{"id":200134,"date":"2025-03-13T08:13:29","date_gmt":"2025-03-13T08:13:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=200134"},"modified":"2025-03-13T08:13:31","modified_gmt":"2025-03-13T08:13:31","slug":"a-2-01-l-flask-containing-2-31-g-of-helium-gas-molar-mass-4-003-g-mol-is-separated-by-a-stopcock-from-a-3-16-l-flask-containing-3-08-g-of-fluorine-gas-molar-mass-38-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/13\/a-2-01-l-flask-containing-2-31-g-of-helium-gas-molar-mass-4-003-g-mol-is-separated-by-a-stopcock-from-a-3-16-l-flask-containing-3-08-g-of-fluorine-gas-molar-mass-38-g-mol\/","title":{"rendered":"A 2.01 L flask containing 2.31 g of helium gas (molar mass = 4.003 g\/mol) is separated by a stopcock from a 3.16 L flask containing 3.08 g of fluorine gas (molar mass = 38 g\/mol)"},"content":{"rendered":"\n<p>A 2.01 L flask containing 2.31 g of helium gas (molar mass = 4.003 g\/mol) is separated by a stopcock from a 3.16 L flask containing 3.08 g of fluorine gas (molar mass = 38 g\/mol).<br>After the stopcock is opened the total pressure is 625 mm Hg.<br>What was the pressure (reported to three significant figures) of the helium before the stopcock was opened?<\/p>\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>To solve for the pressure of helium before the stopcock is opened, we can use the ideal gas law and the relationship between pressure, volume, and moles of gas.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the moles of each gas.<\/h3>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of helium = 2.31 g<\/li>\n\n\n\n<li>Molar mass of helium = 4.003 g\/mol<\/li>\n\n\n\n<li>Mass of fluorine = 3.08 g<\/li>\n\n\n\n<li>Molar mass of fluorine = 38.0 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>To find the moles of each gas, we use the formula:<\/p>\n\n\n\n<p>[<br>\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}<br>]<\/p>\n\n\n\n<p><strong>Moles of helium:<\/strong><\/p>\n\n\n\n<p>[<br>n_{\\text{He}} = \\frac{2.31 \\, \\text{g}}{4.003 \\, \\text{g\/mol}} = 0.577 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p><strong>Moles of fluorine:<\/strong><\/p>\n\n\n\n<p>[<br>n_{\\text{F}_2} = \\frac{3.08 \\, \\text{g}}{38.0 \\, \\text{g\/mol}} = 0.0813 \\, \\text{mol}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the ideal gas law to find the pressures.<\/h3>\n\n\n\n<p>We will now use the ideal gas law, ( PV = nRT ), where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( P ) is pressure,<\/li>\n\n\n\n<li>( V ) is volume,<\/li>\n\n\n\n<li>( n ) is the number of moles,<\/li>\n\n\n\n<li>( R ) is the gas constant (0.0821 L\u00b7atm\/(mol\u00b7K)), and<\/li>\n\n\n\n<li>( T ) is the temperature in Kelvin.<\/li>\n<\/ul>\n\n\n\n<p>First, assume the temperature ( T ) is constant for both gases. Since both gases are in separate flasks before the stopcock is opened, we can treat them independently.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the initial pressures in each flask.<\/h3>\n\n\n\n<p>The total volume of the helium is 2.01 L, and the number of moles is 0.577 mol. Using the ideal gas law:<\/p>\n\n\n\n<p>[<br>P_{\\text{He}} = \\frac{n_{\\text{He}} RT}{V_{\\text{He}}}<br>]<\/p>\n\n\n\n<p>For the fluorine gas, the volume is 3.16 L, and the number of moles is 0.0813 mol:<\/p>\n\n\n\n<p>[<br>P_{\\text{F}<em>2} = \\frac{n<\/em>{\\text{F}<em>2} RT}{V<\/em>{\\text{F}_2}}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Total pressure after opening the stopcock.<\/h3>\n\n\n\n<p>After the stopcock is opened, the total volume of the system is ( V_{\\text{total}} = 2.01 + 3.16 = 5.17 \\, \\text{L} ). The total number of moles is ( n_{\\text{total}} = 0.577 + 0.0813 = 0.658 \\, \\text{mol} ).<\/p>\n\n\n\n<p>The total pressure after the stopcock is opened is given as 625 mm Hg. This can be converted to atm:<\/p>\n\n\n\n<p>[<br>P_{\\text{total}} = 625 \\, \\text{mm Hg} \\times \\frac{1 \\, \\text{atm}}{760 \\, \\text{mm Hg}} = 0.8237 \\, \\text{atm}<br>]<\/p>\n\n\n\n<p>Using the ideal gas law for the total system:<\/p>\n\n\n\n<p>[<br>P_{\\text{total}} = \\frac{n_{\\text{total}} RT}{V_{\\text{total}}}<br>]<\/p>\n\n\n\n<p>Substitute in the known values:<\/p>\n\n\n\n<p>[<br>0.8237 \\, \\text{atm} = \\frac{0.658 \\, \\text{mol} \\times 0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K} \\times T}{5.17 \\, \\text{L}}<br>]<\/p>\n\n\n\n<p>We can solve for ( T ) (temperature) from this equation.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Use the temperature to calculate initial pressure of helium.<\/h3>\n\n\n\n<p>Now that we know the temperature ( T ), we can find the initial pressure of helium before the stopcock was opened.<\/p>\n\n\n\n<p>The answer for the initial pressure of helium before the stopcock is opened is calculated as:<\/p>\n\n\n\n<p>[<br>P_{\\text{He}} = \\frac{n_{\\text{He}} RT}{V_{\\text{He}}}<br>]<\/p>\n\n\n\n<p>Final answer: <strong>P_He<\/strong> = 0.586 atm or 448.5 mm Hg.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A 2.01 L flask containing 2.31 g of helium gas (molar mass = 4.003 g\/mol) is separated by a stopcock from a 3.16 L flask containing 3.08 g of fluorine gas (molar mass = 38 g\/mol).After the stopcock is opened the total pressure is 625 mm Hg.What was the pressure (reported to three significant figures) [&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 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