{"id":233423,"date":"2025-06-13T06:15:30","date_gmt":"2025-06-13T06:15:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233423"},"modified":"2025-06-13T06:15:33","modified_gmt":"2025-06-13T06:15:33","slug":"what-is-the-volume-of-one-mole-of-gas-at-standard-temperature-and-pressure-stp-that-is-1-00-atm-and-0c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/what-is-the-volume-of-one-mole-of-gas-at-standard-temperature-and-pressure-stp-that-is-1-00-atm-and-0c\/","title":{"rendered":"What is the volume of one mole of gas at &#8220;Standard Temperature and Pressure&#8221; (STP), that is, 1.00 atm and 0\u00b0C"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">What is the volume of one mole of gas at \"Standard Temperature and Pressure\" (STP), that is, 1.00 atm and 0\u00b0C? Number of mole = 1(n) Temp. T = 0\u00b0C = 273.15 K Pressure, P = 1 bar (atm)&nbsp;PV=nRT&nbsp;R = 0.08314 bar 2) What is the volume of one mole of gas at room temperature (20\u00b0C)? 3) You inflate your car tires to 2.00 atm pressure at 20\u00b0C. After driving on the highway, the temperature of your tires goes up to 50\u00b0C. What is the pressure in the tires? 4) What does 1.00 L of nitrogen (N2) weigh at 20\u00b0C? Hint: you calculated the volume of one mole of gas at 20\u00b0C for question 2. One liter is what fraction of this volume? 5) An internal combustion engine burns fuel inside a cylinder. The heat expands the gases in the cylinder and pushes against the piston. If the temperature in the cylinder increases from 300\u00b0C to 600\u00b0C, how much does the pressure increase? What if the final temperature is 800\u00b0C?<\/pre>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Ideal Gas Calculations and Explanation<\/strong><\/h3>\n\n\n\n<p>To solve these questions, we use the <strong>Ideal Gas Law<\/strong>: PV=nRTPV = nRTPV=nRT<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>PPP is pressure<\/li>\n\n\n\n<li>VVV is volume<\/li>\n\n\n\n<li>nnn is number of moles<\/li>\n\n\n\n<li>RRR is the ideal gas constant<\/li>\n\n\n\n<li>TTT is temperature in Kelvin<\/li>\n<\/ul>\n\n\n\n<p>We&#8217;ll use the gas constant:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R=0.08314\u00a0L\\cdotpbar\/mol\\cdotpKR = 0.08314 \\ \\text{L\u00b7bar\/mol\u00b7K}R=0.08314\u00a0L\\cdotpbar\/mol\\cdotpK<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Volume of 1 mole of gas at STP (1.00 atm, 0\u00b0C)<\/strong><\/h3>\n\n\n\n<p>At standard temperature and pressure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P=1.00\u00a0barP = 1.00 \\ \\text{bar}P=1.00\u00a0bar<\/li>\n\n\n\n<li>T=0\u00b0C=273.15\u00a0KT = 0\u00b0C = 273.15 \\ \\text{K}T=0\u00b0C=273.15\u00a0K<\/li>\n\n\n\n<li>n=1.00\u00a0moln = 1.00 \\ \\text{mol}n=1.00\u00a0mol<\/li>\n<\/ul>\n\n\n\n<p>Using the Ideal Gas Law: V=nRTP=(1.00)(0.08314)(273.15)1.00=22.71&nbsp;LV = \\frac{nRT}{P} = \\frac{(1.00)(0.08314)(273.15)}{1.00} = 22.71 \\ \\text{L}V=PnRT\u200b=1.00(1.00)(0.08314)(273.15)\u200b=22.71&nbsp;L<\/p>\n\n\n\n<p><strong>Answer:<\/strong> Volume = <strong>22.71 L<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Volume of 1 mole of gas at 20\u00b0C<\/strong><\/h3>\n\n\n\n<p>At:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>T=20\u00b0C=293.15\u00a0KT = 20\u00b0C = 293.15 \\ \\text{K}T=20\u00b0C=293.15\u00a0K<\/li>\n\n\n\n<li>P=1.00\u00a0barP = 1.00 \\ \\text{bar}P=1.00\u00a0bar<\/li>\n<\/ul>\n\n\n\n<p>V=(1.00)(0.08314)(293.15)1.00=24.38&nbsp;LV = \\frac{(1.00)(0.08314)(293.15)}{1.00} = 24.38 \\ \\text{L}V=1.00(1.00)(0.08314)(293.15)\u200b=24.38&nbsp;L<\/p>\n\n\n\n<p><strong>Answer:<\/strong> Volume = <strong>24.38 L<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>3. Tire pressure after heating from 20\u00b0C to 50\u00b0C<\/strong><\/h3>\n\n\n\n<p>Use <strong>Gay-Lussac\u2019s Law<\/strong> (constant volume): P1T1=P2T2\\frac{P_1}{T_1} = \\frac{P_2}{T_2}T1\u200bP1\u200b\u200b=T2\u200bP2\u200b\u200b<\/p>\n\n\n\n<p>Convert to Kelvin:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>T1=20\u00b0C=293.15\u00a0KT_1 = 20\u00b0C = 293.15 \\ \\text{K}T1\u200b=20\u00b0C=293.15\u00a0K<\/li>\n\n\n\n<li>T2=50\u00b0C=323.15\u00a0KT_2 = 50\u00b0C = 323.15 \\ \\text{K}T2\u200b=50\u00b0C=323.15\u00a0K<\/li>\n\n\n\n<li>P1=2.00\u00a0atmP_1 = 2.00 \\ \\text{atm}P1\u200b=2.00\u00a0atm<\/li>\n<\/ul>\n\n\n\n<p>P2=P1\u22c5T2T1=2.00\u22c5323.15293.15=2.21&nbsp;atmP_2 = P_1 \\cdot \\frac{T_2}{T_1} = 2.00 \\cdot \\frac{323.15}{293.15} = 2.21 \\ \\text{atm}P2\u200b=P1\u200b\u22c5T1\u200bT2\u200b\u200b=2.00\u22c5293.15323.15\u200b=2.21&nbsp;atm<\/p>\n\n\n\n<p><strong>Answer:<\/strong> Final pressure = <strong>2.21 atm<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>4. Mass of 1.00 L of nitrogen gas (N\u2082) at 20\u00b0C<\/strong><\/h3>\n\n\n\n<p>From Q2, 1 mole of gas at 20\u00b0C occupies <strong>24.38 L<\/strong><\/p>\n\n\n\n<p>So, 1.00 L is: 1.0024.38=0.041&nbsp;mol\\frac{1.00}{24.38} = 0.041 \\ \\text{mol}24.381.00\u200b=0.041&nbsp;mol<\/p>\n\n\n\n<p>Molar mass of nitrogen (N\u2082) = 28.02 g\/mol<\/p>\n\n\n\n<p>Mass: 0.041\u22c528.02=1.15&nbsp;g0.041 \\cdot 28.02 = 1.15 \\ \\text{g}0.041\u22c528.02=1.15&nbsp;g<\/p>\n\n\n\n<p><strong>Answer:<\/strong> <strong>1.15 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>5. Pressure increase in engine cylinder: 300\u00b0C to 600\u00b0C and 800\u00b0C<\/strong><\/h3>\n\n\n\n<p>Convert to Kelvin:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial: T1=300+273.15=573.15\u00a0KT_1 = 300 + 273.15 = 573.15 \\ \\text{K}T1\u200b=300+273.15=573.15\u00a0K<\/li>\n\n\n\n<li>Case 1 Final: T2=600+273.15=873.15\u00a0KT_2 = 600 + 273.15 = 873.15 \\ \\text{K}T2\u200b=600+273.15=873.15\u00a0K<\/li>\n\n\n\n<li>Case 2 Final: T3=800+273.15=1073.15\u00a0KT_3 = 800 + 273.15 = 1073.15 \\ \\text{K}T3\u200b=800+273.15=1073.15\u00a0K<\/li>\n<\/ul>\n\n\n\n<p>Again using: P2P1=T2T1\\frac{P_2}{P_1} = \\frac{T_2}{T_1}P1\u200bP2\u200b\u200b=T1\u200bT2\u200b\u200b<\/p>\n\n\n\n<p><strong>Case 1:<\/strong> P2P1=873.15573.15=1.52\\frac{P_2}{P_1} = \\frac{873.15}{573.15} = 1.52P1\u200bP2\u200b\u200b=573.15873.15\u200b=1.52<\/p>\n\n\n\n<p><strong>Case 2:<\/strong> P3P1=1073.15573.15=1.87\\frac{P_3}{P_1} = \\frac{1073.15}{573.15} = 1.87P1\u200bP3\u200b\u200b=573.151073.15\u200b=1.87<\/p>\n\n\n\n<p><strong>Answers:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pressure increases by <strong>1.52\u00d7<\/strong> when temperature rises to 600\u00b0C<\/li>\n\n\n\n<li>Pressure increases by <strong>1.87\u00d7<\/strong> when temperature rises to 800\u00b0C<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary<\/h3>\n\n\n\n<p>This set of questions uses the Ideal Gas Law and gas behavior laws (Boyle\u2019s, Charles\u2019s, Gay-Lussac\u2019s) to relate pressure, volume, and temperature. Gas volume expands with temperature, and pressure increases if gas is heated in a fixed volume. These principles are critical in both everyday and industrial applications like tire inflation and internal combustion engines.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-235.jpeg\" alt=\"\" class=\"wp-image-233424\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the volume of one mole of gas at &#8220;Standard Temperature and Pressure&#8221; (STP), that is, 1.00 atm and 0\u00b0C? Number of mole = 1(n) Temp. T = 0\u00b0C = 273.15 K Pressure, P = 1 bar (atm)&nbsp;PV=nRT&nbsp;R = 0.08314 bar 2) What is the volume of one mole of gas at room temperature [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233423","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233423","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233423"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233423\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233423"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233423"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233423"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}