# What is the entropy change at "273 K" when a perfect gas is compressed at "1 bar" to a volume of "0.011 m"^3 at "10 bar"?

Apr 5, 2017

I got $- \text{9.28 J/K}$.

Consider entropy as a function of temperature and pressure. We can write the total differential as:

$\mathrm{dS} \left(T , P\right) = {\left(\frac{\partial S}{\partial T}\right)}_{P} \mathrm{dT} + {\left(\frac{\partial S}{\partial P}\right)}_{T} \mathrm{dP}$

• The first derivative term, ${\left(\frac{\partial S}{\partial T}\right)}_{P}$, describes the change in entropy due to the change in temperature at a constant pressure.
• The second derivative term, ${\left(\frac{\partial S}{\partial P}\right)}_{T}$, describes the change in entropy due to the change in pressure at a constant temperature.

Notice that the problem asks for a change in pressure but only tells you one temperature. That is, the temperature is constant. So, we're searching for how to evaluate the second derivative seen above.

Recall that the Gibbs' free energy is commonly regarded as a function of the temperature and pressure, i.e. $G = G \left(T , P\right)$. So, we should recall the Maxwell Relation:

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$

Since $\Delta G$ is a state function, it follows that order in which second-partial derivatives are taken does not matter ($\left(\frac{{\partial}^{2} F}{\partial x \partial y}\right) = \left(\frac{{\partial}^{2} F}{\partial y \partial x}\right)$), and thus, the cross-derivatives are equal:

${\left(\frac{\partial S}{\partial P}\right)}_{T} = - {\left(\frac{\partial V}{\partial T}\right)}_{P}$

Now that we've established the relationship, we have an expression to use in an equation we know: the ideal gas law.

$P V = n R T$

The partial derivative of the volume with respect to the temperature at constant pressure is then:

$- \frac{\partial}{\partial T} {\left[V\right]}_{P} = - \frac{\partial}{\partial T} {\left[\frac{n R T}{P}\right]}_{P}$

$= - \frac{n R}{P} \frac{\partial}{\partial T} {\left[T\right]}_{P}$

$= - \frac{n R}{P}$

The next thing to do is integrate over the two pressures.

$\Delta S = {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial S}{\partial P}\right)}_{T} \mathrm{dP}$

$= - {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial V}{\partial T}\right)}_{P} \mathrm{dP}$

$= - {\int}_{{P}_{1}}^{{P}_{2}} \frac{n R}{P} \mathrm{dP}$

$= - n R {\int}_{{P}_{1}}^{{P}_{2}} \frac{1}{P} \mathrm{dP}$

The integral of $\frac{1}{x}$ is $\ln | x |$. Therefore:

$\Delta S = - n R \left(\ln | {P}_{2} | - \ln | {P}_{1} |\right)$

$= - n R \ln | {P}_{2} / {P}_{1} |$

Since you are only given the molar volume and not the mols, we have to assume that at $\text{1 bar}$ pressure, the starting molar volume has to be calculated at $\text{273 K}$ so that we can find the mols from the given volume (we know the mols will not change).

$\frac{V}{n} = \frac{R T}{P}$

= (("0.083145 L"cdot"atm/mol"cdot"K")("273 K"))/("1 bar")

$=$ $\text{22.700 L/mol}$

Therefore, the mols of perfect gas present are:

n = "1 mol"/(22.700 cancel"L") xx 0.011 cancel("m"^3) xx ((cancel"1 dm")/(0.1 cancel"m"))^3 xx cancel("1 L")/cancel("1 dm"^3)

$=$ $\text{0.485 mols gas}$

Finally, we can calculate the change in entropy at constant temperature due to a change in pressure:

$\textcolor{b l u e}{\Delta S} = - \left(0.485 \cancel{\text{mols")("8.314472 J/"cancel"mol"cdot"K")ln("10 bar"/"1 bar}}\right)$

$= \textcolor{b l u e}{- \text{9.28 J/K}}$