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    Mathematical Proof: Entropy Change of a Reversible Process

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    I have a question about relations for a process even though I am unsure about the scope of this. Still I wonder about a mathematical relation for the following:

    I wonder about how one defines mathematically that a process is reversible for a process that has both pressure and temperature differences from definition of total entropy.

    Is there a mathematical way to show how much a factor the temperature is and how much a factor the pressure is for an ideal gas and then find a situation where the process is reversible for a state that has both pressure and temperature differences that is that dS_tot=0? That is finding a way to express dS_tot while only using p, T, V and C_v as variables while there are pressure and temperature differnces and showing where the values of T and p are that creates dS_tot=0

    Even if it would only be a strict mathematic definition it could be really helpful!

    I need a computer written answer with mathematical signs. I hope it is ok.

    © BrainMass Inc. brainmass.com December 24, 2021, 11:21 pm ad1c9bdddf
    https://brainmass.com/chemistry/energetics-and-thermodynamics/mathematical-proof-entropy-change-reversible-process-561619

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    Question
    Possible mathematical definition of a process that has both pressure and temperature difference but is reversible
    I have a question about relations for a process even though I am unsure about the scope of this. Still I wonder about a mathematical relation for the following:

    I wonder about how one defines mathematically that a process is reversible for a process that has both pressure and temperature differences from definition of total entropy

    Is there a mathematical way to show how much a factor the temperature is and how much a factor the pressure is for an ideal gas and then find a situation where the process is reversible for a state that has both pressure and temperature differences that is that dS_tot=0? That is finding a way to express dS_tot while only using p, T, V and C_v as variables while there are pressure and temperature differences and showing where the values of T and p are that creates dS_tot=0

    Even if it would only be a strict mathematic definition it could be really helpful!

    I need a computer written answer with mathematical signs. I hope it is ok

    ***********************
    Answer

    Reversible Process
    In thermodynamics, a reversible process, or reversible cycle if the process is cyclic, is a process that can be "reversed" by means of infinitesimal changes in some property of the system without entropy production (i.e. dissipation of energy). A reversible process changes the state of a system in such a way that the net change in the combined entropy of the system and its surroundings is zero.

    Entropy of Reversible Process
    Entropy is defined for a reversible process and for a system that, at all times, can be treated as being at a uniform state and thus at a uniform temperature. Reversibility is an ideal that some real processes approximate and that is often presented in study exercises. For a reversible process, entropy behaves as a conserved quantity and no change occurs in total entropy. More specifically, total entropy is conserved in a reversible process and not conserved in an irreversible process. One has to be careful about system boundaries. For example, in the Carnot cycle, while the heat flow from the hot reservoir to the cold reservoir represents an increase in entropy, the work output, if reversibly and perfectly stored in some energy storage mechanism, represents a decrease in entropy that could be used to operate the heat engine in reverse and return to the previous state, thus the total entropy change is still zero at all times if the entire process is reversible. Any process that does not meet the requirements of a reversible process must be treated as an irreversible process, which is usually a complex task. An irreversible process increases entropy.

    Heat transfer situations require two or more non-isolated systems in thermal contact. In irreversible heat transfer, heat energy is irreversibly transferred from the higher temperature system to the lower temperature system, and the combined entropy of the systems increases. Each system, by definition, must have its own absolute temperature applicable within all areas in each respective system in order to calculate the entropy transfer. Thus, when a system at higher temperature TH transfers heat dQ to a system of lower temperature TC, the former loses entropy dQ/ TH and the latter gains entropy dQ/ TC. The combined entropy change is dQ/ TC -dQ/ TH which is positive, reflecting an increase in the combined entropy. This quantity will be zero for a reversible process.

    Heating and Cooling
    For heating (or cooling) of any system (gas, liquid or solid) at constant pressure from an initial temperature to a final temperature , the entropy change is
    .
    provided that the constant-pressure molar heat capacity (or specific heat) CP is constant and that no phase transition occurs in this temperature interval.
    Similarly at constant volume, the entropy change is
    ,
    where the constant-volume heat capacity Cv is constant and there is no phase change.
    At low temperatures near absolute zero, heat capacities of solids quickly drop off to near zero, so the assumption of constant heat capacity does not apply.
    Since entropy is a state function, the entropy change of any process in which temperature and volume both vary is the same as for a path divided into two steps - heating at constant volume and expansion at constant temperature. For an ideal gas, the total entropy change is
    .
    Similarly if the temperature and pressure of an ideal gas both vary,
    .

    Mathematical Proof
    So you need a mathematical proof that the above mentioned quantity is zero for a reversible process.

    Start by writing the total differential of the entropy with respect to T and P:

    dS = (∂S/∂T)P dT + (∂S/∂P)T dP

    You should know that:

    (∂S/∂T)P = Cp/T

    dS = (Cp/T) dT + (∂S/∂P)T dP

    The remaining partial derivative can be expressed in more tractable terms using a Maxwell's relation

    (∂S/∂P)T = -(∂V/∂T)P

    so

    dS = (Cp /T) dT - (∂V/∂T)P dP

    Assuming 1 mole of an ideal gas:

    V = R*T/P

    (∂V/∂T)P = R/P

    So

    dS = (Cp /T) dT - (R/P) dP

    Integrate:

    So - S = ΔS = Cp *ln(To/T) - R*ln(Po/P)

    The potential temperature (To) is the temperature a parcel of fluid would have if that parcel were brought from some initial arbitrary pressure and temperature (P,T) to some standard pressure (Po) along a reversible adiabatic path. Because dS = dq/T for a reversible process, and dq = 0 for an adiabatic process, ΔS =0 for this process.

    Therefore for one mole of ideal gas undergoing a reversible process:

    ΔS = Cp *ln(To/T) - R*ln(Po/P) = 0

    References:
    http://en.wikipedia.org/wiki/Reversible_process_%28thermodynamics%29
    http://en.wikipedia.org/wiki/Entropy
    http://answers.yahoo.com/question/index?qid=20100907144643AAdwtz4

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:21 pm ad1c9bdddf>
    https://brainmass.com/chemistry/energetics-and-thermodynamics/mathematical-proof-entropy-change-reversible-process-561619

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