{"id":203234,"date":"2025-03-19T03:55:44","date_gmt":"2025-03-19T03:55:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=203234"},"modified":"2025-03-19T03:55:46","modified_gmt":"2025-03-19T03:55:46","slug":"the-rms-speed-of-nitrogen-molecules-in-the-air-at-s-t-p-is-about-490-m-s","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/19\/the-rms-speed-of-nitrogen-molecules-in-the-air-at-s-t-p-is-about-490-m-s\/","title":{"rendered":"The rms speed of nitrogen molecules in the air at S.T.P. is about 490 m\/s"},"content":{"rendered":"\n

The rms speed of nitrogen molecules in the air at S.T.P. is about 490 m\/s. Find their mean free path and the average time between collisions. The radius of a nitrogen molecule can be taken to be 2.0 \u00d7 10-10 m.<\/p>\n\n\n\n

What is the mean free path of a gas molecule (radius 2.5 \u00d7 10-10 m) in an ideal gas at 500 \u00b0C when the pressure is 7.0 \u00d7 10-6 mmHg?<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Solution to the First Problem: Mean Free Path and Time Between Collisions of Nitrogen Molecules at S.T.P.<\/strong><\/h3>\n\n\n\n

Step 1: Mean Free Path Calculation<\/strong><\/h4>\n\n\n\n

The mean free path ( \\lambda ) of gas molecules is given by the equation:
[
\\lambda = \\frac{k_B T}{\\sqrt{2} \\pi d^2 P}
]
where:<\/p>\n\n\n\n

    \n
  • ( k_B ) is Boltzmann\u2019s constant ((1.38 \\times 10^{-23} \\, \\text{J\/K})),<\/li>\n\n\n\n
  • ( T ) is the temperature in Kelvin ((273 \\, K) at S.T.P.),<\/li>\n\n\n\n
  • ( d ) is the diameter of a nitrogen molecule ((2 \\times 2.0 \\times 10^{-10} ) m = (4.0 \\times 10^{-10}) m),<\/li>\n\n\n\n
  • ( P ) is the pressure at S.T.P. ((1.013 \\times 10^5) Pa).<\/li>\n<\/ul>\n\n\n\n

    Plugging in values:
    [
    \\lambda = \\frac{(1.38 \\times 10^{-23}) (273)}{\\sqrt{2} \\pi (4.0 \\times 10^{-10})^2 (1.013 \\times 10^5)}
    ]
    [
    \\lambda \\approx 6.7 \\times 10^{-8} \\text{ m} \\text{ (or 67 nm)}
    ]<\/p>\n\n\n\n

    Step 2: Average Time Between Collisions<\/strong><\/h4>\n\n\n\n

    The average time between collisions (( \\tau )) is given by:
    [
    \\tau = \\frac{\\lambda}{v_{\\text{rms}}}
    ]
    Given that ( v_{\\text{rms}} = 490 ) m\/s:
    [
    \\tau = \\frac{6.7 \\times 10^{-8}}{490}
    ]
    [
    \\tau \\approx 1.37 \\times 10^{-10} \\text{ s}
    ]<\/p>\n\n\n\n

    Solution to the Second Problem: Mean Free Path at 500\u00b0C and Low Pressure<\/strong><\/h3>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • Temperature ( T = 500 + 273 = 773 ) K<\/li>\n\n\n\n
    • Pressure ( P = 7.0 \\times 10^{-6} ) mmHg<\/li>\n\n\n\n
    • Converting to Pascals: ( 1 ) mmHg ( = 133.322 ) Pa<\/li>\n\n\n\n
    • So, ( P = (7.0 \\times 10^{-6}) (133.322) = 9.33 \\times 10^{-4} ) Pa<\/li>\n\n\n\n
    • Molecular radius ( r = 2.5 \\times 10^{-10} ) m<\/li>\n\n\n\n
    • Molecular diameter ( d = 2r = 5.0 \\times 10^{-10} ) m<\/li>\n<\/ul>\n\n\n\n

      Using the same formula for ( \\lambda ):<\/p>\n\n\n\n

      [
      \\lambda = \\frac{(1.38 \\times 10^{-23}) (773)}{\\sqrt{2} \\pi (5.0 \\times 10^{-10})^2 (9.33 \\times 10^{-4})}
      ]<\/p>\n\n\n\n

      [
      \\lambda \\approx 2.46 \\text{ m}
      ]<\/p>\n\n\n\n

      Conclusion:<\/strong><\/h3>\n\n\n\n

      At S.T.P., nitrogen molecules have a very small mean free path ((67 \\, nm)) due to frequent collisions, while at low pressure ((7.0 \\times 10^{-6}) mmHg) and high temperature (500\u00b0C), the mean free path is much larger (about 2.46 m), as gas molecules are farther apart.<\/p>\n","protected":false},"excerpt":{"rendered":"

      The rms speed of nitrogen molecules in the air at S.T.P. is about 490 m\/s. Find their mean free path and the average time between collisions. The radius of a nitrogen molecule can be taken to be 2.0 \u00d7 10-10 m. What is the mean free path of a gas molecule (radius 2.5 \u00d7 10-10 […]<\/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-203234","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203234","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=203234"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203234\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203234"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203234"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203234"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}