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    entropy, heat, and temperature during compression of lead

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    The change in volume is small enough to be ignored (.2% of the initial volume).

    1.) 10^(-3) m^3 of lead is compressed reversibly and isothermally at room temperature from 1 to 1000 atm pressure. Using one of Maxwell's thermodynamic relations to find the following:
    a) the change in entropy
    b) The heat given out
    c) the change in internal energy of the lead
    Isothermal compressibility of lead is ( -(1/V) (V/p)T) = 2.2*10^-6 atm^-1

    Volume Coefficient of Expansion is (1/V) (V/T)p = 8*10^-5 K^-1
    1 atm = 10^5 Pa

    2.) Calculate the change in temperature of 10^(-3) m^3 of lead undergoing a reversible and adiabatic compression from 1 to 1000 atms. The adiabatic compressibility of lead is assumed to be independent of pressure with a value of 2.2*10^-6 atm^-1
    (hint: write down the partial differential coefficient of temperature with respect to pressure in an adiabatic , reversible process and convert this by a Maxwell relation. Re-express the result in terms of measurable quantities.)
    some of the required values are given in problem 1 above.
    Cp for lead is 25J/K and the molar volume is 18.3*10-6 m^3

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    https://brainmass.com/physics/partial-differential-equation/entropy-heat-temperature-compression-lead-165008

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    It shows how to find the entropy, heat, and temperature changes during compression of lead. The solution is detailed and was rated '5/5' by the student who originally posted the questions.

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