When a pure liquid is placed in a closed container, some molecules escape from the liquid and enter the vapor phase. Eventually, a dynamic equilibrium is established in which molecules leave and return to the liquid at equal rates. The pressure exerted by the vapor at equilibrium is called the vapor pressure of the liquid.
Now suppose a nonvolatile solute is dissolved in the liquid. Experimentally, the vapor pressure of the solvent decreases. Thermodynamics explains this behavior through the chemical potential of the solvent.
In an ideal solution, the chemical potential of the solvent is
\[ \mu_A=\mu_A^*+RT\ln x_A \]where \(x_A\) is the mole fraction of the solvent. Because \(x_A<1\) whenever a solute is present,
\[ \ln x_A < 0 \]and therefore the chemical potential of the solvent is lower than that of the pure liquid.
At equilibrium, the chemical potential of the solvent in the liquid phase must equal the chemical potential of the solvent in the vapor phase:
\[ \mu_A(\text{liquid}) = \mu_A(\text{vapor}) \]Assuming the vapor behaves ideally, the chemical potential of the solvent vapor is
\[ \mu_A(\text{vapor}) = \mu_A^\circ(\text{vap}) + RT\ln\!\left(\frac{p_A}{p^\circ}\right) \]Combining these expressions and comparing the solution to the pure solvent leads to
\[ p_A=x_Ap_A^* \]This relationship is known as Raoult's Law.
The law states that the vapor pressure of a solvent above an ideal solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
Raoult's Law predicts that adding a nonvolatile solute lowers the vapor pressure of the solvent because the solvent's mole fraction decreases.
For example, if the mole fraction of the solvent is
\[ x_A=0.90 \]then the vapor pressure of the solvent will be only 90% of the vapor pressure of the pure liquid.
The reduction in vapor pressure depends only on the amount of solute present, not on the chemical identity of the solute. This is why vapor-pressure lowering is classified as a colligative property.
Raoult's Law also provides the foundation for understanding freezing-point depression, boiling-point elevation, and osmotic pressure. All of these effects arise because dissolved solutes lower the chemical potential of the solvent.
Big picture: Raoult's Law connects the composition of a solution to the vapor pressure of its solvent. It provides one of the simplest demonstrations of how chemical potential controls equilibrium and serves as the starting point for understanding the colligative properties of solutions.
A 0.100 M solution of NaCl is prepared in water. If the vapor pressure of pure water at this temperature is 19.0 Torr, estimate the vapor pressure of water above the solution.
Raoult's Law gives the vapor pressure of the solvent above the solution:
\[ p_A=x_Ap_A^* \]Because NaCl dissociates into two ions,
\[ \text{NaCl}(aq)\rightarrow \text{Na}^+(aq)+\text{Cl}^-(aq) \]a 0.100 M NaCl solution contains approximately 0.200 mol of dissolved solute particles per liter of solution.
One liter of water contains approximately
\[ n_{H_2O} = \frac{1000\ \text{g}} {18.015\ \text{g mol}^{-1}} = 55.5\ \text{mol} \]Therefore, the mole fraction of water is approximately
\[ x_{H_2O} = \frac{55.5} {55.5+0.200} = 0.996 \]Now apply Raoult's Law:
\[ p_{H_2O} = (0.996)(19.0\ \text{Torr}) \] \[ p_{H_2O} = 18.9\ \text{Torr} \]Therefore, the vapor pressure above the solution is
\[ \boxed{p_{H_2O}=18.9\ \text{Torr}} \]Physical interpretation: Dissolving NaCl lowers the mole fraction of water, which lowers the chemical potential of liquid water. As a result, the equilibrium vapor pressure of water above the solution is slightly lower than that of pure water.
A liquid solution contains two volatile components, A and B. At this temperature, the vapor pressure of pure A is 200 Torr, and the vapor pressure of pure B is 400 Torr. If the mole fraction of B in the liquid is
\[ x_B=0.50 \]calculate the partial pressures of A and B, the total vapor pressure, and the mole fractions of A and B in the vapor phase.
Since the liquid contains only A and B, the mole fraction of A is
\[ x_A=1-x_B=1-0.50=0.50 \]For an ideal solution, Raoult's Law gives the partial pressure of each volatile component:
\[ p_A=x_Ap_A^* \] \[ p_B=x_Bp_B^* \]The partial pressure of A is therefore
\[ p_A=(0.50)(200\ \text{Torr}) \] \[ p_A=100\ \text{Torr} \]The partial pressure of B is
\[ p_B=(0.50)(400\ \text{Torr}) \] \[ p_B=200\ \text{Torr} \]The total vapor pressure is the sum of the partial pressures:
\[ p_{\text{total}}=p_A+p_B \] \[ p_{\text{total}} = 100\ \text{Torr} + 200\ \text{Torr} = 300\ \text{Torr} \]To find the mole fractions in the vapor phase, use Dalton's Law:
\[ y_A=\frac{p_A}{p_{\text{total}}} \] \[ y_B=\frac{p_B}{p_{\text{total}}} \]Therefore,
\[ y_A= \frac{100\ \text{Torr}} {300\ \text{Torr}} = 0.333 \] \[ y_B= \frac{200\ \text{Torr}} {300\ \text{Torr}} = 0.667 \]Thus, the vapor in equilibrium with the liquid has composition
\[ \boxed{y_A=0.333} \] \[ \boxed{y_B=0.667} \]Physical interpretation: Although A and B are present in equal amounts in the liquid phase, B has the larger pure-component vapor pressure. As a result, B contributes more strongly to the vapor pressure and is enriched in the vapor phase.
A liquid solution contains two volatile components, A and B. Assume the solution is ideal. Use Raoult's Law to calculate the partial pressures, total pressure, and vapor-phase mole fractions.
\[ p_A=x_Ap_A^* \qquad p_B=x_Bp_B^* \]| Quantity | Value |
|---|---|
| \(p_A^*\) | |
| \(p_B^*\) | |
| \(x_B\) |
Enter your calculated values below.
| Quantity | Your answer |
|---|---|
| \(p_A\) | Torr |
| \(p_B\) | Torr |
| \(p_{\text{total}}\) | Torr |
| \(y_A\) | |
| \(y_B\) | |
Big picture: Raoult's Law connects the composition of a liquid solution to the composition of the vapor in equilibrium with it. By showing how mixing lowers the chemical potential of a solvent, it provides the thermodynamic foundation for vapor-pressure lowering, freezing-point depression, boiling-point elevation, and other colligative properties.