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The pressure of a liquid is increased isothermally. The molar volume of the liquid decreases from 50.45 × 10-6 m/ mol  to 48 × 10−6 m3 / mol during this process. The isothermal compressibility of the liquid is 10−9 Pa−1 , which can be assumed to be independent of pressure.
The change in the molar Gibbs free energy of the liquid, rounded to nearest integer, is ___ J/mol. 
    Correct answer is '2511.4'. Can you explain this answer?
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    The pressure of a liquid is increased isothermally. The molar volume o...
    At constant temperature 
    dG = Vdp - SdT ⇒ dG VdP = RT/P dP
    P1 = 109 pa, V1 = 50.45×10-6 m3 / mol
    V2 =  48×10-6 m3 / mol
    ΔG = 50.45×103 In (1.051)
    ΔG = 2511.48 J / mole
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    The pressure of a liquid is increased isothermally. The molar volume o...
    ML/mol to 47.96 mL/mol. Calculate the compressibility factor of the liquid.

    We can use the following equation to calculate the compressibility factor:

    Z = PV/RT

    where P is the pressure, V is the molar volume, R is the gas constant, and T is the temperature. Since the process is isothermal, the temperature remains constant and we can simplify the equation to:

    Z = P(Vm/RT)

    where Vm is the molar volume.

    We can calculate the compressibility factor using the initial and final molar volumes and the given pressure:

    Z = P(Vm/RT) = (P/RT)(Vm)

    Z = [(1 atm)/(0.08206 L·atm/mol·K)(298 K)](50.45 mL/mol - 47.96 mL/mol)

    Z = 0.00199

    Therefore, the compressibility factor of the liquid is approximately 0.00199.
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    The pressure of a liquid is increased isothermally. The molar volume of the liquid decreases from 50.45 × 10-6 m3/ molto 48 × 10−6 m3 / mol during this process. The isothermal compressibility of the liquid is 10−9 Pa−1 , which can be assumed to be independent of pressure.The change in the molar Gibbs free energy of the liquid, rounded to nearest integer, is ___ J/mol.Correct answer is '2511.4'. Can you explain this answer?
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    The pressure of a liquid is increased isothermally. The molar volume of the liquid decreases from 50.45 × 10-6 m3/ molto 48 × 10−6 m3 / mol during this process. The isothermal compressibility of the liquid is 10−9 Pa−1 , which can be assumed to be independent of pressure.The change in the molar Gibbs free energy of the liquid, rounded to nearest integer, is ___ J/mol.Correct answer is '2511.4'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The pressure of a liquid is increased isothermally. The molar volume of the liquid decreases from 50.45 × 10-6 m3/ molto 48 × 10−6 m3 / mol during this process. The isothermal compressibility of the liquid is 10−9 Pa−1 , which can be assumed to be independent of pressure.The change in the molar Gibbs free energy of the liquid, rounded to nearest integer, is ___ J/mol.Correct answer is '2511.4'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The pressure of a liquid is increased isothermally. The molar volume of the liquid decreases from 50.45 × 10-6 m3/ molto 48 × 10−6 m3 / mol during this process. The isothermal compressibility of the liquid is 10−9 Pa−1 , which can be assumed to be independent of pressure.The change in the molar Gibbs free energy of the liquid, rounded to nearest integer, is ___ J/mol.Correct answer is '2511.4'. Can you explain this answer?.
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