{"id":30018,"date":"2025-06-21T15:57:04","date_gmt":"2025-06-21T15:57:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=30018"},"modified":"2025-06-21T15:57:06","modified_gmt":"2025-06-21T15:57:06","slug":"1-000-mol-of-urea-is-slowly-combusted-in-a-container-with-a-mobile-piston","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/1-000-mol-of-urea-is-slowly-combusted-in-a-container-with-a-mobile-piston\/","title":{"rendered":"1.000 mol of urea is slowly combusted in a container with a mobile piston"},"content":{"rendered":"\n

1.000 mol of urea is slowly combusted in a container with a mobile piston. As a result of the combustion, the system expands quasistatically against an external pressure of 2.0 atm as it lifts a light weight. The temperature of the system is maintained at 298.15 K<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To analyze this process, we need to apply the First Law of Thermodynamics<\/strong> and understand the thermochemical reaction<\/strong> involved in the combustion of urea (CO(NH\u2082)\u2082)<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Combustion Reaction of Urea:<\/strong><\/h3>\n\n\n\n

The balanced combustion reaction for 1 mole of urea<\/strong> is approximately: CO(NH2)2(s)+32O2(g)\u2192CO2(g)+N2(g)+2H2O(l)\\text{CO(NH}_2\\text{)}_2 (s) + \\frac{3}{2} O_2 (g) \\rightarrow CO_2 (g) + N_2 (g) + 2 H_2O (l)<\/p>\n\n\n\n

At 298.15 K, water forms in the liquid state.<\/p>\n\n\n\n

\n\n\n\n

Thermodynamic Quantities:<\/strong><\/h3>\n\n\n\n

Let\u2019s denote the key thermodynamic terms:<\/p>\n\n\n\n

    \n
  • qq: heat added to the system<\/li>\n\n\n\n
  • ww: work done by the system<\/li>\n\n\n\n
  • \u0394U\\Delta U: change in internal energy<\/li>\n\n\n\n
  • \u0394H\\Delta H: change in enthalpy<\/li>\n<\/ul>\n\n\n\n

    Because the process occurs quasistatically<\/strong> with constant external pressure<\/strong> and fixed temperature<\/strong>, the system does pressure-volume (PV) work<\/strong>. The work done is: w=\u2212Pext\u0394Vw = -P_{\\text{ext}} \\Delta V<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • Pext=2.0P_{\\text{ext}} = 2.0 atm<\/li>\n\n\n\n
    • \u0394V\\Delta V: volume change due to gaseous products<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Gas Volume Change:<\/strong><\/h3>\n\n\n\n

      From the balanced equation:<\/p>\n\n\n\n

        \n
      • Reactant gases = 1.5 mol O2O_2<\/li>\n\n\n\n
      • Product gases = 1 mol CO2CO_2 + 0.5 mol N2N_2 = 1.5 mol<\/li>\n<\/ul>\n\n\n\n

        So, no net change<\/strong> in moles of gas.<\/p>\n\n\n\n

        Thus, \u0394V=0\\Delta V = 0, and: w=\u2212Pext\u0394V=0w = -P_{\\text{ext}} \\Delta V = 0<\/p>\n\n\n\n

        But the question says the system expands and lifts a light weight<\/strong>, which implies some volume change. If we assume gaseous water forms instead of liquid, then: CO(NH2)2+32O2\u2192CO2+N2+2H2O(g)\\text{CO(NH}_2\\text{)}_2 + \\frac{3}{2} O_2 \\rightarrow CO_2 + N_2 + 2 H_2O(g)<\/p>\n\n\n\n

        Now:<\/p>\n\n\n\n

          \n
        • Reactants: 1.5 mol gas<\/li>\n\n\n\n
        • Products: 1 (CO\u2082) + 1 (N\u2082) + 2 (H\u2082O) = 4 mol gas<\/li>\n<\/ul>\n\n\n\n

          Change in gas moles = 4\u22121.5=2.54 – 1.5 = 2.5 mol<\/p>\n\n\n\n

          Using the ideal gas law: \u0394V=\u0394n\u22c5RTP\\Delta V = \\Delta n \\cdot \\frac{RT}{P} =2.5\u22c50.08206\u22c5298.152.0\u224830.6 L= 2.5 \\cdot \\frac{0.08206 \\cdot 298.15}{2.0} \\approx 30.6 \\text{ L}<\/p>\n\n\n\n

          Now the work becomes: w=\u2212P\u0394V=\u22122.0\u22c530.6=\u221261.2 L\\cdotpatmw = -P \\Delta V = -2.0 \\cdot 30.6 = -61.2 \\text{ L\u00b7atm}<\/p>\n\n\n\n

          Convert to joules: 1 L\\cdotpatm=101.325 J\u21d2w\u2248\u22126200 J1 \\text{ L\u00b7atm} = 101.325 \\text{ J} \\Rightarrow w \\approx -6200 \\text{ J}<\/p>\n\n\n\n

          \n\n\n\n

          Conclusion:<\/strong><\/h3>\n\n\n\n

          The system does approximately 6200 J of work<\/strong> during the combustion. The negative sign indicates energy is lost by the system to do this work. Since temperature is constant and the piston is mobile, this is a quasistatic, isothermal expansion<\/strong> where chemical energy is converted into mechanical work.<\/p>\n","protected":false},"excerpt":{"rendered":"

          1.000 mol of urea is slowly combusted in a container with a mobile piston. As a result of the combustion, the system expands quasistatically against an external pressure of 2.0 atm as it lifts a light weight. The temperature of the system is maintained at 298.15 K The correct answer and explanation is: To analyze […]<\/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-30018","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30018","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=30018"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30018\/revisions"}],"predecessor-version":[{"id":30019,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30018\/revisions\/30019"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=30018"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=30018"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=30018"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}