{"id":19003,"date":"2025-06-13T11:54:42","date_gmt":"2025-06-13T11:54:42","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=19003"},"modified":"2025-06-13T11:54:45","modified_gmt":"2025-06-13T11:54:45","slug":"calculate-the-root-mean-square-speed-of-ne-atoms-at-the-temperature-at-which-their-average-kinetic-energy-is-6-38-kj-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-root-mean-square-speed-of-ne-atoms-at-the-temperature-at-which-their-average-kinetic-energy-is-6-38-kj-mol\/","title":{"rendered":"Calculate the root-mean-square speed of Ne atoms at the temperature at which their average kinetic energy is 6.38 kJ\/mol"},"content":{"rendered":"\n<p>Calculate the root-mean-square speed of Ne atoms at the temperature at which their average kinetic energy is 6.38 kJ\/mol.<\/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>Let&#8217;s break down the problem 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\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Average kinetic energy per mole of Ne atoms, E=6.38\u00a0kJ\/mol=6380\u00a0J\/molE = 6.38 \\text{ kJ\/mol} = 6380 \\text{ J\/mol}<\/li>\n\n\n\n<li>We want to find the root-mean-square (rms) speed of neon atoms at this energy.<\/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\">Step 1: Relationship between kinetic energy and temperature<\/h3>\n\n\n\n<p>The average kinetic energy per <strong>molecule<\/strong> is given by \u27e8E\u27e9=32kBT\\langle E \\rangle = \\frac{3}{2} k_B T<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>kB=1.38\u00d710\u221223\u2009J\/Kk_B = 1.38 \\times 10^{-23} \\, \\text{J\/K} is Boltzmann&#8217;s constant,<\/li>\n\n\n\n<li>TT is temperature in kelvin.<\/li>\n<\/ul>\n\n\n\n<p>For 1 mole of atoms, multiply by Avogadro&#8217;s number NA=6.022\u00d71023N_A = 6.022 \\times 10^{23}: Emol=32NAkBT=32RTE_{\\text{mol}} = \\frac{3}{2} N_A k_B T = \\frac{3}{2} R T<\/p>\n\n\n\n<p>where R=NAkB=8.314\u2009J\/mol\\cdotpKR = N_A k_B = 8.314 \\, \\text{J\/mol\u00b7K} is the ideal gas constant.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the temperature<\/h3>\n\n\n\n<p>Rearranging for TT: T=23EmolRT = \\frac{2}{3} \\frac{E_{\\text{mol}}}{R}<\/p>\n\n\n\n<p>Plugging in values: T=23\u00d763808.314=23\u00d7767.0=511.3&nbsp;KT = \\frac{2}{3} \\times \\frac{6380}{8.314} = \\frac{2}{3} \\times 767.0 = 511.3 \\text{ K}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the root-mean-square speed<\/h3>\n\n\n\n<p>The root-mean-square speed vrmsv_{\\text{rms}} for a gas molecule is: vrms=3RTMv_{\\text{rms}} = \\sqrt{\\frac{3RT}{M}}<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R=8.314\u2009J\/mol\\cdotpKR = 8.314 \\, \\text{J\/mol\u00b7K},<\/li>\n\n\n\n<li>T=511.3\u2009KT = 511.3 \\, \\text{K},<\/li>\n\n\n\n<li>MM is molar mass in kg\/mol.<\/li>\n<\/ul>\n\n\n\n<p>Neon atomic mass is about 20.18 g\/mol = 0.02018 kg\/mol.<\/p>\n\n\n\n<p>Substitute: vrms=3\u00d78.314\u00d7511.30.02018v_{\\text{rms}} = \\sqrt{\\frac{3 \\times 8.314 \\times 511.3}{0.02018}}<\/p>\n\n\n\n<p>Calculate numerator: 3\u00d78.314\u00d7511.3=12747.73 \\times 8.314 \\times 511.3 = 12747.7<\/p>\n\n\n\n<p>Then vrms=12747.70.02018=631785=795.7&nbsp;m\/sv_{\\text{rms}} = \\sqrt{\\frac{12747.7}{0.02018}} = \\sqrt{631785} = 795.7 \\text{ m\/s}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>vrms\u2248796\u2009m\/s\\boxed{v_{\\text{rms}} \\approx 796 \\, \\text{m\/s}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (about 300 words):<\/h3>\n\n\n\n<p>The root-mean-square (rms) speed of gas atoms or molecules is a measure of the average speed that particles in a gas have due to their thermal energy. It is derived from the kinetic theory of gases, which relates the macroscopic properties of gases, such as temperature and pressure, to the microscopic motions of their particles.<\/p>\n\n\n\n<p>Each atom or molecule in a gas has kinetic energy related to the temperature. For an ideal monatomic gas like neon, the average kinetic energy per molecule is given by 32kBT\\frac{3}{2}k_B T, where kBk_B is Boltzmann\u2019s constant and TT is the absolute temperature in kelvin.<\/p>\n\n\n\n<p>In the problem, instead of being given the temperature, we are given the average kinetic energy per mole of neon atoms: 6.38 kJ\/mol. Since 1 mole contains Avogadro\u2019s number of atoms, the total kinetic energy of 1 mole is related to temperature by the formula E=32RTE = \\frac{3}{2}RT, where RR is the gas constant. Rearranging this lets us find the temperature corresponding to that energy.<\/p>\n\n\n\n<p>Once we have the temperature, the rms speed can be calculated. The rms speed formula comes from equating the kinetic energy expression 12mv2\\frac{1}{2} m v^2 (where mm is the mass of one atom) to the thermal energy and rearranging. Expressed in terms of molar quantities, it becomes vrms=3RTMv_{\\text{rms}} = \\sqrt{\\frac{3RT}{M}}, where MM is the molar mass.<\/p>\n\n\n\n<p>Plugging in the values for neon\u2019s molar mass and the calculated temperature, we find the rms speed of neon atoms is approximately 796 m\/s. This speed reflects the typical velocity of neon atoms due to their thermal energy at the temperature that corresponds to an average kinetic energy of 6.38 kJ\/mol.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the root-mean-square speed of Ne atoms at the temperature at which their average kinetic energy is 6.38 kJ\/mol. The correct answer and explanation is: Let&#8217;s break down the problem step-by-step: Given: Step 1: Relationship between kinetic energy and temperature The average kinetic energy per molecule is given by \u27e8E\u27e9=32kBT\\langle E \\rangle = \\frac{3}{2} k_B [&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-19003","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19003","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=19003"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19003\/revisions"}],"predecessor-version":[{"id":19004,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19003\/revisions\/19004"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=19003"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=19003"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=19003"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}