{"id":18697,"date":"2025-06-13T08:37:11","date_gmt":"2025-06-13T08:37:11","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=18697"},"modified":"2025-06-13T08:37:15","modified_gmt":"2025-06-13T08:37:15","slug":"a-cylinder-contains-40-0-g-of-he-56-0-g-of-n2-and-42-0-g-of-ar","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-cylinder-contains-40-0-g-of-he-56-0-g-of-n2-and-42-0-g-of-ar\/","title":{"rendered":"A cylinder contains 40.0 g of He, 56.0 g of N2, and 42.0 g of Ar"},"content":{"rendered":"\n<p>A cylinder contains 40.0 g of He, 56.0 g of N2, and 42.0 g of Ar. a. Calculate the mol fraction of each gas in the mixture? mol fraction of He mol fraction of N2 mol fraction of Ar b. If the cylinder is of a 25.0L volume and at a temperature of 25.0 \u00b0C. Calculate the total pressure of the mixture. c. What is the partial pressure of Ar in the mixture?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this problem, we will go step by step:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of He = 40.0 g<\/li>\n\n\n\n<li>Mass of N\u2082 = 56.0 g<\/li>\n\n\n\n<li>Mass of Ar = 42.0 g<\/li>\n\n\n\n<li>Volume (V) = 25.0 L<\/li>\n\n\n\n<li>Temperature (T) = 25.0\u00b0C = <strong>298.15 K<\/strong><\/li>\n\n\n\n<li>Ideal gas constant (R) = 0.0821 L\u00b7atm\/mol\u00b7K<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>a. Mole fraction of each gas<\/strong><\/h2>\n\n\n\n<p>First, calculate <strong>moles<\/strong> of each gas using molar mass:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of He = 4.00 g\/mol \u2192<br>nHe=40.04.00=10.0\u2009moln_{He} = \\frac{40.0}{4.00} = 10.0 \\, \\text{mol}<\/li>\n\n\n\n<li>Molar mass of N\u2082 = 28.0 g\/mol \u2192<br>nN2=56.028.0=2.0\u2009moln_{N_2} = \\frac{56.0}{28.0} = 2.0 \\, \\text{mol}<\/li>\n\n\n\n<li>Molar mass of Ar = 39.95 g\/mol \u2192<br>nAr=42.039.95\u22481.051\u2009moln_{Ar} = \\frac{42.0}{39.95} \\approx 1.051 \\, \\text{mol}<\/li>\n\n\n\n<li>Total moles ntotal=10.0+2.0+1.051=13.051\u2009moln_{total} = 10.0 + 2.0 + 1.051 = 13.051 \\, \\text{mol}<\/li>\n<\/ul>\n\n\n\n<p><strong>Mole fractions:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>XHe=10.013.051\u22480.766X_{He} = \\frac{10.0}{13.051} \\approx 0.766<\/li>\n\n\n\n<li>XN2=2.013.051\u22480.153X_{N_2} = \\frac{2.0}{13.051} \\approx 0.153<\/li>\n\n\n\n<li>XAr=1.05113.051\u22480.081X_{Ar} = \\frac{1.051}{13.051} \\approx 0.081<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>b. Total Pressure of the Gas Mixture<\/strong><\/h2>\n\n\n\n<p>Use the ideal gas law:<br>P=nRTVP = \\frac{nRT}{V}<\/p>\n\n\n\n<p>P=13.051\u00d70.0821\u00d7298.1525.0\u224812.72\u2009atmP = \\frac{13.051 \\times 0.0821 \\times 298.15}{25.0} \\approx 12.72 \\, \\text{atm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>c. Partial Pressure of Ar<\/strong><\/h2>\n\n\n\n<p>PAr=XAr\u00d7Ptotal=0.081\u00d712.72\u22481.03\u2009atmP_{Ar} = X_{Ar} \\times P_{total} = 0.081 \\times 12.72 \\approx 1.03 \\, \\text{atm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (\u2248300 words):<\/strong><\/h3>\n\n\n\n<p>In a mixture of gases, the <strong>mole fraction<\/strong> expresses the proportion of moles of a given gas to the total moles present. Calculating the number of moles for each gas is the first step. For helium (He), nitrogen (N\u2082), and argon (Ar), we used their respective molar masses to convert from grams to moles. Once we had all the moles, we added them together to find the total moles in the system.<\/p>\n\n\n\n<p>With this, the mole fraction of each gas is straightforward \u2014 it\u2019s the ratio of the moles of one gas to the total moles. Helium makes up about 76.6% of the mixture, nitrogen 15.3%, and argon 8.1%.<\/p>\n\n\n\n<p>To find the <strong>total pressure<\/strong>, we used the <strong>ideal gas law<\/strong>, PV=nRTPV = nRT, solving for pressure. The total pressure depends on the total number of moles, the temperature (converted to Kelvin), and the volume of the cylinder. Plugging in the values, the total pressure of the gas mixture is approximately <strong>12.72 atm<\/strong>.<\/p>\n\n\n\n<p>Finally, <strong>partial pressure<\/strong> refers to the contribution of a single gas to the total pressure. It can be found by multiplying the mole fraction of the gas by the total pressure. For argon, with a mole fraction of 0.081, the partial pressure comes out to <strong>1.03 atm<\/strong>. This method is grounded in <strong>Dalton\u2019s Law of Partial Pressures<\/strong>, which states each gas in a mixture exerts pressure independently of the others.<\/p>\n\n\n\n<p>This approach assumes ideal gas behavior, which is generally valid under the given conditions (moderate temperature and pressure).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A cylinder contains 40.0 g of He, 56.0 g of N2, and 42.0 g of Ar. a. Calculate the mol fraction of each gas in the mixture? mol fraction of He mol fraction of N2 mol fraction of Ar b. If the cylinder is of a 25.0L volume and at a temperature of 25.0 \u00b0C. [&hellip;]<\/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-18697","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18697","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=18697"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18697\/revisions"}],"predecessor-version":[{"id":18698,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18697\/revisions\/18698"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18697"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18697"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18697"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}