# 21.1: Entropy Increases With Increasing Temperature


##### 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:

$dS=\frac{C_V}{T}dT \nonumber$

Integrating:

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

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:

$dS=\frac{C_P}{T}dT \nonumber$

Integrating:

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

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.