Liquid chlorobenzene goes through a process where there is negligible heat flow and no work done on the surroundings. initially it is at 330

Question

Liquid chlorobenzene goes through a process where there is negligible heat flow and no work done on the surroundings. initially it is at 330 k and 3,500 kpa. at the end of the process the final pressure is 2,000kpa. estimate the temperature change and the entropy change of chlorobenzene.

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Eirian 4 years 2021-08-10T20:26:47+00:00 1 Answers 26 views 0

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  1. thuthao
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    2021-08-10T20:28:19+00:00
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    Answer:

    The change in temperature is -141.428K

    The entropy change is -60.75J/mole K Kg

    Explanation:

    Given that,

    Initial Temperature, T₁ = 330k

    Initial Pressure, P₁ = 3500kpa

    Final pressure, P₂ = 2000kpa

    Since the work done on the surrounding is zero

    hence, the change in volume of the system will be zero and the process will be constant

    Calculate the Temperature of the system

    \frac{P_1}{P_2} = \frac{T_1}{T_2}

    Substitute the value in the above equation

    \frac{3500}{2000} = \frac{330}{T_2} \\\\T_2 = 188.57K

    Calculate the change in temperature

    ΔT = T₂ – T₁

         = 188.57K – 330K

          = -141.428K

    The change in temperature is -141.428K

    Expression of change in entropy

    ds =\frac{dQ}{T}

    ds is the entropy change

    equation for dQ = c_vdT

    Substitute for dQ in the expression of change in entropy and integrate

    ds = \frac{C_vdT}{T}

    \int\limits^ {s_2 }_ {s_1 } \, ds = \int\limits^ { T_2}_ {T_1 } \frac{C_vdT}{T}

    s_2-s_1=C_vIn\frac{T_2}{T_1}

    substitute the values

    \Delta s = \frac{152}{1.4} In\frac{188.57}{330} \\\\\Delta s = -60.75J/Mole-K -Kg

    The entropy change is -60.75J/mole K Kg

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