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21.1: Entropy Increases With Increasing Temperature

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    Learning Objectives
    • Define entropy and its relation to energy flow.

    Entropy versus temperature

    We can put together the first and the second law for a reversible process with no other work than volume (\(PV\)) work and obtain:

    \[dU= δq_{rev} + δw_{rev} \nonumber \]

    Entropy is the dispersal of energy and is related to heat:

    \[δq_{rev}= TdS \nonumber \]

    Work is related to the change in volume:

    \[δw_{rev}= -PdV \nonumber \]

    Plugging these into our expression for \(dU\) for reversible changes:

    \[dU= TdS -PdV \nonumber \]

    We no longer have any path functions in the expression, as \(U\), \(S\) and \(V\) are all state functions. This means this expression must be an exact differential. We can generalize the expression to hold for irreversible processes, but then the expression becomes an inequality:

    \[dU≤ TdS - PdV \nonumber \]

    This equality expresses \(U\) as a function of two variables, entropy and volume: \(U(S,V)\). \(S\) and \(V\) are the natural variables of \(U\).

    Entropy and heat capacity

    At constant volume, \(dU\) becomes:

    \[dU=TdS \nonumber \]

    Recall that internal energy is related to constant volume heat capacity, \(C_V\):

    \[C_V=\left(\frac{dU}{dT}\right)_V \nonumber \]

    Combining these two expressions, we obtain:

    \[dS=\frac{C_V}{T}dT \nonumber \]


    \[\Delta S=\int_{T_1}^{T_2}{\frac{C_V(T)}{T}dT} \nonumber \]

    If we know how \(C_V\) changes with temperature, we can calculate the change in entropy, \(\Delta S\). Since heat capacity is always a positive value, entropy must increase as the temperature increases. There is nothing to stop us from expressing \(U\) in other variables, e.g. \(T\) and \(V\). In fact, we can derive some interesting relationships if we do.

    Example 21.1.1
    1. Write \(U\) as a function of \(T\) and \(V\).
    2. Write \(U\) as a function of its natural variables.
    3. Rearrange (2) to find an expression for \(dS\).
    4. Substitute (1) into (3) and rearrange. This is the definition of \(C_V\).
    5. Write out \(S\) as a function of \(T\) and \(V\).

    We can also derive an expression for the change in entropy as a function of constant pressure heat capacity, \(C_P\). To start, we need to change from internal energy, \(U\), to enthalpy, \(H\):

    \[\begin{align*} H &= U + PV \\[4pt] dH &= dU +d(PV) \\[4pt] &= dU + PdV + VdP \end{align*} \nonumber \]

    For reversible processes:

    \[\begin{align*} dH &= dU + PdV + VdP \\[4pt] &= TdS -PdV + PdV + VP \\[4pt] &= TdS + VdP\end{align*} \nonumber \]

    The natural variables of the enthalpy are \(S\) and \(P\) (not: \(V\)). A similar derivation as above shows that the temperature change of entropy is related to the constant pressure heat capacity:

    \[dH=TdS+VdP \nonumber \]

    At constant pressure:

    \[dH=TdS+VdP \nonumber \]

    Recall that:

    \[C_P=\frac{dH}{dT} \nonumber \]

    Combining, we obtain:

    \[dS=\frac{C_P}{T}dT \nonumber \]


    \[\Delta S=\int_{T_1}^{T_2}{\frac{C_P(T)}{T}dT} \nonumber \]

    This means that if we know the heat capacities as a function of temperature we can calculate how the entropy changes with temperature. Usually it is easier to obtain data under constant \(P\) conditions than for constant \(V\), so that the route with \(C_p\) is the more common one.

    21.1: Entropy Increases With Increasing Temperature is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.