{"id":43462,"date":"2025-06-30T06:27:37","date_gmt":"2025-06-30T06:27:37","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43462"},"modified":"2025-06-30T06:27:39","modified_gmt":"2025-06-30T06:27:39","slug":"you-have-a-200-0-l-vessel-at-200-0-k-that-contains-15-0-g-of-ch4-gas-and-15-0-g-of-co-gas","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/you-have-a-200-0-l-vessel-at-200-0-k-that-contains-15-0-g-of-ch4-gas-and-15-0-g-of-co-gas\/","title":{"rendered":"You have a 200.0 L vessel at 200.0 K that contains 15.0 g of CH4 gas and 15.0 g of CO gas"},"content":{"rendered":"\n<p>You have a 200.0 L vessel at 200.0 K that contains 15.0 g of CH4 gas and 15.0 g of CO gas, what is the total pressure of the system?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the total pressure of the system, we can use the ideal gas law for each gas individually and then add their pressures together, since the gases are non-reactive and behave ideally. The ideal gas law is:<\/p>\n\n\n\n<p>PV=nRTPV = nRT<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>PP = pressure (in atmospheres)<\/li>\n\n\n\n<li>VV = volume (in liters)<\/li>\n\n\n\n<li>nn = number of moles of gas<\/li>\n\n\n\n<li>RR = ideal gas constant = 0.0821 L\u00b7atm\/(mol\u00b7K)<\/li>\n\n\n\n<li>TT = temperature (in Kelvin)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate moles of CH\u2084 and CO<\/h3>\n\n\n\n<p>To find the number of moles, we use the formula: n=massmolar&nbsp;massn = \\frac{\\text{mass}}{\\text{molar mass}}<\/p>\n\n\n\n<p>For CH\u2084 (methane):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CH\u2084 = 16.04 g\/mol<\/li>\n\n\n\n<li>Moles of CH\u2084 = 15.0\u2009g16.04\u2009g\/mol=0.935\u2009mol\\frac{15.0 \\, \\text{g}}{16.04 \\, \\text{g\/mol}} = 0.935 \\, \\text{mol}<\/li>\n<\/ul>\n\n\n\n<p>For CO (carbon monoxide):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CO = 28.01 g\/mol<\/li>\n\n\n\n<li>Moles of CO = 15.0\u2009g28.01\u2009g\/mol=0.536\u2009mol\\frac{15.0 \\, \\text{g}}{28.01 \\, \\text{g\/mol}} = 0.536 \\, \\text{mol}<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the partial pressures<\/h3>\n\n\n\n<p>Now, use the ideal gas law for each gas:<\/p>\n\n\n\n<p>For CH\u2084: PCH\u2084=nRTV=(0.935\u2009mol)(0.0821\u2009L\\cdotpatm\/mol\\cdotpK)(200.0\u2009K)200.0\u2009LP_{\\text{CH\u2084}} = \\frac{nRT}{V} = \\frac{(0.935 \\, \\text{mol})(0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K})(200.0 \\, \\text{K})}{200.0 \\, \\text{L}} PCH\u2084=15.3200.0=0.0765\u2009atmP_{\\text{CH\u2084}} = \\frac{15.3}{200.0} = 0.0765 \\, \\text{atm}<\/p>\n\n\n\n<p>For CO: PCO=nRTV=(0.536\u2009mol)(0.0821\u2009L\\cdotpatm\/mol\\cdotpK)(200.0\u2009K)200.0\u2009LP_{\\text{CO}} = \\frac{nRT}{V} = \\frac{(0.536 \\, \\text{mol})(0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K})(200.0 \\, \\text{K})}{200.0 \\, \\text{L}} PCO=8.8200.0=0.0440\u2009atmP_{\\text{CO}} = \\frac{8.8}{200.0} = 0.0440 \\, \\text{atm}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Total pressure<\/h3>\n\n\n\n<p>The total pressure is the sum of the partial pressures of the two gases: Ptotal=PCH\u2084+PCO=0.0765\u2009atm+0.0440\u2009atm=0.1205\u2009atmP_{\\text{total}} = P_{\\text{CH\u2084}} + P_{\\text{CO}} = 0.0765 \\, \\text{atm} + 0.0440 \\, \\text{atm} = 0.1205 \\, \\text{atm}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The total pressure of the system is <strong>0.1205 atm<\/strong>.<\/p>\n\n\n\n<p>This calculation assumes ideal gas behavior, which is a good approximation at the given conditions (200.0 K and 200.0 L vessel). If the gases were interacting or deviating from ideal behavior, corrections would be necessary.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>You have a 200.0 L vessel at 200.0 K that contains 15.0 g of CH4 gas and 15.0 g of CO gas, what is the total pressure of the system? The correct answer and explanation is: To calculate the total pressure of the system, we can use the ideal gas law for each gas individually [&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-43462","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43462","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=43462"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43462\/revisions"}],"predecessor-version":[{"id":43463,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43462\/revisions\/43463"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43462"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43462"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43462"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}