Subfield: Thermodynamics

Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or $\mathrm{B})$ of their saturated vapors obey, to a good approximation, the formula:

$$
\ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B}
$$

where $p_{o}$ denotes the normal atmospheric pressure, $T$ - the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}$ ( $i=\mathrm{A}$ or $\left.\mathrm{B}\right)$ - certain constants depending on the liquid. (The symbol $\ln$ denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots)$

The values of the ratio $p_{i} / p_{0}$ for the liquids $\mathrm{A}$ and $\mathrm{B}$ at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1.

Table 1.1

| $t\left[{ }^{\circ} \mathrm{C}\right]$ | $i=\mathrm{A}$ | $p_{i} / p_{0}$ |
| :---: | :---: | :---: |
|  | 0.284 | 0.07278 |
| 40 | 1.476 | 0.6918 |
| 90 |  |  |

The errors of these values are negligible.
Context question:
A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$.
Context answer:
\boxed{$76.8$ , $100$}


Question: B. The liquids A and B were poured into a vessel in which the layers shown in Fig. 1.1 were formed. The surface of the liquid B has been covered with a thin layer of a non-volatile liquid $\mathrm{C}$, which is insoluble in the liquids $\mathrm{A}$ and $\mathrm{B}$ and vice versa, thereby preventing any free evaporation from the upper surface of the liquid $B$, The ratio of molecular masses of the liquids A and B (in the gaseous phase) is:

$$
\gamma=\mu_{A} / \mu_{B}=8
$$

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Fig. 1.1

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Fig. 1.2

The masses of the liquids A and B were initially the same, each equal to $m=100 \mathrm{~g}$. The heights of the layers of the liquids in the vessel and the densities of the liquids are small enough to make the assumption that the pressure in any point in the vessel is practically equal to the normal atmospheric pressure $p_{0}$.

The system of liquids in the vessel is slowly, but continuously and uniformly, heated. It was established that the temperature $t$ of the liquids changed with time $\tau$ as shown schematically in the Fig. 1.2.

Determine the temperatures $t_{1}$ and $t_{2}$ corresponding to the horizontal parts of the diagram and the masses of the liquids $\mathrm{A}$ and $\mathrm{B}$ at the time $\tau_{1}$. The temperatures should be rounded to the nearest degree (in ${ }^{\circ} \mathrm{C}$ ) and the masses of the liquids should be determined to one-tenth of gram.

REMARK: Assume that the vapors of the liquids, to a good approximation,

(1) obey the Dalton law stating that the pressure of a mixture of gases is equal to the sum of the partial pressures of the gases forming the mixture and

(2) can be treated as perfect gases up to the pressures corresponding to the saturated vapors

Solution:
PART B

As the liquids are in thermal contact with each other, their temperatures increase in time in the same way.

At the beginning of the heating, what corresponds to the left sloped part of the diagram, no evaporation can occur. The free evaporation from the upper surface of the liquid B cannot occur - it is impossible due to the layer of the non-volatile liquid $\mathrm{C}$. The evaporation from the inside of the system is considered below.



Let us consider a bubble formed in the liquid $\mathrm{A}$ or in the liquid $\mathrm{B}$ or on the surface that separates these liquids. Such a bubble can be formed due to fluctuations or for many other reasons, which will not be analyzed here.

The bubble can get out of the system only when the pressure inside it equals to the external pressure $p_{0}$ (or when it is a little bit higher than $p_{0}$ ). Otherwise, the bubble will collapse.

The pressure inside the bubble formed in the volume of the liquid A or in the volume of the liquid B equals to the pressure of the saturated vapor of the liquid A or B, respectively. However, the pressure inside the bubble formed on the surface separating the liquids A and B is equal to the sum of the pressures of the saturated vapors of both these liquids, as then the bubble is in a contact with the liquids $\mathrm{A}$ and $\mathrm{B}$ at the same time. In the case considered the pressure inside the bubble is greater than the pressures of the saturated vapors of each of the liquids $\mathrm{A}$ and $\mathrm{B}$ (at the same temperature).

Therefore, when the system is heated, the pressure $p_{0}$ is reached first in the bubbles that were formed on the surface separating the liquids. Thus, the temperature $t_{1}$ corresponds to a kind of common boiling of both liquids that occurs in the region of their direct contact. The temperature $t_{1}$ is for sure lower than the boiling temperatures of the liquids $\mathrm{A}$ and $\mathrm{B}$ as then the pressures of the saturated vapors of the liquids $\mathrm{A}$ and $\mathrm{B}$ are less then $p_{0}$ (their sum equals to $p_{0}$ and each of them is greater than zero).

In order to determine the value of $t_{1}$ with required accuracy, we can calculate the values of the sum of the saturated vapors of the liquids A and B for several values of the temperature $t$ and look when one gets the value $p_{0}$.

From the formula given in the text of the problem, we have:

$$
\begin{aligned}
& \frac{p_{A}}{p_{0}}=e^{\frac{\alpha_{A}}{T}+\beta_{A}}, \\
& \frac{p_{B}}{p_{0}}=e^{\frac{\alpha_{B}}{T}+\beta_{B}} .
\end{aligned}
$$

$p_{A}+p_{B}$ equals to $p_{0}$ if

$$
\frac{p_{A}}{p_{0}}+\frac{p_{B}}{p_{0}}=1
$$

Thus, we have to calculate the values of the following function:

$$
y(x)=e^{\frac{\alpha_{A}}{t+t_{0}}+\beta_{A}}+e^{\frac{\alpha_{B}}{t+t_{0}}+\beta_{B}}
$$

(where $t_{0}=273.15^{\circ} \mathrm{C}$ ) and to determine the temperature $t=t_{1}$, at which $y(t)$ equals to 1 . When calculating the values of the function $y(t)$ we can divide the intervals of the temperatures $t$ by 2 (approximately) and look whether the results are greater or less than 1 .

We have:



Table 1.2

| $t$ | $y(t)$ |
| :--- | :--- |
| $40^{\circ} \mathrm{C}$ | $<1($ see Tab. 1.1$)$ |
| $77^{\circ} \mathrm{C}$ | $>1\left(\right.$ as $t_{1}$ is less than $\left.t_{b A}\right)$ |
| $59^{\circ} \mathrm{C}$ | $0.749<1$ |
| $70^{\circ} \mathrm{C}$ | $1.113>1$ |
| $66^{\circ} \mathrm{C}$ | $0.966<1$ |
| $67^{\circ} \mathrm{C}$ | $1.001>1$ |
| $66.5^{\circ} \mathrm{C}$ | $0.983<1$ |

Therefore, $t_{1} \approx 67^{\circ} \mathrm{C}$ (with required accuracy).

Now we calculate the pressures of the saturated vapors of the liquids $A$ and $B$ at the temperature $t_{1} \approx 67^{\circ} \mathrm{C}$, i.e. the pressures of the saturated vapors of the liquids $\mathrm{A}$ and $\mathrm{B}$ in each bubble formed on the surface separating the liquids. From the equations (1) and (2), we get:

$$
\begin{aligned}
& p_{A} \approx 0.734 p_{0}, \\
& p_{B} \approx 0.267 p_{0}, \\
& \left(p_{A}+p_{B}=1.001 p_{0} \approx p_{0}\right) .
\end{aligned}
$$

These pressures depend only on the temperature and, therefore, they remain constant during the motion of the bubbles through the liquid $\mathrm{B}$. The volume of the bubbles during this motion also cannot be changed without violation of the relation $p_{A}+p_{B}=p_{0}$. It follows from the above remarks that the mass ratio of the saturated vapors of the liquids A and B in each bubble is the same. This conclusion remains valid as long as both liquids are in the system. After total evaporation of one of the liquids the temperature of the system will increase again (second sloped part of the diagram). Then, however, the mass of the system remains constant until the temperature reaches the value $t_{2}$ at which the boiling of the liquid (remained in the vessel) starts. Therefore, the temperature $t_{2}$ (the higher horizontal part of the diagram) corresponds to the boiling of the liquid remained in the vessel.

The mass ratio $m_{A} / m_{B}$ of the saturated vapors of the liquids $A$ and $B$ in each bubble leaving the system at the temperature $t_{1}$ is equal to the ratio of the densities of these vapors $\rho_{A} / \rho_{B}$. According to the assumption 2, stating that the vapors can be treated as ideal gases, the last ratio equals to the ratio of the products of the pressures of the saturated vapors by the molecular masses:

$$
\frac{m_{A}}{m_{B}}=\frac{\rho_{A}}{\rho_{B}}=\frac{p_{A} \mu_{A}}{p_{B} \mu_{B}}=\frac{p_{A}}{p_{B}} \mu
$$

Thus,

$$
\frac{m_{A}}{m_{B}} \approx 22.0
$$

We see that the liquid A evaporates 22 times faster than the liquid B. The evaporation of $100 \mathrm{~g}$ of the liquid A during the "surface boiling" at the temperature $t_{1}$ is associated with the



evaporation of $100 \mathrm{~g} / 22 \approx 4.5 \mathrm{~g}$ of the liquid B. Thus, at the time $\tau_{1}$ the vessel contains 95.5 $\mathrm{g}$ of the liquid $\mathrm{B}$ (and no liquid $\mathrm{A}$ ). The temperature $t_{2}$ is equal to the boiling temperature of the liquid $\mathrm{B}: t_{2}=100^{\circ} \mathrm{C}$.

Final answer: $67$ , $100$ $^{\circ}\mathrm{C}$