\/div>\n\n<h3>Solving a Charles&#8217;s Law Problem in 5 Steps<\/h3>\n\n<ol>\n<li><strong>Convert both temperatures to kelvin.<\/strong> Add 273.15 to each Celsius value. Do this before you do anything else.<\/li>\n<li><strong>Check the conditions.<\/strong> Is the pressure constant? Is the amount of gas fixed? If either answer is no, you need a different gas law.<\/li>\n<li><strong>Write V\u2081\/T\u2081 = V\u2082\/T\u2082 and label the knowns.<\/strong> Three of the four quantities will be given.<\/li>\n<li><strong>Rearrange for the unknown first, then substitute.<\/strong> Solving for volume gives V\u2082 = V\u2081 \u00d7 T\u2082\/T\u2081. Solving for temperature gives T\u2082 = T\u2081 \u00d7 V\u2082\/V\u2081.<\/li>\n<li><strong>Sanity-check the direction.<\/strong> Hotter must mean bigger; colder must mean smaller. If your answer disagrees, you have inverted the ratio.<\/li>\n<\/ol>\n\n<p>Step 5 catches more errors than the other four combined. In practice, a student who inverts T\u2082\/T\u2081 produces an answer that is wrong by a plausible-looking factor, and nothing on the page objects \u2014 except the physics.<\/p>\n\n<p>Prefer to check your working, or skip the arithmetic altogether? Our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/charles-law\">Charles&#8217;s Law Calculator<\/a> solves for any of the four quantities \u2014 enter the three you know and it rearranges the ratio and converts your temperatures to kelvin automatically.<\/p>\n\n<h3>The Ratio, Checked Numerically<\/h3>\n\n<p>Take 2.00 L of gas at 300 K and warm it at constant pressure. Watch what stays fixed.<\/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;border:1px solid #D9CFB8;text-align:left;\">Temperature T (K)<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Temperature (\u00b0C)<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Volume V (L)<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">V \/ T (L\/K)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">300<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">26.85<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">2.00<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.00667<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">400<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">126.85<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">2.67<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.00667<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">500<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">226.85<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">3.33<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.00667<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">600<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">326.85<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">4.00<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.00667<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">V\/T holds at 0.00667 L\/K throughout (3 s.f.). Notice that the Celsius column does no such thing.<\/p>\n\n<h2>How Charles&#8217;s Law Works: Why Heating a Gas Pushes the Piston Out<\/h2>\n\n<p>Zoom in far enough and a gas is just molecules, flying, colliding, rebounding off the walls. Temperature is a measure of how fast they are going \u2014 specifically, of their average kinetic energy.<\/p>\n\n<p>Heat the gas and every molecule speeds up. Each wall collision now lands harder, and collisions arrive more often. Pressure is force per unit area, so if the container could not move, the pressure would climb.<\/p>\n\n<p>But under Charles&#8217;s law the container <em>can<\/em> move. The piston is loaded with a fixed weight, so the outside pressure never changes. The faster molecules shove the piston outward, the gas volume grows, and each square centimetre of wall now receives fewer hits per second because the molecules have farther to travel between them.<\/p>\n\n<p>The piston stops when the hit rate per unit area has fallen back to exactly what it was before. That happens when the volume has grown in the same ratio as the absolute temperature. Which is the law.<\/p>\n\n<svg viewBox=\"0 0 700 400\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" role=\"img\" aria-label=\"Two cylinders illustrating Charles law: identical weighted pistons, but the hotter gas on the right occupies twice the volume\" style=\"display:block;width:100%;height:auto;max-width:700px;margin:32px auto;\">\n<rect width=\"700\" height=\"400\" fill=\"#0A1628\" rx=\"4\"><\/rect>\n<text x=\"350\" y=\"30\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"17\" font-weight=\"600\" fill=\"#FAF6EE\">Constant pressure: the piston moves, the pressure does not<\/text>\n<text x=\"350\" y=\"52\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" fill=\"#C5D0DC\">Same gas, same number of molecules, same weight on the piston<\/text>\n\n<line x1=\"120\" y1=\"70\" x2=\"120\" y2=\"315\" stroke=\"#D9CFB8\" stroke-width=\"3\"><\/line>\n<line x1=\"250\" y1=\"70\" x2=\"250\" y2=\"315\" stroke=\"#D9CFB8\" stroke-width=\"3\"><\/line>\n<rect x=\"112\" y=\"315\" width=\"146\" height=\"10\" fill=\"#D9CFB8\" rx=\"2\"><\/rect>\n<rect x=\"123\" y=\"220\" width=\"124\" height=\"95\" fill=\"#142139\"><\/rect>\n<rect x=\"120\" y=\"208\" width=\"130\" height=\"12\" fill=\"#C5D0DC\" rx=\"2\"><\/rect>\n<rect x=\"155\" y=\"168\" width=\"60\" height=\"40\" fill=\"#7A1F2B\" rx=\"3\"><\/rect>\n<text x=\"185\" y=\"194\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" font-weight=\"600\" fill=\"#FAF6EE\">P<\/text>\n<circle cx=\"145\" cy=\"245\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"145\" y1=\"245\" x2=\"152\" y2=\"241\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"178\" cy=\"236\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"178\" y1=\"236\" x2=\"171\" y2=\"241\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"212\" cy=\"252\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"212\" y1=\"252\" x2=\"219\" y2=\"248\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"233\" cy=\"272\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"233\" y1=\"272\" x2=\"227\" y2=\"277\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"140\" cy=\"288\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"140\" y1=\"288\" x2=\"147\" y2=\"292\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"172\" cy=\"299\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"172\" y1=\"299\" x2=\"165\" y2=\"295\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"206\" cy=\"292\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"206\" y1=\"292\" x2=\"213\" y2=\"296\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<circle cx=\"234\" cy=\"233\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"234\" y1=\"233\" x2=\"228\" y2=\"229\" stroke=\"#C8932A\" stroke-width=\"1\"><\/line>\n<text x=\"185\" y=\"348\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#FAF6EE\">T\u2081 = 300 K<\/text>\n<text x=\"185\" y=\"370\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"600\" fill=\"#C8932A\">V\u2081 = 2.0 L<\/text>\n\n<line x1=\"290\" y1=\"215\" x2=\"400\" y2=\"215\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n<polygon points=\"400,206 420,215 400,224\" fill=\"#C8932A\"><\/polygon>\n<text x=\"352\" y=\"200\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">add heat<\/text>\n<text x=\"352\" y=\"240\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">V\/T is unchanged<\/text>\n<text x=\"352\" y=\"292\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">n and P unchanged<\/text>\n\n<line x1=\"450\" y1=\"70\" x2=\"450\" y2=\"315\" stroke=\"#D9CFB8\" stroke-width=\"3\"><\/line>\n<line x1=\"580\" y1=\"70\" x2=\"580\" y2=\"315\" stroke=\"#D9CFB8\" stroke-width=\"3\"><\/line>\n<rect x=\"442\" y=\"315\" width=\"146\" height=\"10\" fill=\"#D9CFB8\" rx=\"2\"><\/rect>\n<rect x=\"453\" y=\"132\" width=\"124\" height=\"183\" fill=\"#142139\"><\/rect>\n<rect x=\"450\" y=\"120\" width=\"130\" height=\"12\" fill=\"#C5D0DC\" rx=\"2\"><\/rect>\n<rect x=\"485\" y=\"80\" width=\"60\" height=\"40\" fill=\"#7A1F2B\" rx=\"3\"><\/rect>\n<text x=\"515\" y=\"106\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" font-weight=\"600\" fill=\"#FAF6EE\">P<\/text>\n<circle cx=\"472\" cy=\"152\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"472\" y1=\"152\" x2=\"486\" y2=\"145\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"540\" cy=\"168\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"540\" y1=\"168\" x2=\"526\" y2=\"176\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"505\" cy=\"205\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"505\" y1=\"205\" x2=\"519\" y2=\"198\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"563\" cy=\"228\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"563\" y1=\"228\" x2=\"550\" y2=\"236\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"466\" cy=\"252\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"466\" y1=\"252\" x2=\"480\" y2=\"259\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"530\" cy=\"278\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"530\" y1=\"278\" x2=\"516\" y2=\"270\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"488\" cy=\"300\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"488\" y1=\"300\" x2=\"502\" y2=\"293\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<circle cx=\"566\" cy=\"146\" r=\"3\" fill=\"#C8932A\"><\/circle><line x1=\"566\" y1=\"146\" x2=\"552\" y2=\"153\" stroke=\"#C8932A\" stroke-width=\"1.2\"><\/line>\n<text x=\"515\" y=\"348\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#FAF6EE\">T\u2082 = 600 K<\/text>\n<text x=\"515\" y=\"370\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"600\" fill=\"#C8932A\">V\u2082 = 4.0 L<\/text>\n<\/svg>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Eight molecules on the left, eight on the right \u2014 the amount of gas never changes. Doubling the kelvin temperature doubles the volume.<\/p>\n\n<p>Push the sliders below and watch the ratio hold. Set T\u2082 to twice T\u2081 and the volume should land on exactly twice V\u2081 \u2014 no more, no less.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Charles&#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\/charles-law.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>Why Charles&#8217;s Law Only Works in Kelvin<\/h2>\n\n<p>Here is the single most expensive mistake in this topic, and it costs marks in every exam season.<\/p>\n\n<p>Celsius has an arbitrary zero. It was pinned to the freezing point of water because water was convenient, not because anything special happens to molecular motion there. A gas at 0 \u00b0C has plenty of energy left in it.<\/p>\n\n<p>Kelvin has an <em>absolute<\/em> zero. According to <a href=\"https:\/\/www.nist.gov\/pml\/owm\/si-units-temperature\" target=\"_blank\" rel=\"noopener\">NIST&#8217;s definition of the kelvin<\/a>, 0 K is absolute zero and 0 \u00b0C sits at 273.15 K. Only from that origin does &#8220;twice the temperature&#8221; mean &#8220;twice the molecular kinetic energy&#8221; \u2014 and therefore twice the volume.<\/p>\n\n<p>Try it in Celsius and the absurdity is immediate. Going from 1 \u00b0C to 2 \u00b0C would double the volume. Going from 0 \u00b0C to 10 \u00b0C would multiply it by infinity. Going from \u221210 \u00b0C to \u221220 \u00b0C would somehow produce a positive volume from a negative one.<\/p>\n\n<p>Plot volume against absolute temperature and you get a straight line aimed squarely at the origin. That is the whole law in one picture.<\/p>\n\n<svg viewBox=\"0 0 700 450\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" role=\"img\" aria-label=\"Charles law graph: gas volume plotted against absolute temperature gives a straight line passing through zero kelvin\" style=\"display:block;width:100%;height:auto;max-width:700px;margin:32px auto;\">\n<rect width=\"700\" height=\"450\" fill=\"#0A1628\" rx=\"4\"><\/rect>\n<text x=\"350\" y=\"30\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"17\" font-weight=\"600\" fill=\"#FAF6EE\">Volume against absolute temperature<\/text>\n<text x=\"350\" y=\"52\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" fill=\"#C5D0DC\">Fixed mass of gas at constant pressure<\/text>\n\n<rect x=\"90\" y=\"70\" width=\"540\" height=\"280\" fill=\"#142139\"><\/rect>\n<rect x=\"90\" y=\"70\" width=\"81\" height=\"280\" fill=\"#7A1F2B\" fill-opacity=\"0.30\"><\/rect>\n<text x=\"130\" y=\"240\" transform=\"rotate(-90 130 240)\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#FAF6EE\">real gases liquefy here<\/text>\n\n<line x1=\"270\" y1=\"70\" x2=\"270\" y2=\"350\" stroke=\"#D9CFB8\" stroke-opacity=\"0.14\" stroke-width=\"1\"><\/line>\n<line x1=\"450\" y1=\"70\" x2=\"450\" y2=\"350\" stroke=\"#D9CFB8\" stroke-opacity=\"0.14\" stroke-width=\"1\"><\/line>\n\n<line x1=\"90\" y1=\"215\" x2=\"360\" y2=\"215\" stroke=\"#C8932A\" stroke-width=\"1\" stroke-dasharray=\"4 4\" stroke-opacity=\"0.55\"><\/line>\n<line x1=\"360\" y1=\"215\" x2=\"360\" y2=\"350\" stroke=\"#C8932A\" stroke-width=\"1\" stroke-dasharray=\"4 4\" stroke-opacity=\"0.55\"><\/line>\n<line x1=\"90\" y1=\"80\" x2=\"630\" y2=\"80\" stroke=\"#C8932A\" stroke-width=\"1\" stroke-dasharray=\"4 4\" stroke-opacity=\"0.55\"><\/line>\n<line x1=\"630\" y1=\"80\" x2=\"630\" y2=\"350\" stroke=\"#C8932A\" stroke-width=\"1\" stroke-dasharray=\"4 4\" stroke-opacity=\"0.55\"><\/line>\n\n<line x1=\"90\" y1=\"350\" x2=\"225\" y2=\"282.5\" stroke=\"#C8932A\" stroke-width=\"2.5\" stroke-dasharray=\"7 5\"><\/line>\n<line x1=\"225\" y1=\"282.5\" x2=\"630\" y2=\"80\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n\n<line x1=\"90\" y1=\"350\" x2=\"645\" y2=\"350\" stroke=\"#C5D0DC\" stroke-width=\"1.5\"><\/line>\n<line x1=\"90\" y1=\"62\" x2=\"90\" y2=\"350\" stroke=\"#C5D0DC\" stroke-width=\"1.5\"><\/line>\n\n<circle cx=\"90\" cy=\"350\" r=\"4\" fill=\"#FAF6EE\"><\/circle>\n<circle cx=\"360\" cy=\"215\" r=\"5\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/circle>\n<circle cx=\"630\" cy=\"80\" r=\"5\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/circle>\n\n<text x=\"290\" y=\"126\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">Double the temperature in kelvin,<\/text>\n<text x=\"290\" y=\"146\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"600\" fill=\"#C8932A\">and the volume doubles.<\/text>\n<text x=\"300\" y=\"333\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">dashed line = extrapolation, not measurement<\/text>\n\n<text x=\"80\" y=\"220\" text-anchor=\"end\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">V<\/text>\n<text x=\"80\" y=\"85\" text-anchor=\"end\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">2V<\/text>\n\n<line x1=\"270\" y1=\"350\" x2=\"270\" y2=\"357\" stroke=\"#C5D0DC\"><\/line>\n<line x1=\"450\" y1=\"350\" x2=\"450\" y2=\"357\" stroke=\"#C5D0DC\"><\/line>\n<line x1=\"630\" y1=\"350\" x2=\"630\" y2=\"357\" stroke=\"#C5D0DC\"><\/line>\n<text x=\"90\" y=\"372\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">0 K<\/text>\n<text x=\"270\" y=\"372\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">200 K<\/text>\n<text x=\"450\" y=\"372\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">400 K<\/text>\n<text x=\"630\" y=\"372\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">600 K<\/text>\n<text x=\"90\" y=\"391\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">\u2212273 \u00b0C<\/text>\n<text x=\"270\" y=\"391\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">\u221273 \u00b0C<\/text>\n<text x=\"450\" y=\"391\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">127 \u00b0C<\/text>\n<text x=\"630\" y=\"391\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">327 \u00b0C<\/text>\n\n<text x=\"360\" y=\"416\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">Absolute temperature T<\/text>\n<text x=\"350\" y=\"437\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11\" fill=\"#C5D0DC\">0 K is absolute zero, which is \u2212273.15 \u00b0C<\/text>\n<text x=\"26\" y=\"210\" transform=\"rotate(-90 26 210)\" text-anchor=\"middle\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">Volume V<\/text>\n<\/svg>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Extend the measured line backwards and it hits zero volume at 0 K. No gas ever gets there \u2014 it liquefies first \u2014 but the extrapolation is how absolute zero was first pinned down.<\/p>\n\n<h2>Real-World Examples of Charles&#8217;s Law<\/h2>\n\n<p><strong>1. Hot-air balloons.<\/strong> A burner heats the air inside the envelope from roughly 15 \u00b0C to about 100 \u00b0C. At constant pressure the air cannot expand sideways \u2014 the envelope is already full \u2014 so it expands out of the mouth, leaving fewer molecules inside. The trapped air is now about 23% less dense than the air outside, and the balloon rises.<\/p>\n\n<p><strong>2. A balloon dunked in liquid nitrogen.<\/strong> The classic lecture demo. Drop an inflated balloon into nitrogen at 77 K and it shrivels to a fraction of its size within seconds. Lift it out and it reinflates as the air warms back to room temperature. Nothing escaped; volume simply tracked T.<\/p>\n\n<p><strong>3. Rescuing a dented ping-pong ball.<\/strong> Roll it into hot water and the dent pops out. The air inside warms, tries to occupy more volume, and the thin celluloid gives way at the weakest point \u2014 the dent.<\/p>\n\n<p><strong>4. Bread and cakes rising in the oven.<\/strong> Yeast and baking powder make the bubbles; the oven makes them big. Every trapped pocket of gas expands as it heats, and the dough sets around the enlarged bubbles. This is why an oven door opened too early gives you a sunken cake.<\/p>\n\n<p><strong>5. Thermals and rising warm air.<\/strong> At constant pressure a gas&#8217;s density is inversely proportional to its absolute temperature \u2014 the same proportionality, written the other way up. Sun-warmed air over a ploughed field is less dense than the air above it, so it rises. Gliders and buzzards both make a living from this.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/IMG_4446.webp\"\n\n       alt=\"Hot-air balloon rising, a real-world example of Charles law\"\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;\">The burner does not add lift directly. It heats the air, the air expands, and the density drop does the rest.<\/figcaption>\n\n<\/figure>\n\n<h2>Charles&#8217;s Law vs the Other Gas Laws<\/h2>\n\n<p>Four simple gas laws exist, and each one freezes two variables to watch the other two dance. Students mix them up constantly. The fix is to ask one question first: <em>what is being held constant?<\/em><\/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;border:1px solid #D9CFB8;text-align:left;\">Gas law<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Relationship<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Held constant<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Two-state equation<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Graph shape<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Boyle&#8217;s law<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P \u221d 1\/V<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">T, n<\/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 (P against V)<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Charles&#8217;s law<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V \u221d T<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P, n<\/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 through the origin (V against T)<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Gay-Lussac&#8217;s law<\/strong> (pressure law)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P \u221d T<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V, n<\/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 through the origin (P against T)<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Avogadro&#8217;s law<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V \u221d n<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P, T<\/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 the origin (V against n)<\/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;\">PV\/T is constant<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">n only<\/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;\">Surface, not a curve<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Ideal gas law<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">PV = nRT<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">R = 8.314 J\/(mol\u00b7K)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">PV = nRT<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Surface, not a curve<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>One warning worth carrying into an exam. Some textbooks call the pressure\u2013temperature relationship &#8220;Gay-Lussac&#8217;s law&#8221;, others call it &#8220;Amontons&#8217;s law&#8221;, and a handful use &#8220;Gay-Lussac&#8217;s law&#8221; to mean Charles&#8217;s law itself. Read the equation, not the name.<\/p>\n\n<h2>Common Misconceptions About Charles&#8217;s Law<\/h2>\n\n<h3>Misconception 1: Doubling the Celsius temperature doubles the volume<\/h3>\n\n<p>It does not, and the error is enormous. Heating a gas from 20 \u00b0C to 40 \u00b0C looks like a doubling, but in kelvin it is 293.15 K to 313.15 K \u2014 a rise of just 6.8%.<\/p>\n\n<p>The volume grows by 6.8%, not by 100%. A student who forgets to convert overstates the answer by nearly fifteen-fold.<\/p>\n\n<h3>Misconception 2: Charles&#8217;s law applies to any sealed container<\/h3>\n\n<p>Only if the container can change volume. Heat a sealed steel gas cylinder and its volume barely budges; the pressure rockets instead. That is the pressure law, not Charles&#8217;s law.<\/p>\n\n<p>Ask yourself whether the walls can move. Balloon, piston, syringe, flexible bag \u2014 Charles&#8217;s law. Rigid tin, tyre, aerosol can \u2014 not Charles&#8217;s law.<\/p>\n\n<h3>Misconception 3: Gas volume really reaches zero at absolute zero<\/h3>\n\n<p>The straight line says so; nature does not oblige. Every real gas condenses into a liquid long before it gets near 0 K \u2014 nitrogen at 77 K, helium at 4.2 K.<\/p>\n\n<p>Below that point there is no gas left to obey the law. The extrapolation is a mathematical device, and a spectacularly useful one: it is how absolute zero was located in the first place.<\/p>\n\n<h3>Misconception 4: Heating a balloon works by raising the pressure inside<\/h3>\n\n<p>Barely. A rubber balloon holds its internal pressure at atmospheric plus a small excess from the stretched skin. Heat it and that pressure stays almost unchanged.<\/p>\n\n<p>What changes is the volume. The gas has an escape route \u2014 the balloon simply grows \u2014 so pressure never gets the chance to build. Take that escape route away and you have a different law entirely.<\/p>\n\n<h2>How Charles&#8217;s Law Relates to the Ideal Gas Law and Thermodynamics<\/h2>\n\n<p>Charles&#8217;s law is not a standalone rule. It is a special case, quietly hiding inside the ideal gas law.<\/p>\n\n<div class=\"pf-formula\">P\u00b7V = n\u00b7R\u00b7T<\/div>\n\n<p>Rearrange for V\/T and you get V\/T = nR\/P. Hold n and P fixed and the entire right-hand side is a constant \u2014 which is precisely the statement V\u2081\/T\u2081 = V\u2082\/T\u2082. Charles&#8217;s law falls out in one line.<\/p>\n\n<p>Boyle&#8217;s law and Avogadro&#8217;s law drop out the same way, each by freezing a different pair of variables. This is worth internalising, because it means you only ever need to remember one equation.<\/p>\n\n<p>Underneath all of it sits kinetic theory, where the absolute temperature of a gas is directly proportional to the average <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a> of its molecules. That single link explains why the kelvin scale is the only one that works: it counts energy from zero, not from the freezing point of a puddle.<\/p>\n\n<p>The distinction between energy transferred and temperature reached matters here too. If the difference between <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/heat-vs-temperature\/\">heat and temperature<\/a> feels blurry, that is worth clearing up before you tackle gas laws in earnest.<\/p>\n\n<p>How much heat you need to raise the temperature depends on the material&#8217;s <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/specific-heat-capacity\/\">specific heat capacity<\/a>. And the constant-pressure expansion described by Charles&#8217;s law is exactly the process in which a gas does work on its surroundings \u2014 a first-law idea explored in the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/laws-of-thermodynamics\/\">laws of thermodynamics<\/a>.<\/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 balloon holds 2.50 L of air at 27 \u00b0C. It is warmed to 87 \u00b0C at constant pressure. What is its new volume?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Charles&#8217;s law applies (constant pressure, sealed balloon). V\u2081\/T\u2081 = V\u2082\/T\u2082, so V\u2082 = V\u2081 \u00d7 T\u2082\/T\u2081.<\/p>\n<p>Step 2: Convert to kelvin. T\u2081 = 27 + 273.15 = 300.15 K. T\u2082 = 87 + 273.15 = 360.15 K.<\/p>\n<p>Step 3: V\u2082 = 2.50 L \u00d7 (360.15 K \/ 300.15 K) = 2.50 L \u00d7 1.1999 = 3.00 L.<\/p>\n<p><strong>Answer: V\u2082 = 3.00 L<\/strong><\/p>\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 1.20 m\u00b3 at 350 K. It is cooled to 280 K at constant pressure. Find the new volume.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: V\u2082 = V\u2081 \u00d7 T\u2082\/T\u2081. Both temperatures are already in kelvin.<\/p>\n<p>Step 2: V\u2082 = 1.20 m\u00b3 \u00d7 (280 K \/ 350 K).<\/p>\n<p>Step 3: V\u2082 = 1.20 m\u00b3 \u00d7 0.800 = 0.960 m\u00b3.<\/p>\n<p>Sanity check: the gas got colder, so the volume shrank. Correct direction.<\/p>\n<p><strong>Answer: V\u2082 = 0.960 m\u00b3<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A syringe holds 60.0 mL of air at 20.0 \u00b0C. The plunger is free to move. The air is warmed until the volume reads 66.0 mL. What is the final temperature in \u00b0C?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Solving for temperature, T\u2082 = T\u2081 \u00d7 V\u2082\/V\u2081.<\/p>\n<p>Step 2: T\u2081 = 20.0 + 273.15 = 293.15 K. V\u2082\/V\u2081 = 66.0 mL \/ 60.0 mL = 1.100.<\/p>\n<p>Step 3: T\u2082 = 293.15 K \u00d7 1.100 = 322.47 K.<\/p>\n<p>Step 4: Convert back. \u03b8\u2082 = 322.47 \u2212 273.15 = 49.3 \u00b0C.<\/p>\n<p><strong>Answer: T\u2082 = 322.5 K, or 49.3 \u00b0C<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A student claims that heating a gas from 25 \u00b0C to 50 \u00b0C at constant pressure will double its volume. By what percentage does the volume actually increase?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: The claim uses Celsius. Convert. T\u2081 = 298.15 K, T\u2082 = 323.15 K.<\/p>\n<p>Step 2: V\u2082\/V\u2081 = T\u2082\/T\u2081 = 323.15 \/ 298.15 = 1.0839.<\/p>\n<p>Step 3: Percentage increase = (1.0839 \u2212 1) \u00d7 100% = 8.39%.<\/p>\n<p>The volume grows by about 8.4%, not 100%. Doubling the Celsius reading is not doubling the temperature.<\/p>\n<p><strong>Answer: about 8.4%<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A gas is cooled at constant pressure from an unknown temperature down to 250 K. Its volume falls from 4.50 L to 3.75 L. What was the initial temperature?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Rearrange V\u2081\/T\u2081 = V\u2082\/T\u2082 for T\u2081, giving T\u2081 = T\u2082 \u00d7 V\u2081\/V\u2082.<\/p>\n<p>Step 2: T\u2081 = 250 K \u00d7 (4.50 L \/ 3.75 L).<\/p>\n<p>Step 3: T\u2081 = 250 K \u00d7 1.20 = 300 K.<\/p>\n<p>Sanity check: the gas was hotter to begin with, because it started bigger. 300 K &gt; 250 K. Correct.<\/p>\n<p><strong>Answer: T\u2081 = 300 K (26.85 \u00b0C)<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">Air has a density of 1.29 kg\/m\u00b3 at 0 \u00b0C and atmospheric pressure. What is its density at 100 \u00b0C, at the same pressure?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Density is mass per unit volume, \u03c1 = m\/V. With mass fixed and V \u221d T, density is inversely proportional to T: \u03c1\u2082 = \u03c1\u2081 \u00d7 T\u2081\/T\u2082.<\/p>\n<p>Step 2: T\u2081 = 273.15 K, T\u2082 = 373.15 K.<\/p>\n<p>Step 3: \u03c1\u2082 = 1.29 kg\/m\u00b3 \u00d7 (273.15 K \/ 373.15 K) = 1.29 \u00d7 0.7320 = 0.944 kg\/m\u00b3.<\/p>\n<p>In practice this is why a chimney draws: hot flue gas is roughly 27% lighter than the cold air around it.<\/p>\n<p><strong>Answer: \u03c1\u2082 = 0.944 kg\/m\u00b3<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A hot-air balloon envelope holds 2800 m\u00b3. The ambient air is at 15 \u00b0C with density 1.225 kg\/m\u00b3. The air inside is heated to 100 \u00b0C at atmospheric pressure. Find the lift, in kilograms and in newtons. Ignore the mass of the envelope and basket.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Find the density inside using \u03c1\u2082 = \u03c1\u2081 \u00d7 T\u2081\/T\u2082, with T\u2081 = 288.15 K and T\u2082 = 373.15 K.<\/p>\n<p>Step 2: \u03c1_inside = 1.225 kg\/m\u00b3 \u00d7 (288.15 \/ 373.15) = 1.225 \u00d7 0.7722 = 0.946 kg\/m\u00b3.<\/p>\n<p>Step 3: Density difference \u0394\u03c1 = 1.225 \u2212 0.946 = 0.279 kg\/m\u00b3.<\/p>\n<p>Step 4: Lifted mass = \u0394\u03c1 \u00d7 V = 0.279 kg\/m\u00b3 \u00d7 2800 m\u00b3 = 781 kg.<\/p>\n<p>Step 5: Lift force = 781 kg \u00d7 9.81 m\/s\u00b2 = 7.66 \u00d7 10\u00b3 N.<\/p>\n<p>Sanity check: a real envelope, basket, burner and fuel weigh several hundred kilograms, which is why balloons this size carry only a handful of passengers.<\/p>\n<p><strong>Answer: 781 kg of lift, or 7.66 kN<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">In a constant-pressure gas thermometer, a fixed mass of gas occupies 25.00 mL at 0.00 \u00b0C and 34.15 mL at 100.0 \u00b0C. Extrapolate the volume\u2013temperature line to zero volume to estimate absolute zero, in \u00b0C.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Volume varies linearly with Celsius temperature: V = V\u2080 + m\u03b8, where V\u2080 = 25.00 mL is the volume at 0.00 \u00b0C.<\/p>\n<p>Step 2: Find the gradient. m = (34.15 \u2212 25.00) mL \/ (100.0 \u2212 0.00) \u00b0C = 9.15 \/ 100.0 = 0.0915 mL\/\u00b0C.<\/p>\n<p>Step 3: Set V = 0 and solve for \u03b8. 0 = 25.00 + 0.0915\u03b8, so \u03b8 = \u221225.00 \/ 0.0915.<\/p>\n<p>Step 4: \u03b8 = \u2212273.2 \u00b0C.<\/p>\n<p>This is how absolute zero was first estimated \u2014 not by reaching it, but by drawing a line towards it. The accepted value is \u2212273.15 \u00b0C.<\/p>\n<p><strong>Answer: absolute zero \u2248 \u2212273.2 \u00b0C<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is Charles&#039;s law in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\nCharles&#8217;s law says that a fixed amount of gas at constant pressure expands in direct proportion to its absolute temperature. Heat it and it grows; cool it and it shrinks. Double the temperature in kelvin and you double the volume. The ratio V\/T stays the same throughout, which is why the law is written V\u2081\/T\u2081 = V\u2082\/T\u2082.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for Charles&#039;s law?<\/summary><div class=\"pf-faq-item-answer\">\nThe formula is V\u2081\/T\u2081 = V\u2082\/T\u2082, where V\u2081 and V\u2082 are the initial and final volumes and T\u2081 and T\u2082 are the initial and final absolute temperatures in kelvin. Rearranged to find a new volume it becomes V\u2082 = V\u2081 \u00d7 T\u2082\/T\u2081. The two volumes may use any unit as long as it is the same unit for both.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why does Charles&#039;s law use Kelvin and not Celsius?<\/summary><div class=\"pf-faq-item-answer\">\nKelvin measures temperature from absolute zero, so a kelvin value is directly proportional to the average kinetic energy of the gas molecules. Celsius starts at the freezing point of water, an arbitrary point where molecules still carry plenty of energy. Using Celsius produces nonsense: 0 \u00b0C would imply zero volume, and negative temperatures would imply negative volumes.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is Charles&#039;s law the same as Gay-Lussac&#039;s law?<\/summary><div class=\"pf-faq-item-answer\">\nNo, though the naming is genuinely confusing. Charles&#8217;s law relates volume to temperature at constant pressure. Gay-Lussac&#8217;s law, as most textbooks use the term, relates pressure to temperature at constant volume. Gay-Lussac published Charles&#8217;s volume law in 1802 and credited Charles for it, which is why his name is attached to both.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does Charles&#039;s law work for real gases?<\/summary><div class=\"pf-faq-item-answer\">\nReal gases follow Charles&#8217;s law closely at low pressure and well above their boiling points, which covers most everyday conditions. Deviations grow near condensation, where molecular attractions and molecular volume stop being negligible. Air at room temperature and atmospheric pressure obeys the law to within a fraction of a per cent.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What happens to Charles&#039;s law at absolute zero?<\/summary><div class=\"pf-faq-item-answer\">\nThe extrapolated line predicts zero volume at 0 K, but no real gas ever reaches it. Every gas liquefies first \u2014 nitrogen at 77 K, helium at 4.2 K \u2014 and liquids do not obey gas laws. The prediction is still valuable: extending the volume\u2013temperature line backwards is how absolute zero was originally located.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How does a hot-air balloon use Charles&#039;s law?<\/summary><div class=\"pf-faq-item-answer\">\nThe burner heats the air inside the envelope, and because the envelope is open at the bottom the pressure stays atmospheric. The hot air expands, some spills out, and what remains is less dense than the surrounding air. That density difference generates buoyant lift. Heating air from 15 \u00b0C to 100 \u00b0C drops its density by roughly 23%.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Leave a balloon in a hot car and it strains at the knot, taut and glossy. Bring it out into a cold night and within minutes it sags, wrinkled, looking half-empty. Nothing leaked. The same air sits inside, pressed by the same atmosphere \u2014 only the temperature moved, and the volume obediently followed. That is &#8230; <a title=\"Charles&#8217;s Law Explained\" class=\"read-more\" href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/charles-law\/\" aria-label=\"Read more about Charles&#8217;s Law Explained\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":465,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[290,288,283,282,289,28],"class_list":["post-464","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics","tag-absolute-temperature","tag-charles-law","tag-gas-laws","tag-ideal-gas-law","tag-kelvin","tag-thermodynamics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/464","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=464"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/464\/revisions"}],"predecessor-version":[{"id":469,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/464\/revisions\/469"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/465"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}