# What mass of water is present as a liquid when equilibrium is established?

$1.00~\mathrm{g}$ of water is introduced into a $5.00~\mathrm{L}$ flask at $50~^\circ\mathrm{C}$. What mass of water is present as a liquid when equilibrium is established? Also you know $92.5~\mathrm{mmHg}$ (vapor pressure).

1. $0.083~\mathrm{g}$
2. $0.41~\mathrm{g}$
3. $0.59~\mathrm{g}$
4. $0.91~\mathrm{g}$

I need thorough explanation for each step because I'm new to this. This question is in the gases section of the book, so I think I need to apply the ideal gas law, but I don't know.

With the help from the comment I came up with the following solution:

\begin{align} \frac{92.5}{760}~\mathrm{atm} \cdot 5.00~\mathrm{L} &= n \cdot 0.08206 \cdot 323~\mathrm{K}\\ n &= 0.023~\mathrm{mol} \end{align}

Since $1~\mathrm{mol}$ of $\ce{H2O}$ is $18~\mathrm{g}$, the answer is $0.414~\mathrm{g}$, or (2) $0.41~\mathrm{g}$.

The given temperature of $$T=50\ \mathrm{^\circ C}$$ and the given pressure of $$p=92.5\ \mathrm{mmHg}$$ approximately correspond to equilibrium conditions for liquid water and steam. (Strictly speaking, the corresponding saturation pressure for a given temperature of $$T=50.000\ \mathrm{^\circ C}$$ is $$p=92.647\ \mathrm{mmHg}$$, which could be rounded to $$p=92.6\ \mathrm{mmHg}$$ and not $$p=92.5\ \mathrm{mmHg}$$; in the opposite direction, however, the corresponding saturation temperature for a given pressure of $$p=92.500\ \mathrm{mmHg}$$ is $$T=49.968\ \mathrm{^\circ C}$$, which could be rounded to $$T=50\ \mathrm{^\circ C}$$.) The question does not explain whether these values describe the initial or final state; however, since the values correspond to equilibrium conditions, we may assume that these values apply to the final state when equilibrium is established (e.g. this can be achieved by keeping the closed system at a constant temperature of $$T=50\ \mathrm{^\circ C}$$).

Strictly speaking, the question doesn’t even explain whether the added water initially is liquid or steam; however, this is not relevant for the final state. If the water is added as liquid, a part of it evaporates until the equilibrium state is reached in the closed container; if the water is introduced as steam, a part of it condenses until the equilibrium state is reached. Therefore, when the defined equilibrium is established, the container contains certain amounts of liquid water and steam irrespective of the initial conditions.

The question does not mention any air in the container. For simplicity’s sake, we may assume that the container has been evacuated before the experiment and contains only liquid water and vapour.

The available volume of the container is reduced by the volume of the liquid water. However, since the density of liquid water at the given temperature and pressure is about $$\rho=988\ \mathrm{kg\ m^{-3}}$$, the given mass of $$m=1.00\ \mathrm g$$ corresponds to a maximum volume of $$V_\text{liquid}=1.01\ \mathrm{ml}=0.00101\ \mathrm l$$, assuming that no water has evaporated. Considering that the total volume is given as $$V=5.00\ \mathrm l$$ (note the number of significant digits), the difference caused by the liquid water is not significant and may be neglected.

Note that the use of the non-SI unit “conventional millimetre of mercury” (unit symbol: mmHg) is deprecated; the use of SI units is to be preferred. $$1\ \mathrm{mmHg}\approx 133.3224\ \mathrm{Pa}$$ You might want to convert the given values to coherent SI units, or find a value for the molar gas constant $$R$$ that is expressed in a suitable unit.

The following data are given:

• Temperature $$T=50\ \mathrm{^\circ C}=323.15\ \mathrm K$$
• Pressure $$p=92.5\ \mathrm{mmHg}=12332.322\ \mathrm{Pa}$$
• Volume $$V=5.00\ \mathrm l=0.00500\ \mathrm{m^3}$$

The value of the molar gas constant is $$R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$$.[source]

The amount of vapour $$n$$ in the container may be estimated using the ideal gas law \begin{align} p\cdot V&=n\cdot R\cdot T\\[6pt] n&=\frac{p\cdot V}{R\cdot T}\\[6pt] &=\frac{12332.322\ \mathrm{Pa}\times0.00500\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times323.15\ \mathrm K}\\[6pt] &=0.022949674\ \mathrm{mol} \end{align} (Note that this intermediate result is given with an excessive number of digits. It is deliberately not rounded in order to avoid round-off errors in subsequent calculations.)

Since the molar mass of water is $$M=18.01528\ \mathrm{g\ mol^{-1}}$$, the corresponding mass of vapour is \begin{align} m&=M\cdot n\\[6pt] &=18.01528\ \mathrm{g\ mol^{-1}}\times0.022949674\ \mathrm{mol}\\[6pt] &=0.413444803\ \mathrm g\\[6pt] &\approx0.41\ \mathrm g \end{align} (Note that this final result is rounded to two significant digits. The number of significant digits is estimated in view of the fact that the mass of water $$m=1.00\ \mathrm g$$, the volume $$V=5.00\ \mathrm l$$, and the pressure $$p=92.5\ \mathrm{mmHg}$$ are given with three significant digits, but the temperature $$T=50\ \mathrm{^\circ C}$$ is only given with two significant digits.)

This result is the mass of vapour $$m_\text{vapour}$$ in the container, whereas the question is about the mass of liquid water $$m_\text{liquid}$$, which can be calculated from the given total mass of water $$m_\text{total}=1.00\ \mathrm{g}$$. \begin{align} m_\text{total}&=m_\text{vapour}+m_\text{liquid}\\[6pt] m_\text{liquid}&=m_\text{total}-m_\text{vapour}\\[6pt] &=1.00\ \mathrm g-0.41\ \mathrm g\\[6pt] &=0.59\ \mathrm g \end{align}

Therefore, the correct answer is option 3) $$0.59\ \mathrm g$$.

By way of comparison, precise engineering calculations for water and steam usually do not rely on the ideal gas law but use so-called steam tables. We may use such steam tables to check the accuracy of our estimate that has been based on the ideal gas law.

For example, for a temperature of $$T=50\ \mathrm{^\circ C}$$ at equilibrium, we find the following values in REFPROP – NIST Standard Reference Database 23, Version 9.0:

• Pressure $$p=12352\ \mathrm{Pa}$$
• Liquid density $$\rho_\text{liquid}=988.00\ \mathrm{kg\ m^{-3}}=988.00\ \mathrm{g\ l^{-1}}$$
• Vapour density $$\rho_\text{vapour}=0.083147\ \mathrm{kg\ m^{-3}}=0.083147\ \mathrm{g\ l^{-1}}$$

Furthermore, the database includes various other thermodynamic and transport properties, which are not required for the following calculations.

If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $$T=50\ \mathrm{^\circ C}$$ at equilibrium can be obtained using the input “water boiling at 50 °C”:

• Pressure $$p=12352\ \mathrm{Pa}$$
• Liquid density $$\rho_\text{liquid}=988\ \mathrm{kg\ m^{-3}}=988\ \mathrm{g\ l^{-1}}$$
• Vapour density $$\rho_\text{vapour}=0.08315\ \mathrm{kg\ m^{-3}}=0.08315\ \mathrm{g\ l^{-1}}$$ Note: The indicated values for STP are not in accordance with IUPAC recommendations.

We know that the mass balance in the container is $$m_\text{total}=m_\text{vapour}+m_\text{liquid}\tag{1}$$ where $$m_\text{total}=1.0000\ \mathrm g$$ is the total mass of water (liquid and vapour) in the container (in this example, the precision of this value is arbitrarily increased to five significant digits).

Since density is defined as $$\rho=\frac mV$$ the corresponding volume balance can also be expressed in terms of mass: \begin{align} V_\text{total}&=V_\text{vapour}+V_\text{liquid}\\[6pt] &=\frac{m_\text{vapour}}{\rho_\text{vapour}}+\frac{m_\text{liquid}}{\rho_\text{liquid}}\tag{2} \end{align} where $$V_\text{total}=5.0000\ \mathrm l$$ is the total volume of the container (in this example, the precision of this value is arbitrarily increased to five significant digits).

Solving the system of the equations $$\text(1)$$ and $$\text(2)$$ yields the solutions

\begin{align} m_\text{vapour}&=\frac{\rho_\text{vapour}\cdot\left(V_\text{total}\cdot\rho_\text{liquid}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt] &=\frac{0.083147\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times988.00\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt] &=0.41569\ \mathrm g \end{align}

and

\begin{align} m_\text{liquid}&=-\frac{\rho_\text{liquid}\cdot\left(V_\text{total}\cdot\rho_\text{vapour}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt] &=-\frac{988.00\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times0.083147\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt] &=0.58431\ \mathrm g \end{align}

These results confirm that the above-mentioned estimates $$m_\text{vapour}\approx0.41\ \mathrm g$$ and $$m_\text{liquid}\approx0.59\ \mathrm g$$, which have been obtained using the ideal gas law and neglecting the volume of the liquid water, are reasonable and sufficient in order to answer the question.