Question

Consider liquid water in equilibrium with its vapor at \(100{ }^{\circ} \mathrm{C}\). Estimate the number of water molecules per \(\mathrm{cm}^{3}\) in (a) the liquid (density \(=0.958 \mathrm{~g} / \mathrm{cm}^{3}\) at \(\left.100{ }^{\circ} \mathrm{C}\right)\); (b) the vapor.

Short answer

Liquid: \(3.21 \times 10^{22} \text{ molecules/cm}^3\), Vapor: \(1.94 \times 10^{19} \text{ molecules/cm}^3\).

Step-by-step solution

Calculate the number of water molecules in liquid

To determine the number of water molecules in 1 cm³ of liquid water, start with its density and convert it to molecules. The density is given as 0.958 g/cm³. The molar mass of water is 18 g/mol, so the number of moles in 0.958 g is \( \frac{0.958}{18} \text{ mol} \). Use Avogadro's number \( 6.022 \times 10^{23} \text{ molecules/mol} \) to find the number of molecules: \[ \frac{0.958}{18} \times 6.022 \times 10^{23} \approx 3.21 \times 10^{22} \text{ molecules/cm}^3 \]

Use Ideal Gas Law for the vapor

For the vapor, we use the ideal gas law \( PV = nRT \). First, convert the temperature to Kelvin: \( T = 100 ^\circ C + 273.15 = 373.15 \text{ K} \). At 100 °C, the vapor pressure is 1 atm, or 101325 Pa. The gas constant \( R \) is 8.314 J/(mol K). Calculate \( n/V \), the number of moles per cubic meter: \( n/V = \frac{P}{RT} = \frac{101325}{8.314 \times 373.15} \approx 0.0322 \text{ mol/m}^3\). Convert this to cm³ and molecules: \[ \frac{0.0322}{1000000} \times 6.022 \times 10^{23} \approx 1.94 \times 10^{19} \text{ molecules/cm}^3 \]

Conclusion

We now have the number of water molecules per cm³ in both the liquid and vapor phases. The liquid has significantly more molecules, approximately \(3.21 \times 10^{22} \) molecules/cm³ compared to the vapor's \(1.94 \times 10^{19} \) molecules/cm³. This reflects the much higher density of the liquid phase.

Key concepts

Water Molecule Calculation

To understand how we calculate the number of water molecules in a given volume, we start with the density of water. Imagine you have 1 cm³ of liquid water. Its density at 100°C is 0.958 g/cm³.

First, you need to know the molar mass of water, which is 18 g/mol. This tells us how much one mole of water weighs. Next, we need to figure out how many moles are in that 0.958 g of water. You find this by dividing the mass by the molar mass: \[ \text{moles of water} = \frac{0.958}{18} \approx 0.0532 \text{ mol} \]Let's count molecules:

• The number of molecules in one mole of any substance is given by Avogadro's number, \(6.022 \times 10^{23}\) molecules/mol.
• Multiplying the number of moles by Avogadro's number gives us the number of molecules: \[ \text{molecules} = 0.0532 \times 6.022 \times 10^{23} \approx 3.21 \times 10^{22} \text{ molecules/cm}^3 \]

This gives you how densely packed the water molecules are in liquid form.

Ideal Gas Law Application

When we apply the ideal gas law to find the number of water molecules in vapor at 100°C, we are essentially looking at how gases behave under different conditions. Here's a simple breakdown of the ideal gas law: \[ PV = nRT \]Understanding the variables:

• P is the pressure (in this case, vapor pressure at 100°C, which is 1 atm or 101325 Pa).
• V is volume in cubic meters.
• n is the number of moles.
• R is the ideal gas constant, 8.314 J/(mol·K).
• T is the temperature in Kelvin, so 100°C becomes 373.15 K.

To find the number of moles per volume (n/V), we rearrange the formula: \[ \frac{n}{V} = \frac{P}{RT} = \frac{101325}{8.314 \times 373.15} \approx 0.0322 \text{ mol/m}^3 \]\

Then convert this moles per cubic meter to moles per cubic centimeter and then to molecules using Avogadro's number:\[ \text{molecules per cm}^3 \approx 0.0322 \times \frac{6.022 \times 10^{23}}{1000000} \approx 1.94 \times 10^{19} \]In the vapor phase, molecules are much more spread out, resulting in a lower density compared to the liquid phase.

Phase Equilibrium

Phase equilibrium occurs when the rates of evaporation and condensation are equal. This means that the amount of water turning into vapor is equal to the amount turning back into liquid. At this point, both phases are balanced, and the system is stable.

What does it look like?

• Even though the two phases are different in density — liquid being dense and vapor being sparse — they coexist.
• The temperature plays a crucial role. At 100°C, water and its vapor reach this equilibrium state naturally.
• The presence of a vapor pressure (1 atm at 100°C) above the liquid allows this dynamic balance.

Why is this important? Well, it helps us understand real-world phenomena, such as boiling, when the liquid-gas phase change occurs throughout the liquid. Understanding phase equilibrium provides insights into how different states of matter interact in thermal environments.