Chapter 11, Problem 143AP

A closed vessel of volume 9.6 L contains 2.0 g of water. Calculate the temperature $\left(\text{in ºC}\right)$ at which only half of the water remains in the liquid phase. (See Table 10.5 for vapor pressures of water at different temperatures.)

Interpretation Introduction

Interpretation:

The temperature, at which half of the water remains in the liquid phase in a closed container, is to be calculated.

Concept introduction:

The ideal gas equation is as follows:

$P=\frac{n\text{R}T}{V}$

Here, $P$

is the pressure, $\text{R}$

is the ideal gas constant, $T$

is the temperature, and $V$

is the volume.

Answer to Problem 143AP

Solution: $\text{55}°\text{C}$ or $\text{328 K}$.

Explanation of Solution

Given information: Volume of the vessel is $\text{9}\text{.6 L}$, mass of water is $\text{2}\text{.0 g}$.

According to the ideal gas equation;

$P=\frac{nRT}{V}$

The initial mass of water is $\text{2}\text{.0 g}$. Mass of water in the liquid state is $\text{1}\text{.0 g}$. Thus, water in the gaseous state is $\text{1}\text{.0 g}$.

Convert $\text{1}\text{.0 g}$

mass to moles by multiplying $\text{1}\text{.0 g}$

with the conversion factor, $\frac{\text{1 mol}}{\text{18}\text{.02 g}}$, and substitute it for $n$

;$0.0821\text{L}\text{.atm}/\text{K}\text{.mol}$

for $\text{R}$

; and $\text{9}\text{.6 L}$

for $V$

in the ideal gas equation as follows:

$\begin{array}{c}P=\frac{0.0821\cancel{\text{L}}\text{.atm}/\cancel{\text{K}}\text{.}\cancel{\text{mol}}\left(\text{1}\text{.0 }\cancel{\text{g}}\right)\times \left(\frac{\text{1 }\cancel{\text{mol}}}{\text{18}\text{.02 }\cancel{\text{g}}}\right)\times T\left(\cancel{\text{K}}\right)}{9.6\cancel{\text{L}}}\\ P=4.7\times {10}^{-4}T.\text{atm}\end{array}$

As, $\text{1 atm=760 mm Hg}$

Multiply $\left(4.7\times {10}^{-4}\right)T\text{atm}$

with the conversion factor $\frac{\text{760 mm Hg}}{\text{1 atm}}$ as follows:

$\left(4.7\times {10}^{-4}\right)T\cancel{\text{atm}}\times \left(\frac{\text{760 mm Hg}}{\text{1 }\cancel{\text{atm}}}\right)=0.36T\text{mm Hg}$

Assume temperature equal to $\text{40}°\text{C}$ or $313\text{ K}$, $45°\text{C}$

or $318\text{ K}$, $\text{50}°\text{C}$

or $\text{323 K}$, $\text{55}°\text{C}$ or $\text{328 K}$, and $60°\text{C}$ or $\text{333 K}$, $65°\text{C}$ or $338\text{ K}$, the vapour pressure is as follows:

$\begin{array}{l}P=\text{0}\text{.36}\times \text{313 mmHg}=\text{112}\text{.7 mmHg}\\ P=\text{0}\text{.36}\times \text{318 mmHg}=\text{114}\text{.5 mmHg}\\ P=\text{0}\text{.36}\times \text{323 mmHg}=\text{116}\text{.3 mmHg}\end{array}$

$\begin{array}{l}P=\text{0}\text{.36}\times \text{328 mmHg}=\text{118}\text{.1 mmHg}\\ P=\text{0}\text{.36}\times \text{333 mmHg}=\text{119}\text{.9 mmHg}\\ P=\text{0}\text{.36}\times \text{338 mmHg}=\text{121}\text{.7 mmHg}\end{array}$

Compare this with the actual vapour pressure at these temperatures, as follows:

$\begin{array}{l}P=55.\text{3 mmHg at 40}°\text{C}\\ P=71.9\text{ mmHg at 45}°\text{C}\\ P=\text{92}\text{.5 mmHg at 50}°\text{C}\end{array}$

$\begin{array}{l}P=118.0\text{ mmHg at 55}°\text{C}\\ P=149.4\text{ mmHg at 60}°\text{C}\\ P=187.5\text{ mmHg at 65}°\text{C}\end{array}$

The calculated pressure at $\text{55}°\text{C}$ or $\text{328 K}$, is close to the actual pressure at $\text{55}°\text{C}$ or $\text{328 K}$.

Conclusion

The temperature at which half of the water remains in the liquid phase in a closed container containing $\text{2}\text{.0 g}$

water, is $\text{55}°\text{C}$ or $\text{328 K}$.