0.23 Free energy and thermodynamic equilibrium  (Page 4/6)

Now let’s determine the pressure at 25 ˚C at which the liquid and vapor are in equilibrium. If we cool the liquid-vapor equilibrium from 100 ºC to 25 ºC while maintaining equilibrium, we observe that the vapor pressure of the water drops to 23.8 torr. This means that, when the temperature is 25 ºC and the pressure is 23.8 torr, ∆G = 0 and ∆S - ∆H/T=0. We have already calculated that 8.55 kJ/mol, and therefore is not zero, so apparently ∆G depends on the pressure. Since ∆G = ∆H – T∆S, then either ∆S or ∆H changes when we change the pressure. Experiments tell us that pressure has little effect on ∆H. This makes sense: in the gas phase, the molecules do not interact, so unless we increase the pressure enormously, changing the separation of the gas particles does not affect their energy. ∆S, it turns out, does depend strongly on pressure.

Why does entropy depend so strongly on pressure? From Boyle’s Law, we know that 1 mole of water vapor occupies a much larger volume at a low pressure like 23.8 torr than it does at the considerably higher vapor pressure of 1 atm. In that much larger volume, the water molecules have a much larger space to move in, so the number of microstates for the water molecules must be larger in a larger volume. Therefore, the entropy of one mole of water vapor is larger when the volume is larger and the pressure is lower. This means that ∆S for the evaporation of one mole of water is greater when that one mole evaporates to a lower pressure.

We need to calculate how much the entropy of one mole of a gas increases as we decrease the pressure. To do so, we will use our equation S = k ln W. We can reasonably assume that the number of microstates W for the gas molecules is proportional to the volume V, because the larger the volume, the more places there are for the molecules to be. From the previous study, the entropy is given by S = k ln W, which means that S must also be proportional to ln V. Therefore, we can say that

S(V ₂ ) – S(V ₁ ) = R ln W ₂ – R ln W ₁ = R ln W ₂ / W ₁ = R ln V ₂ /V ₁

Since we are interested in the variation of S with pressure rather than volume, we can use Boyle’s Law, which states that for a fixed temperature, the volume of one mole of gas is inversely proportional to its pressure. This tells us how entropy varies with pressure:

S(P ₂ ) – S(P ₁ ) = R ln P ₁ /P ₂ = – R ln P ₂ /P ₁

Let’s apply this to water vapor. We know that the entropy of one mole of water vapor at 1.00 atm pressure is S˚ = 188.8 J/K. For a pressure of 23.8 torr = 0.0313 atm, Equation (5) shows that S(23.8 torr) = 217.6 J/K for one mole of water vapor. S changes quite a lot as P changes!

From this, we can calculate ΔS for the evaporation of one mole of water vapor to 23.8 torr at 25 ˚C: ∆S = S(water vapor at 23.8 torr) - S(liquid water) = 217.6 J/mol·K – 69.9 J/mol·K = 147.6 J/mol·K.

At equilibrium, ∆S - ∆H/T = 0. When we plug in this value of ∆S and the previous value of ∆H as well as the temperature 298.15 K, we get 0.00 J/mol·K. Or we can calculate ΔG = ΔH - TΔS = 44.0 kJ – (298.15K)(147.6 J/K) = 0.00 kJ. Either of these is the condition we expected for equilibrium. This confirms our observations.