{"id":46210,"date":"2025-07-01T21:07:44","date_gmt":"2025-07-01T21:07:44","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=46210"},"modified":"2025-07-01T21:07:45","modified_gmt":"2025-07-01T21:07:45","slug":"the-specific-gravity-of-mercury-is-13-6","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-specific-gravity-of-mercury-is-13-6\/","title":{"rendered":"The specific gravity of mercury is 13.6"},"content":{"rendered":"\n

The specific gravity of mercury is 13.6, When the barometric pressure is 720 mm of Hg, what is the atmospheric pressure in kPa? 106 kPa 100 kPa 96.1 kPa<\/p>\n\n\n\n

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

To find the atmospheric pressure in kPa, we first need to understand the relationship between the given quantities: specific gravity of mercury, barometric pressure, and the required atmospheric pressure.<\/p>\n\n\n\n

Step 1: Understanding Specific Gravity<\/h3>\n\n\n\n

Specific gravity is the ratio of the density of a substance to the density of water. For mercury (Hg), the specific gravity is 13.6, which means mercury is 13.6 times denser than water.<\/p>\n\n\n\n

Step 2: Calculating the Height of the Mercury Column<\/h3>\n\n\n\n

The barometric pressure is given as 720 mm of Hg. This height is the column of mercury in a barometer and is used to measure atmospheric pressure. The pressure exerted by a liquid column is given by the formula:P=\u03c1ghP = \\rho g hP=\u03c1gh<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • PPP is the pressure (in Pascals or Pa),<\/li>\n\n\n\n
  • \u03c1\\rho\u03c1 is the density of the fluid (mercury in this case),<\/li>\n\n\n\n
  • ggg is the acceleration due to gravity,<\/li>\n\n\n\n
  • hhh is the height of the fluid column (720 mm of Hg).<\/li>\n<\/ul>\n\n\n\n

    The density of mercury (\u03c1Hg\\rho_{\\text{Hg}}\u03c1Hg\u200b) can be calculated using the specific gravity. The density of water is approximately 1000\u2009kg\/m31000 \\, \\text{kg\/m}^31000kg\/m3, so:\u03c1Hg=13.6\u00d71000\u2009kg\/m3=13600\u2009kg\/m3\\rho_{\\text{Hg}} = 13.6 \\times 1000 \\, \\text{kg\/m}^3 = 13600 \\, \\text{kg\/m}^3\u03c1Hg\u200b=13.6\u00d71000kg\/m3=13600kg\/m3<\/p>\n\n\n\n

    Now, we convert the height of the mercury column to meters:h=720\u2009mm=0.72\u2009mh = 720 \\, \\text{mm} = 0.72 \\, \\text{m}h=720mm=0.72m<\/p>\n\n\n\n

    Step 3: Finding the Pressure in Pascals<\/h3>\n\n\n\n

    Using the formula P=\u03c1ghP = \\rho g hP=\u03c1gh, where g=9.81\u2009m\/s2g = 9.81 \\, \\text{m\/s}^2g=9.81m\/s2:P=13600\u2009kg\/m3\u00d79.81\u2009m\/s2\u00d70.72\u2009mP = 13600 \\, \\text{kg\/m}^3 \\times 9.81 \\, \\text{m\/s}^2 \\times 0.72 \\, \\text{m}P=13600kg\/m3\u00d79.81m\/s2\u00d70.72mP=96,019.2\u2009Pa\u224896.02\u2009kPaP = 96,019.2 \\, \\text{Pa} \\approx 96.02 \\, \\text{kPa}P=96,019.2Pa\u224896.02kPa<\/p>\n\n\n\n

    Step 4: Conclusion<\/h3>\n\n\n\n

    The atmospheric pressure, based on a barometric reading of 720 mm of Hg, is approximately 96.1 kPa.<\/p>\n\n\n\n

    So, the correct answer is 96.1 kPa<\/strong>.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The specific gravity of mercury is 13.6, When the barometric pressure is 720 mm of Hg, what is the atmospheric pressure in kPa? 106 kPa 100 kPa 96.1 kPa The Correct Answer and Explanation is: To find the atmospheric pressure in kPa, we first need to understand the relationship between the given quantities: specific gravity […]<\/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-46210","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46210","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=46210"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46210\/revisions"}],"predecessor-version":[{"id":46212,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46210\/revisions\/46212"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46210"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46210"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46210"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}