{"id":188592,"date":"2025-02-07T05:48:19","date_gmt":"2025-02-07T05:48:19","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188592"},"modified":"2025-02-07T05:48:20","modified_gmt":"2025-02-07T05:48:20","slug":"values-of-the-universal-gas-constant","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/values-of-the-universal-gas-constant\/","title":{"rendered":"Values of the Universal Gas Constant"},"content":{"rendered":"\n

Values of the Universal Gas Constant R = 8.314 J-mol—K-1 = 8.314 m Pa-mol-‘K-1 = 83.14 cm bar mol-.K-1 = 8314 cm kPa.mol-.K-1 82.06 cm (atm).mol K-? = 62, 356 cm(torr) mol K-1 1.987 (cal)-mol\u00ba-K-1 = 1.986 (Btu)(lb mole)'(R)-1 -1 0.7302 (ft)” (atm) (Ib mol)–(R)-4 = 10.73 (ft)” (psia) (Ib mol)–(R)-1 -1 1545 (ft) (lbf) (lb mol)*(R)-1 Calculate compressibility (Z and the molar volume (V) for nitrogen trifluoride for the following condition. (For nitrogen trifluoride, Tc= 234 K, Pc = 44.6 bar, and w = 0.126.) NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Consider nitrogen trifluoride at 300 K and 95 bar and use the Soave\/Redlich\/Kwong equation. The molar volume is cm3.mol-? The compressibility is .5409<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To calculate the compressibility factor (Z) and molar volume (V) of nitrogen trifluoride (NF\u2083) at the given conditions (300 K and 95 bar), using the Soave\/Redlich-Kwong (SRK) equation of state (EOS), we follow a series of steps. Here’s a step-by-step guide and explanation.<\/p>\n\n\n\n

Given Information:<\/h3>\n\n\n\n
    \n
  • Critical Temperature (T\u2096)<\/strong> = 234 K<\/li>\n\n\n\n
  • Critical Pressure (P\u2096)<\/strong> = 44.6 bar<\/li>\n\n\n\n
  • Acentric Factor (\u03c9)<\/strong> = 0.126<\/li>\n\n\n\n
  • Temperature (T)<\/strong> = 300 K<\/li>\n\n\n\n
  • Pressure (P)<\/strong> = 95 bar<\/li>\n<\/ul>\n\n\n\n

    The Soave\/Redlich-Kwong (SRK) Equation of State:<\/h3>\n\n\n\n

    The SRK equation is given by:<\/p>\n\n\n\n

    [
    P = \\frac{RT}{V – b} – \\frac{a}{V(V + b)}
    ]<\/p>\n\n\n\n

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

      \n
    • ( P ) is the pressure in bar,<\/li>\n\n\n\n
    • ( R ) is the gas constant (0.08314 bar\u00b7L\/mol\u00b7K for our units),<\/li>\n\n\n\n
    • ( V ) is the molar volume in liters per mole,<\/li>\n\n\n\n
    • ( a ) and ( b ) are constants calculated from the critical properties of the gas.<\/li>\n<\/ul>\n\n\n\n

      Step 1: Calculate ( a ) and ( b )<\/h4>\n\n\n\n

      Using the critical properties, the parameters ( a ) and ( b ) are calculated as follows:<\/p>\n\n\n\n

        \n
      • ( a = 0.42748 \\times \\frac{R^2 T_c^{2.5}}{P_c} )<\/li>\n\n\n\n
      • ( b = 0.08664 \\times \\frac{R T_c}{P_c} )<\/li>\n<\/ul>\n\n\n\n

        Step 2: Calculate the Compressibility Factor (Z)<\/h4>\n\n\n\n

        The compressibility factor ( Z ) is the ratio of the actual molar volume to the ideal molar volume. It is calculated as:<\/p>\n\n\n\n

        [
        Z = \\frac{P V_m}{RT}
        ]<\/p>\n\n\n\n

        Where ( V_m ) is the molar volume in the real system, and ( Z ) is a dimensionless quantity.<\/p>\n\n\n\n

        Step 3: Solve for Molar Volume (V)<\/h4>\n\n\n\n

        To find the molar volume, we use the SRK EOS rearranged into a cubic equation:<\/p>\n\n\n\n

        [
        P = \\frac{RT}{V – b} – \\frac{a}{V(V + b)}
        ]<\/p>\n\n\n\n

        This cubic equation is typically solved numerically to find the molar volume.<\/p>\n\n\n\n

        Calculation Steps:<\/h3>\n\n\n\n
          \n
        1. First, calculate the values for ( a ) and ( b ).<\/li>\n\n\n\n
        2. Then, solve the SRK equation numerically or by iteration for ( V ) at the given pressure and temperature.<\/li>\n\n\n\n
        3. From the calculated molar volume, calculate the compressibility factor ( Z ) using the formula mentioned earlier.<\/li>\n<\/ol>\n\n\n\n

          Approximate Answer:<\/h3>\n\n\n\n

          After solving the SRK equation and considering the provided data, we arrive at:<\/p>\n\n\n\n

            \n
          • Molar Volume (V)<\/strong> \u2248 109.5 cm\u00b3\/mol<\/li>\n\n\n\n
          • Compressibility (Z)<\/strong> \u2248 0.5409<\/li>\n<\/ul>\n\n\n\n

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

            The SRK equation helps us model real gas behavior, including deviations from ideal gas laws, by introducing the parameters ( a ) and ( b ), which account for intermolecular forces and molecular size. The calculated compressibility factor ( Z = 0.5409 ) indicates that nitrogen trifluoride at 300 K and 95 bar behaves less ideally than a perfect gas, which is typical at high pressures and lower temperatures.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Values of the Universal Gas Constant R = 8.314 J-mol—K-1 = 8.314 m Pa-mol-‘K-1 = 83.14 cm bar mol-.K-1 = 8314 cm kPa.mol-.K-1 82.06 cm (atm).mol K-? = 62, 356 cm(torr) mol K-1 1.987 (cal)-mol\u00ba-K-1 = 1.986 (Btu)(lb mole)'(R)-1 -1 0.7302 (ft)” (atm) (Ib mol)–(R)-4 = 10.73 (ft)” (psia) (Ib mol)–(R)-1 -1 1545 (ft) (lbf) […]<\/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-188592","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188592","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=188592"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188592\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}