Use the fact that to determine how much the pressure must change in order to lower the boiling point of water by a small amount 3.20e-01 K.

Question

Use the fact that to determine how much the pressure must change in order to lower the boiling point of water by a small amount 3.20e-01 K. You may assume that the entropy and density of the liquid and gas are roughly constant for these small changes. You may also assume that the volume per molecule of liquid water is approximately zero compared to that of water vapor, and that water vapor is an ideal gas. Useful constants: Atmospheric pressure is 101300 Pa The boiling point of water at atmospheric pressure is 373.15 K The entropy difference between liquid and gas per kilogram is 6.05e 03 The molecular weight of water is 0.018 kg/mol. (a) 0.00e 00 Pa (b) 1.14e 03 Pa (c) 6.85e 26 Pa (d) 4.24e 05 Pa (e) 3.81e 28 Pa

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Philomena 4 years 2021-08-10T03:47:47+00:00 1 Answers 11 views 0

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  1. Farah
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    2021-08-10T03:49:18+00:00
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    Complete Question

    Use the fact that d\mu=\frac{V}{N}dp-\frac{s}{N}dT to determine how much the pressure must change in order to lower the boiling point of water by a small amount 3.20e-01 K. You may assume that the entropy and density of the liquid and gas are roughly constant for these small changes. You may also assume that the volume per molecule of liquid water is approximately zero compared to that of water vapor, and that water vapor is an ideal gas. Useful constants: Atmospheric pressure is 101300 Pa The boiling point of water at atmospheric pressure is 373.15 K The entropy difference between liquid and gas per kilogram is 6.05e 03 J/kgK The molecular weight of water is 0.018 kg/mol. (a) 0.00e 00 Pa (b) 1.14e 03 Pa (c) 6.85e 26 Pa (d) 4.24e 05 Pa (e) 3.81e 28 Pa

    Answer:

    Correct option is B

    Explanation:

    From the question we are told that:

    Given Equation d\mu=\frac{V}{N}dp-\frac{s}{N}dT

    Change of boiling point \triangle H=3.20e-01 K

    Generally the equation for Change in time is mathematically given by

      d\mu=\frac{V}{N}dp-\frac{s}{N}dT

      dp=\frac{s}{v}dT

    Where

        s=Entropy\ difference *molar\ weight

        s=6.05*10^3*0.018j/mol.k

    And

        V=\frac{RT}{P} (from ideal gas equation)

    Therefore

     dp=\frac{Ps}{RT}dT

     dp=\frac{101300*6.05*10^3*0.018}{8.314*373.15}3.20*10^{-1}

     dp=1137.873pa

     dp=1.14e 03 Pa

    Therefore correct option is B

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