{"id":187941,"date":"2025-02-06T08:28:09","date_gmt":"2025-02-06T08:28:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187941"},"modified":"2025-02-06T08:28:11","modified_gmt":"2025-02-06T08:28:11","slug":"methyl-alcohol-has-a-specific-gravity-of-0-789-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/methyl-alcohol-has-a-specific-gravity-of-0-789-2\/","title":{"rendered":"Methyl alcohol has a specific gravity of 0.789"},"content":{"rendered":"\n<ol start=\"5\" class=\"wp-block-list\">\n<li>Methyl alcohol has a specific gravity of 0.789. Calculate its density and its specific weight.<\/li>\n\n\n\n<li>A cylindrical container is 150 mm in diameter and weighs 2.25 N when empty. When filled to a depth of 200 mm with a certain oil, it weighs 35.4. Calculate the specific gravity of the oil.<\/li>\n\n\n\n<li>A storage vessel for gasoline (sg = 0.68) is a vertical cylinder 10 m in diameter. If it is filled to a depth of 6.75 m, calculate the weight and mass of the gasoline.<\/li>\n\n\n\n<li>Compute the pressure change required to cause to decrease in the volume of ethyl alcohol by 1%.<\/li>\n<\/ol>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s address each question in turn:<\/p>\n\n\n\n<p><strong>5. Methyl alcohol has a specific gravity of 0.789. Calculate its density and its specific weight.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Density<\/strong>: pecific gravity is the ratio of the substance&#8217;s density to the density of water. Since the density of water is approximately 1000 kg\/m\u00b3, the density of methyl alcohol is:<br>\ue200calculator\ue202turn0calculator0\ue201<\/li>\n\n\n\n<li><strong>Specific Weight<\/strong>: pecific weight is the weight per unit volume, calculated as density multiplied by the acceleration due to gravity (9.81 m\/s\u00b2):<br>\ue200calculator\ue202turn0calculator1\ue201<\/li>\n<\/ul>\n\n\n\n<p><strong>6. A cylindrical container is 150 mm in diameter and weighs 2.25 N when empty. When filled to a depth of 200 mm with a certain oil, it weighs 35.4 N. Calculate the specific gravity of the oil.<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the weight of the oil<\/strong>: ubtract the empty weight of the container from the filled weight:<br>\ue200calculator\ue202turn0calculator2\ue201<\/li>\n\n\n\n<li><strong>Calculate the volume of the oil<\/strong>: he volume ( V ) of a cylinder is given by ( V = \\pi r^2 h ), where ( r ) is the radius and ( h ) is the height. Converting dimensions to meters:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Radius ( r = 150\\, \\text{mm} \/ 2 = 75\\, \\text{mm} = 0.075\\, \\text{m} )<\/li>\n\n\n\n<li>Height ( h = 200\\, \\text{mm} = 0.2\\, \\text{m} ) hus, the volume is:<br>[ V = \\pi \\times (0.075\\, \\text{m})^2 \\times 0.2\\, \\text{m} \\approx 0.003534\\, \\text{m}^3 ]<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the density of the oil<\/strong>: ensity ( \\rho ) is mass per unit volume. First, find the mass by dividing the weight by gravity:<br>[ \\text{Mass} = \\frac{33.15\\, \\text{N}}{9.81\\, \\text{m\/s}^2} \\approx 3.38\\, \\text{kg} ]<br>hen, density:<br>[ \\rho = \\frac{3.38\\, \\text{kg}}{0.003534\\, \\text{m}^3} \\approx 956\\, \\text{kg\/m}^3 ]<\/li>\n\n\n\n<li><strong>Calculate the specific gravity<\/strong>: pecific gravity is the ratio of the oil&#8217;s density to the density of water:<br>[ \\text{Specific Gravity} = \\frac{956\\, \\text{kg\/m}^3}{1000\\, \\text{kg\/m}^3} = 0.956 ]<br><strong>7. A storage vessel for gasoline (sg = 0.68) is a vertical cylinder 10 m in diameter. If it is filled to a depth of 6.75 m, calculate the weight and mass of the gasoline.<\/strong><\/li>\n\n\n\n<li><strong>Calculate the volume of gasoline<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Radius ( r = 10\\, \\text{m} \/ 2 = 5\\, \\text{m} )<\/li>\n\n\n\n<li>Height ( h = 6.75\\, \\text{m} ) [ V = \\pi \\times (5\\, \\text{m})^2 \\times 6.75\\, \\text{m} \\approx 530.14\\, \\text{m}^3 ]<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the density of gasoline<\/strong>: sing the specific gravity:<br>[ \\rho = 0.68 \\times 1000\\, \\text{kg\/m}^3 = 680\\, \\text{kg\/m}^3 ]<\/li>\n\n\n\n<li><strong>Calculate the mass of gasoline<\/strong>: [ \\text{Mass} = \\rho \\times V = 680\\, \\text{kg\/m}^3 \\times 530.14\\, \\text{m}^3 \\approx 360,495\\, \\text{kg} ]<\/li>\n\n\n\n<li><strong>Calculate the weight of gasoline<\/strong>: [ \\text{Weight} = \\text{Mass} \\times g = 360,495\\, \\text{kg} \\times 9.81\\, \\text{m\/s}^2 \\approx 3,537,454\\, \\text{N} ]<br><strong>8. Compute the pressure change required to cause a decrease in the volume of ethyl alcohol by 1%.<\/strong><\/li>\n<\/ol>\n\n\n\n<p>he bulk modulus ( K ) is defined as:<br>[ K = -V \\frac{dP}{dV} ]<br>earranging to solve for the pressure change ( dP ):<br>[ dP = -K \\frac{dV}{V} ]<br>iven:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( \\frac{dV}{V} = -0.01 ) (a 1% decrease in volume)<\/li>\n\n\n\n<li>( K ) for ethyl alcohol is approximately ( 1.06 \\times 10^9\\, \\text{Pa} ) \ue200cite\ue202turn0search0\ue201<br>alculating ( dP ):<br>[ dP = 1.06 \\times 10^9\\, \\text{Pa} \\times 0.01 = 1.06 \\times 10^7\\, \\text{Pa} ]<br>onverting to kilopascals:<br>[ dP = 1.06 \\times 10^7\\, \\text{Pa} = 10,600\\, \\text{kPa} ]<br>herefore, a pressure increase of approximately 10,600 kPa is required to achieve a 1% decrease in the volume of ethyl alcohol.<br><strong>Explanation:<\/strong><\/li>\n<\/ul>\n\n\n\n<p>he bulk modulus of a substance quantifies its resistance to uniform compression; it is defined<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is : Let&#8217;s address each question in turn: 5. Methyl alcohol has a specific gravity of 0.789. Calculate its density and its specific weight. 6. A cylindrical container is 150 mm in diameter and weighs 2.25 N when empty. When filled to a depth of 200 mm with a certain [&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-187941","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187941","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=187941"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187941\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187941"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187941"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187941"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}