{"id":231375,"date":"2025-06-11T03:55:28","date_gmt":"2025-06-11T03:55:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231375"},"modified":"2025-06-11T03:55:30","modified_gmt":"2025-06-11T03:55:30","slug":"calculate-the-height-of-a-column-of-methanol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/calculate-the-height-of-a-column-of-methanol\/","title":{"rendered":"Calculate the height of a column of methanol"},"content":{"rendered":"\n

Calculate the height of a column of methanol
that would be supported by atmospheric pressure. The density of methanol is 0.787 g\/cm
. To receive full credit for you answer, you must show your calculations (include the units in all of your conversions), and report the correct number of significant figures. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Calculation of Methanol Column Height<\/strong><\/h3>\n\n\n\n

Final Answer:<\/strong>
The height of a column of methanol supported by standard atmospheric pressure is 13.1 m<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Explanation and Calculations<\/strong><\/h3>\n\n\n\n

The pressure exerted by the atmosphere at sea level can support a column of a liquid in an inverted tube, a device known as a barometer. The height of this liquid column is inversely proportional to the liquid’s density. The standard for measuring atmospheric pressure is the mercury barometer, where 1 atmosphere (atm) of pressure supports a column of mercury (Hg) exactly 760 mm high.<\/p>\n\n\n\n

The pressure exerted by a fluid column is calculated using the formula:<\/p>\n\n\n\n

P = h \u00d7 d \u00d7 g<\/p>\n\n\n\n

where:<\/p>\n\n\n\n

    \n
  • P is the pressure<\/li>\n\n\n\n
  • h is the height of the fluid column<\/li>\n\n\n\n
  • d is the density of the fluid<\/li>\n\n\n\n
  • g is the acceleration due to gravity<\/li>\n<\/ul>\n\n\n\n

    At equilibrium, the pressure exerted by the methanol column must equal the pressure exerted by the standard mercury column (i.e., atmospheric pressure).<\/p>\n\n\n\n

    Pmethanol=PmercuryP<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b=P<\/em>m<\/em>erc<\/em>u<\/em>ry<\/em>\u200b<\/code><\/pre>\n\n\n\n
    Therefore, we can set their pressure equations equal to each other:<\/p>\n\n\n\n
    hmethanol\u00d7dmethanol\u00d7g=hHg\u00d7dHg\u00d7gh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b\u00d7d<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b\u00d7g<\/em>=h<\/em>H<\/em>g<\/em>\u200b\u00d7d<\/em>H<\/em>g<\/em>\u200b\u00d7g<\/em><\/code><\/pre>\n\n\n\n
    Since the acceleration due to gravity (g) is the same for both columns, it cancels out, simplifying the relationship to:<\/p>\n\n\n\n
    hmethanol\u00d7dmethanol=hHg\u00d7dHgh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b\u00d7d<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b=h<\/em>H<\/em>g<\/em>\u200b\u00d7d<\/em>H<\/em>g<\/em>\u200b<\/code><\/pre>\n\n\n\n
    We can now solve for the unknown height of the methanol column (<\/p>\n\n\n\n
    hmethanolh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b<\/code><\/pre>\n\n\n\n
    ).<\/p>\n\n\n\n
    Known Values:<\/strong><\/p>\n\n\n\n
      \n
    • Height of mercury (hHgh<\/em>H<\/em>g<\/em>\u200b<\/code>): 760 mm<\/li>\n\n\n\n
    • Density of mercury (dHgd<\/em>H<\/em>g<\/em>\u200b<\/code>): 13.6 g\/cm\u00b3<\/li>\n\n\n\n
    • Density of methanol (dmethanold<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b<\/code>): 0.787 g\/cm\u00b3<\/li>\n<\/ul>\n\n\n\n
      Calculation:<\/strong><\/p>\n\n\n\n
        \n
      1. Rearrange the formula to solve for\u00a0hmethanolh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b<\/code>:
        hmethanol=hHg\u00d7dHgdmethanolh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b=d<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200bh<\/em>H<\/em>g<\/em>\u200b\u00d7d<\/em>H<\/em>g<\/em>\u200b\u200b<\/code><\/li>\n\n\n\n
      2. Substitute the known values into the equation:
        hmethanol=(760\u00a0mm)\u00d7(13.6\u00a0g\/cm3)0.787\u00a0g\/cm3h<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b=0.787\u00a0g\/cm3(760\u00a0mm)\u00d7(13.6\u00a0g\/cm3)\u200b<\/code><\/li>\n\n\n\n
      3. Calculate the result. The units of g\/cm\u00b3 cancel, leaving the height in mm:
        hmethanol=10336\u00a0mm0.787h<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b=0.78710336\u00a0mm\u200b<\/code>
        hmethanol=13133.418...\u00a0mmh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b=13133.418...\u00a0mm<\/code><\/li>\n\n\n\n
      4. Determine the correct number of significant figures. The densities (13.6 and 0.787) both have three significant figures, which is the least number of significant figures in the calculation. Therefore, the answer must be rounded to three significant figures.
        hmethanol\u224813100\u00a0mmh<\/em>m<\/em>e<\/em>t<\/em>han<\/em>o<\/em>l<\/em>\u200b\u224813100\u00a0mm<\/code><\/li>\n\n\n\n
      5. For practicality, convert the height from millimeters (mm) to meters (m), as this is a more appropriate unit for such a large height.
        13100\u00a0mm\u00d71\u00a0m1000\u00a0mm=13.1\u00a0m13100\u00a0mm\u00d71000\u00a0mm1\u00a0m\u200b=13.1\u00a0m<\/strong><\/code><\/li>\n<\/ol>\n\n\n\n
        The final result, reported to three significant figures, is 13.1 meters. This demonstrates that because methanol is far less dense than mercury, a much taller column is required to exert the same pressure as the atmosphere.<\/p>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
        Calculate the height of a column of methanol that would be supported by atmospheric pressure. The density of methanol is 0.787 g\/cm. To receive full credit for you answer, you must show your calculations (include the units in all of your conversions), and report the correct number of significant figures. For the toolbar, press ALT+F10 […]<\/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-231375","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231375","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=231375"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231375\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231375"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231375"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231375"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}