{"id":217648,"date":"2025-05-22T09:19:13","date_gmt":"2025-05-22T09:19:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217648"},"modified":"2025-05-22T09:19:15","modified_gmt":"2025-05-22T09:19:15","slug":"the-specific-heat-capacity-of-carbon-dioxide-which-has-a-molecular-weight-of-44-u-is-849-j-kg-1-k-1-at-constant-pressure","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/the-specific-heat-capacity-of-carbon-dioxide-which-has-a-molecular-weight-of-44-u-is-849-j-kg-1-k-1-at-constant-pressure\/","title":{"rendered":"The specific heat capacity of carbon dioxide, which has a molecular weight of 44 u, is 849 J kg-1 K-1 at constant pressure"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">(a) The specific heat capacity of carbon dioxide, which has a molecular weight of 44 u, is 849 J kg-1 K-1 at constant pressure. How many degrees of freedom does the carbon dioxide molecule have at room temperature? (b).Calculate the volume occupied by one mole of an ideal gas at a pressure of 92 atm and a temperature of 457 ?C (conditions appropriate to the surface of Venus). Hence estimate the density of the Venusian atmosphere at its surface, assuming a purely carbon dioxide composition. By which factor does this exceed the density of the Earth\u2019s atmosphere at sea level? (c) The NovaLynx 400-8210 pilot balloon has a mass of 10 g and a standard inflated diameter of 46 cm. This balloon is not designed to carry a payload (they\u2019re intended to be tracked from the ground to investigate wind conditions). If the balloon is filled with helium and launched on a day when the pressure is 101 kPa and the temperature is 15 ?C, what is its acceleration as it lifts off?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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<p class=\"wp-block-paragraph\">Let\u2019s solve each part step by step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(a) Degrees of Freedom of CO\u2082 at Room Temperature<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Specific heat at constant pressure, $c_p = 849 \\, \\text{J\/kg\u00b7K}$<\/li>\n\n\n\n<li>Molecular mass of CO\u2082, $M = 44 \\, \\text{g\/mol} = 0.044 \\, \\text{kg\/mol}$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">We know from kinetic theory:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>C_p = \\left(1 + \\frac{f}{2}\\right) R<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$f$ is the number of degrees of freedom,<\/li>\n\n\n\n<li>$R = 8.314 \\, \\text{J\/mol\u00b7K}$ is the gas constant.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">We convert the specific heat per kg to per mole:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>C_p^{\\text{mol}} = c_p \\cdot M = 849 \\cdot 0.044 = 37.356 \\, \\text{J\/mol\u00b7K}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now solve for $f$:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>37.356 = \\left(1 + \\frac{f}{2}\\right) \\cdot 8.314<br>\\Rightarrow \\frac{f}{2} = \\frac{37.356}{8.314} &#8211; 1 = 3.49<br>\\Rightarrow f \\approx 7<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer (a):<\/strong> <strong>7 degrees of freedom<\/strong> \u2014 3 translational + 2 rotational + 2 vibrational (at room temperature, CO\u2082 exhibits vibrational modes due to being a linear molecule).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(b) Volume and Density on Venus<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$P = 92 \\, \\text{atm} = 92 \\times 1.013 \\times 10^5 = 9.3196 \\times 10^6 \\, \\text{Pa}$<\/li>\n\n\n\n<li>$T = 457^\\circ C = 730 \\, \\text{K}$<\/li>\n\n\n\n<li>$n = 1 \\, \\text{mol}$<\/li>\n\n\n\n<li>$R = 8.314 \\, \\text{J\/mol\u00b7K}$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Using Ideal Gas Law:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>PV = nRT \\Rightarrow V = \\frac{nRT}{P} = \\frac{1 \\cdot 8.314 \\cdot 730}{9.3196 \\times 10^6} = 6.51 \\times 10^{-4} \\, \\text{m}^3<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate density $\\rho$:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\rho = \\frac{\\text{mass}}{\\text{volume}} = \\frac{0.044 \\, \\text{kg}}{6.51 \\times 10^{-4}} \\approx 67.6 \\, \\text{kg\/m}^3<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Earth\u2019s atmospheric density at sea level \u2248 1.225 kg\/m\u00b3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Density factor} = \\frac{67.6}{1.225} \\approx 55.2<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer (b):<\/strong> Volume \u2248 <strong>0.000651 m\u00b3<\/strong>, Density \u2248 <strong>67.6 kg\/m\u00b3<\/strong>, about <strong>55 times denser<\/strong> than Earth\u2019s atmosphere.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(c) Acceleration of the Balloon at Lift-off<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of balloon: $m = 0.01 \\, \\text{kg}$<\/li>\n\n\n\n<li>Diameter: $d = 0.46 \\, \\text{m} \\Rightarrow r = 0.23 \\, \\text{m}$<\/li>\n\n\n\n<li>Volume: $V = \\frac{4}{3} \\pi r^3 = \\frac{4}{3} \\pi (0.23)^3 \\approx 0.051 \\, \\text{m}^3$<\/li>\n\n\n\n<li>Helium molar mass \u2248 4 g\/mol = 0.004 kg\/mol<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">At $T = 15^\\circ C = 288 \\, \\text{K}$, $P = 101 \\times 10^3 \\, \\text{Pa}$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using ideal gas law:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>n = \\frac{PV}{RT} = \\frac{101000 \\cdot 0.051}{8.314 \\cdot 288} \\approx 2.15 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Mass of helium: $m_{\\text{He}} = 2.15 \\cdot 0.004 = 0.0086 \\, \\text{kg}$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Total mass = 0.01 + 0.0086 = <strong>0.0186 kg<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Buoyant force: $F_b = \\rho_{\\text{air}} \\cdot V \\cdot g = 1.225 \\cdot 0.051 \\cdot 9.8 \\approx 0.612 \\, \\text{N}$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Weight: $W = m_{\\text{total}} \\cdot g = 0.0186 \\cdot 9.8 \\approx 0.182 \\, \\text{N}$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Net force: $F_{\\text{net}} = 0.612 &#8211; 0.182 = 0.43 \\, \\text{N}$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Acceleration:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>a = \\frac{F_{\\text{net}}}{m_{\\text{total}}} = \\frac{0.43}{0.0186} \\approx 23.1 \\, \\text{m\/s}^2<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answer (c):<\/strong> <strong>Acceleration \u2248 23.1 m\/s\u00b2<\/strong><\/p>\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<ul class=\"wp-block-list\">\n<li>(a) CO\u2082 has <strong>7 degrees of freedom<\/strong> at room temp.<\/li>\n\n\n\n<li>(b) One mole occupies <strong>0.000651 m\u00b3<\/strong>, Venus\u2019s atmosphere is <strong>55\u00d7 denser<\/strong> than Earth\u2019s.<\/li>\n\n\n\n<li>(c) Balloon\u2019s lift-off acceleration is <strong>\u2248 23.1 m\/s\u00b2<\/strong><\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-30.png\" alt=\"\" class=\"wp-image-217649\"\/><\/figure>\n\n\n\n<ul class=\"wp-block-list\"><\/ul>\n","protected":false},"excerpt":{"rendered":"<p>(a) The specific heat capacity of carbon dioxide, which has a molecular weight of 44 u, is 849 J kg-1 K-1 at constant pressure. How many degrees of freedom does the carbon dioxide molecule have at room temperature? (b).Calculate the volume occupied by one mole of an ideal gas at a pressure of 92 atm [&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-217648","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217648","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=217648"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217648\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217648"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217648"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217648"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}