Question

Sketch qualitatively accurate graphs of $G$vs.$T$for the three phases of$H_{2}O$(ice, water, and steam) at atmospheric pressure. Put all three graphs on the same set of axes, and label the temperatures$0°C$and $100°C$. How would the graphs differ at a pressure of$0.001$bar?

Short answer

The graphs differ at a pressure of$0.001$bar is

Step-by-step solution

Given Infomration

We have to sketch qualitatively accurate graphs of $G$ vs. $T$ for the three phases of$H_{2}O$ (ice, water, and steam) at atmospheric pressure.

Simplify

The Gibbs free energy thermodynamic potential is as follows:

$dG=-SdT+VdP+\mu dN$

$dN=0$and $dT=0$, at constant number $N$and constant temperature $T$, we get:

${\left(\frac{\partial G}{\partial P}\right)}_{T,N}=V\left(1\right)$

$dN=0$and $dP=0$, at constant number $N$and constant pressure $P$, we get:

${\left(\frac{\partial G}{\partial T}\right)}_{P,N}=-S\left(2\right)$

According to this equation, the slope of a curve between $G$and $T$is a negative sign of entropy, and the entropy varies from phase to phase; the steam has the highest inclination, followed by the liquid, and finally the ice; at pressures of $1$bar, the boiling point occurs at $100°C$, so the Gibbs free energy for the steam and the liquid is the same, implying that the two curves intersect at this point; and the freezing point occurs at

Explanation

The water is never found in liquid form at $P=0.001$bar, and the ice sublimes directly to steam at a temperature less than $0^o$. Relation $\left(1\right)$shows that the Gibbs free energy for each phase will decrease, but the gas phase will decrease the fastest because the volume of the gas is much greater, so the graph looks like this: