{"id":21579,"date":"2025-06-15T13:21:54","date_gmt":"2025-06-15T13:21:54","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=21579"},"modified":"2025-06-15T13:21:56","modified_gmt":"2025-06-15T13:21:56","slug":"unheated-heated-the-unheated-gas-in-the-above-system-has-a-volume-of-20-0l-at-a-temperature-of-25-0c-and-a-pressure-of-1-00-atm","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/unheated-heated-the-unheated-gas-in-the-above-system-has-a-volume-of-20-0l-at-a-temperature-of-25-0c-and-a-pressure-of-1-00-atm\/","title":{"rendered":"unheated heated The unheated gas in the above system has a volume of 20.0L at a temperature of 25.0\u00b0C and a pressure of 1.00 atm"},"content":{"rendered":"\n<p>unheated heated The unheated gas in the above system has a volume of 20.0L at a temperature of 25.0\u00b0C and a pressure of 1.00 atm. The gas is heated to a temperature of 100.0 \u00b0C while the volume remains constant. What is the pressure of the heated gas? Ignore the block labeled &#8220;A&#8221;.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this problem, we can use <strong>Gay-Lussac&#8217;s Law<\/strong>, which relates the pressure and temperature of a gas at <strong>constant volume<\/strong>: P1T1=P2T2\\frac{P_1}{T_1} = \\frac{P_2}{T_2}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P1P_1 = initial pressure = 1.00 atm<\/li>\n\n\n\n<li>T1T_1 = initial temperature = 25.0\u00b0C = 298.15 K<\/li>\n\n\n\n<li>P2P_2 = final pressure = ?<\/li>\n\n\n\n<li>T2T_2 = final temperature = 100.0\u00b0C = 373.15 K<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step Calculation:<\/h3>\n\n\n\n<p>Convert temperatures to Kelvin (K): T1=25.0+273.15=298.15\u2009KT_1 = 25.0 + 273.15 = 298.15\\,K T2=100.0+273.15=373.15\u2009KT_2 = 100.0 + 273.15 = 373.15\\,K<\/p>\n\n\n\n<p>Now apply Gay-Lussac\u2019s Law: P1T1=P2T2\u21d2P2=P1\u22c5T2T1\\frac{P_1}{T_1} = \\frac{P_2}{T_2} \\Rightarrow P_2 = P_1 \\cdot \\frac{T_2}{T_1} P2=1.00\u2009atm\u22c5373.15298.15\u22481.25\u2009atmP_2 = 1.00\\,\\text{atm} \\cdot \\frac{373.15}{298.15} \\approx 1.25\\,\\text{atm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>The pressure of the heated gas is <strong>approximately 1.25 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>Explanation (300 words):<\/strong><\/h3>\n\n\n\n<p>This problem involves a gas in a rigid container (constant volume) that is heated from 25.0\u00b0C to 100.0\u00b0C. Since the volume doesn\u2019t change, we apply <strong>Gay-Lussac\u2019s Law<\/strong>, which states that the pressure of a gas is directly proportional to its absolute temperature when volume is constant.<\/p>\n\n\n\n<p>Temperature in gas laws must always be in <strong>Kelvin<\/strong>, not Celsius. That\u2019s because the Kelvin scale starts at absolute zero, where particles theoretically stop moving. In this case, we convert 25.0\u00b0C and 100.0\u00b0C to 298.15 K and 373.15 K, respectively.<\/p>\n\n\n\n<p>Using the law P1\/T1=P2\/T2P_1\/T_1 = P_2\/T_2, we find that as the gas is heated, its pressure increases because the kinetic energy of the gas particles increases. Faster-moving particles collide with the walls of the container more forcefully and more frequently, leading to increased pressure.<\/p>\n\n\n\n<p>This principle is important in real-world applications like pressure cookers, car engines, and aerosol cans, where gases heat up and pressure builds. It also highlights the need for careful pressure regulation in enclosed gas systems to avoid explosions or mechanical failure.<\/p>\n\n\n\n<p>In summary, when temperature rises and volume remains fixed, the pressure must increase in proportion. That\u2019s why the heated gas ends up at a higher pressure of <strong>1.25 atm<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>unheated heated The unheated gas in the above system has a volume of 20.0L at a temperature of 25.0\u00b0C and a pressure of 1.00 atm. The gas is heated to a temperature of 100.0 \u00b0C while the volume remains constant. What is the pressure of the heated gas? Ignore the block labeled &#8220;A&#8221;. The correct [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-21579","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/21579","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=21579"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/21579\/revisions"}],"predecessor-version":[{"id":21580,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/21579\/revisions\/21580"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=21579"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=21579"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=21579"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}