5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol

Question

5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol(b)Evaporation of one mole of liquid carbon dioxide at its boiling point, 216.6 K. ΔHvap = 15.326 kJ/mol

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Sapo 4 months 2021-09-05T01:38:30+00:00 1 Answers 0 views 0

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    2021-09-05T01:39:49+00:00

    Answer:

    a) ΔS = 14.46 J/k

    b) ΔS = 70.76 J/k

    Explanation:

    The general formula to calculate the entropy change accompanied with a process is:

    ΔS = ΔQ/T

    where,

    ΔS = entropy change for the process

    ΔQ = Heat Transfer during the process

    T = Absolute Temperature during the process

    a)

    In this case the heat transfer will be given as:

    ΔQ = (ΔHfus)(N)

    where,

    ΔHfus = Molar Heat of Fusion of Tin = 7.029 KJ/mol

    N = No. of moles of tin = 1 mol

    Therefore,

    ΔQ = (7.029 KJ/mol)(1 mol)

    ΔQ = 7.029 KJ = 7029 J

    and the absolute temperature is:

    T = 213°C +273 = 486 k

    using these values in the entropy formula, we get:

    ΔS = 7029 J/486 k

    ΔS = 14.46 J/k

    b)

    In this case the heat transfer will be given as:

    ΔQ = (ΔHvap)(N)

    where,

    ΔHvap = Molar Heat of Vaporization of Carbon Dioxide = 15.326 KJ/mol

    N = No. of moles of Carbon Dioxide = 1 mol

    Therefore,

    ΔQ = (15.326 KJ/mol)(1 mol)

    ΔQ = 15.326 KJ = 15326 J

    and the absolute temperature is:

    T = 216.6 k

    using these values in the entropy formula, we get:

    ΔS = 15326 J/216.6 k

    ΔS = 70.76 J/k

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