{"id":204672,"date":"2025-03-22T12:32:31","date_gmt":"2025-03-22T12:32:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=204672"},"modified":"2025-03-22T12:32:33","modified_gmt":"2025-03-22T12:32:33","slug":"according-to-charles-law-the-pressure-p-in-pascals-of-a-fixed-volume-of-a-gas-is-linearly-related-to-the-temperature-t-in-degrees-celsius","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/22\/according-to-charles-law-the-pressure-p-in-pascals-of-a-fixed-volume-of-a-gas-is-linearly-related-to-the-temperature-t-in-degrees-celsius\/","title":{"rendered":"According to Charles&#8217; Law, the pressure P in pascals of a fixed volume of a gas is linearly related to the temperature T in degrees Celsius"},"content":{"rendered":"\n<p>According to Charles&#8217; Law, the pressure P in pascals of a fixed volume of a gas is linearly related to the temperature T in degrees Celsius. In an experiment, it was found that when T = 40, P = 90 and when T = 80, P = 100.<\/p>\n\n\n\n<p>a. What is the slope of the line containing these points?<\/p>\n\n\n\n<p>b. Explain the meaning of the slope in this context.<\/p>\n\n\n\n<p>c. Graph the function<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s work through this problem step-by-step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (a): Finding the slope of the line<\/h3>\n\n\n\n<p>The formula for the slope (m) of a line between two points ((T_1, P_1)) and ((T_2, P_2)) is given by:<\/p>\n\n\n\n<p>[<br>m = \\frac{P_2 &#8211; P_1}{T_2 &#8211; T_1}<br>]<\/p>\n\n\n\n<p>From the given data, we know:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>When (T = 40), (P = 90) (point 1: (T_1 = 40), (P_1 = 90))<\/li>\n\n\n\n<li>When (T = 80), (P = 100) (point 2: (T_2 = 80), (P_2 = 100))<\/li>\n<\/ul>\n\n\n\n<p>Now, substitute these values into the slope formula:<\/p>\n\n\n\n<p>[<br>m = \\frac{100 &#8211; 90}{80 &#8211; 40} = \\frac{10}{40} = 0.25<br>]<\/p>\n\n\n\n<p>So, the <strong>slope<\/strong> of the line is <strong>0.25<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (b): Meaning of the slope<\/h3>\n\n\n\n<p>The slope (0.25) in this context represents the rate of change of pressure (P) with respect to temperature (T). In other words, for every increase of 1 degree Celsius in temperature, the pressure increases by 0.25 pascals. This slope is a direct indication of how sensitive the pressure is to changes in temperature for the fixed volume of gas, according to Charles&#8217; Law.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (c): Graphing the function<\/h3>\n\n\n\n<p>The linear relationship between pressure and temperature can be written as:<\/p>\n\n\n\n<p>[<br>P = mT + b<br>]<\/p>\n\n\n\n<p>where (m = 0.25) is the slope, and (b) is the y-intercept. To find the value of (b), we can use one of the points. Let&#8217;s use the point ((T_1 = 40, P_1 = 90)) and substitute into the equation:<\/p>\n\n\n\n<p>[<br>90 = 0.25(40) + b<br>]<\/p>\n\n\n\n<p>Simplifying:<\/p>\n\n\n\n<p>[<br>90 = 10 + b<br>]<\/p>\n\n\n\n<p>[<br>b = 90 &#8211; 10 = 80<br>]<\/p>\n\n\n\n<p>Thus, the equation of the line is:<\/p>\n\n\n\n<p>[<br>P = 0.25T + 80<br>]<\/p>\n\n\n\n<p>This is the equation that relates pressure and temperature for this gas.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Graph of the function<\/h3>\n\n\n\n<p>To graph this equation, we will plot the points ((40, 90)) and ((80, 100)) on a coordinate plane and draw a straight line through them. The line will have a slope of (0.25) and will intersect the pressure axis (P-axis) at (P = 80).<\/p>\n\n\n\n<p>Let&#8217;s summarize the key points:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Slope: (0.25) means for each 1\u00b0C increase in temperature, pressure increases by 0.25 pascals.<\/li>\n\n\n\n<li>Equation: The relationship between pressure and temperature is linear and can be described by (P = 0.25T + 80).<\/li>\n\n\n\n<li>Graph: The line will pass through the points ((40, 90)) and ((80, 100)), with a slope of 0.25 and y-intercept at (P = 80).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>According to Charles&#8217; Law, the pressure P in pascals of a fixed volume of a gas is linearly related to the temperature T in degrees Celsius. In an experiment, it was found that when T = 40, P = 90 and when T = 80, P = 100. a. What is the slope of the [&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-204672","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204672","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=204672"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204672\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204672"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204672"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204672"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}