{"id":226558,"date":"2025-06-05T11:11:12","date_gmt":"2025-06-05T11:11:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226558"},"modified":"2025-06-05T11:11:14","modified_gmt":"2025-06-05T11:11:14","slug":"determine-the-specific-gravity-n2-gas-at-75-f-and-765-torr-compared-to-air-at-a-standard-temperature-and-pressure-b-75-f-and-765-torr","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/determine-the-specific-gravity-n2-gas-at-75-f-and-765-torr-compared-to-air-at-a-standard-temperature-and-pressure-b-75-f-and-765-torr\/","title":{"rendered":"Determine the specific gravity N2 gas at 75 \u00b0F and 765 torr compared to air at a) standard temperature and pressure b) 75 \u00b0F and 765 torr"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Determine the specific gravity N2 gas at 75 \u00b0F and 765 torr compared to air at a) standard temperature and pressure b) 75 \u00b0F and 765 torr<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the <strong>specific gravity (SG)<\/strong> of nitrogen gas (N\u2082) compared to air, we use the formula: Specific&nbsp;Gravity&nbsp;(SG)=Density&nbsp;of&nbsp;N2Density&nbsp;of&nbsp;air\\text{Specific Gravity (SG)} = \\frac{\\text{Density of N}_2}{\\text{Density of air}}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Specific gravity is <strong>dimensionless<\/strong> and typically calculated at the <strong>same conditions<\/strong> (temperature and pressure) for both substances.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Molar Masses<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of <strong>N\u2082<\/strong> = 28.02 g\/mol<\/li>\n\n\n\n<li>Molar mass of <strong>air<\/strong> \u2248 28.97 g\/mol (approximate mixture of N\u2082, O\u2082, Ar, CO\u2082, etc.)<\/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\">a) SG of N\u2082 at 75 \u00b0F and 765 torr <strong>compared to air at STP (0 \u00b0C, 1 atm)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This is a comparison at <strong>different conditions<\/strong>, so we must calculate the actual gas densities using the <strong>ideal gas law<\/strong>: \u03c1=PMRT\\rho = \\frac{PM}{RT}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c1\\rho = density (g\/L)<\/li>\n\n\n\n<li>PP = pressure (atm)<\/li>\n\n\n\n<li>MM = molar mass (g\/mol)<\/li>\n\n\n\n<li>RR = ideal gas constant = 0.08206 L\u00b7atm\/mol\u00b7K<\/li>\n\n\n\n<li>TT = temperature in Kelvin<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Convert units<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>75 \u00b0F = (75 \u2013 32) \u00d7 5\/9 + 273.15 = 297.04 K<\/li>\n\n\n\n<li>765 torr = 765 \/ 760 = 1.00658 atm<\/li>\n\n\n\n<li>STP: T = 273.15 K, P = 1 atm<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Density of N\u2082 at 75 \u00b0F and 765 torr<\/strong>: \u03c1N2=(1.00658)(28.02)(0.08206)(297.04)=1.148\u2009g\/L\\rho_{N_2} = \\frac{(1.00658)(28.02)}{(0.08206)(297.04)} = 1.148 \\, \\text{g\/L}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Density of air at STP<\/strong>: \u03c1air=(1.0)(28.97)(0.08206)(273.15)=1.293\u2009g\/L\\rho_{\\text{air}} = \\frac{(1.0)(28.97)}{(0.08206)(273.15)} = 1.293 \\, \\text{g\/L} SGa=1.1481.293\u22480.89\\text{SG}_{a} = \\frac{1.148}{1.293} \\approx \\boxed{0.89}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">b) SG of N\u2082 at 75 \u00b0F and 765 torr <strong>compared to air at 75 \u00b0F and 765 torr<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Since both gases are at the <strong>same conditions<\/strong>, we can simply use molar masses: SGb=28.0228.97\u22480.97\\text{SG}_{b} = \\frac{28.02}{28.97} \\approx \\boxed{0.97}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Specific gravity (SG) compares the density of a substance to that of a reference\u2014in this case, nitrogen gas (N\u2082) compared to air. It&#8217;s important to ensure both substances are under the same conditions when determining SG, as gas density changes with temperature and pressure.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In part (a), we compare N\u2082 at 75 \u00b0F and 765 torr to air at standard temperature and pressure (STP: 0 \u00b0C and 1 atm). Since the gases are under different conditions, we must calculate their actual densities using the ideal gas law. The density formula \u03c1=PMRT\\rho = \\frac{PM}{RT} tells us how pressure, temperature, and molar mass influence density. Calculations show N\u2082 has a density of about 1.148 g\/L at 75 \u00b0F and 765 torr, while air has a density of 1.293 g\/L at STP. The specific gravity is the ratio of these densities: SG \u2248 0.89. This indicates that under these conditions, N\u2082 is lighter than air at STP.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In part (b), both gases are at the same conditions (75 \u00b0F and 765 torr), allowing us to simply compare their molar masses. Nitrogen\u2019s molar mass is 28.02 g\/mol, while air&#8217;s is approximately 28.97 g\/mol. The SG is then 28.02 \/ 28.97 \u2248 0.97. This means that at identical conditions, nitrogen is slightly less dense than air, but very close in value\u2014just 3% lighter.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In summary, specific gravity varies with comparison conditions. When comparing gases at different states, their physical densities must be calculated. But when at the same state, SG simplifies to a ratio of molar masses.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-454.jpeg\" alt=\"\" class=\"wp-image-226559\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the specific gravity N2 gas at 75 \u00b0F and 765 torr compared to air at a) standard temperature and pressure b) 75 \u00b0F and 765 torr The Correct Answer and Explanation is: To determine the specific gravity (SG) of nitrogen gas (N\u2082) compared to air, we use the formula: Specific&nbsp;Gravity&nbsp;(SG)=Density&nbsp;of&nbsp;N2Density&nbsp;of&nbsp;air\\text{Specific Gravity (SG)} = \\frac{\\text{Density [&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-226558","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226558","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=226558"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226558\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226558"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226558"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226558"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}