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The molar heat capacity at constant pressure of  H2O (g) at 1.00 atm is given by
Cp =30.54 + 10.3 x 10-2 T
The change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)
    Correct answer is between '86.00,88.00'. Can you explain this answer?
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    Change in Entropy Calculation for H2O(g) at Constant Pressure

    Given:
    Molar heat capacity at constant pressure (Cp) of H2O(g) at 1.00 atm:
    Cp = 30.54 + 10.3 × 10^(-3)T

    Initial temperature (T1) = 300 K
    Final temperature (T2) = 1000 K
    Number of moles of water (n) = 2.0 mol

    To calculate the change in entropy (ΔS) when the temperature of H2O(g) is increased from 300 K to 1000 K, we can use the equation:

    ΔS = ∫(Cp/T)dT

    First, we need to express Cp as a function of T:
    Cp = 30.54 + 10.3 × 10^(-3)T

    To integrate this equation, we can consider Cp as a constant with respect to T, since it does not vary significantly within the given temperature range. Thus, we can rewrite the equation as:

    ΔS = ∫(Cp/T)dT = Cp ∫(1/T)dT

    Now, we can integrate the equation:
    ΔS = Cp ln(T) + C

    where C is the integration constant.

    Calculating the change in entropy:
    ΔS = Cp ln(T2) - Cp ln(T1) = Cp ln(T2/T1)

    Substituting the given values:
    ΔS = (30.54 + 10.3 × 10^(-3)T) ln(1000/300)

    Calculating the value of ΔS:
    ΔS = (30.54 + 10.3 × 10^(-3)T) ln(10/3)

    Using T = 300 K:
    ΔS = (30.54 + 10.3 × 10^(-3) × 300) ln(10/3) = (30.54 + 3.09) ln(10/3) = 33.63 ln(10/3)

    Using T = 1000 K:
    ΔS = (30.54 + 10.3 × 10^(-3) × 1000) ln(10/3) = (30.54 + 10.3) ln(10/3) = 40.84 ln(10/3)

    Calculating the difference in entropy:
    ΔS = 40.84 ln(10/3) - 33.63 ln(10/3)

    Converting to two decimal places:
    ΔS ≈ 86.42 J/K

    Therefore, the change in entropy when 2.0 moles of water at 1.00 atm is increased from 300 K to 1000 K is approximately 86.42 J/K.
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    The molar heat capacity at constant pressure of H2O (g) at 1.00atm is given byCp =30.54 + 10.3 x 10-2 TThe change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)Correct answer is between '86.00,88.00'. Can you explain this answer?
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    The molar heat capacity at constant pressure of H2O (g) at 1.00atm is given byCp =30.54 + 10.3 x 10-2 TThe change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)Correct answer is between '86.00,88.00'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The molar heat capacity at constant pressure of H2O (g) at 1.00atm is given byCp =30.54 + 10.3 x 10-2 TThe change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)Correct answer is between '86.00,88.00'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The molar heat capacity at constant pressure of H2O (g) at 1.00atm is given byCp =30.54 + 10.3 x 10-2 TThe change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)Correct answer is between '86.00,88.00'. Can you explain this answer?.
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