Problem 80 - Chemistry and Chemical Reactivity

Question

If you have a sample of water in a closed container, some of the water will evaporate until the pressure of the water vapor, at \(25^{\circ} \mathrm{C},\) is \(23.8 \mathrm{mm}\) Hg. How many molecules of water per cubic centimeter exist in the vapor phase?

Step-by-Step Solution

Answer

The vapor phase contains approximately \(7.71 \times 10^{17}\) water molecules per cubic centimeter.

Step 1: Understanding the Problem

We need to find the number of water molecules in the vapor phase per cubic centimeter, given that the vapor pressure at 25 degrees Celsius is 23.8 mm Hg.

Step 2: Converting Pressure Units

Convert the given vapor pressure from mm Hg to atm using the conversion factor: 1 atm = 760 mm Hg. \[P = \frac{23.8}{760} \approx 0.0313\, \text{atm}\]

Step 3: Using the Ideal Gas Law

Apply the ideal gas law to relate the volume, pressure, temperature, and amount of gas: \[PV = nRT\]where \(P\) is pressure in atm, \(V\) is volume in liters, \(n\) is moles, \(R\) is the ideal gas constant \(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\), and \(T\) is temperature in Kelvin (\(T = 25 + 273.15 = 298.15 \, \text{K}\)).Set \(V = 1 \, \text{L}\) for simplicity. Then, solve for \(n\) (moles of water vapor): \[ n = \frac{PV}{RT} = \frac{0.0313 \times 1}{0.0821 \times 298.15} \approx 0.00128 \, \text{mol} \]

Step 4: Calculating Number of Molecules

Convert moles to number of molecules using Avogadro's number, \(6.022 \times 10^{23}\, \text{molecules/mol}\): \[ n\, = 0.00128 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 7.71 \times 10^{20} \, \text{molecules}\, \text{per cm}^3\]

Step 5: Converting to Per Cubic Centimeter

Since we used 1 liter to simplify calculations, we need to express the result in molecules per cm³ by dividing by 1000 (1 L = 1000 cm³):\[ \frac{7.71 \times 10^{20}}{1000} = 7.71 \times 10^{17} \text{ molecules/cm}^3\]

Key Concepts

Avogadro's Number

Avogadro's Number is a crucial constant in chemistry that links the macroscopic world we see to the molecular world we can't. It is defined as the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. This number is a fundamental constant and is approximately \(6.022 \times 10^{23}\). It allows scientists to count particles by weighting them, which is essential in chemical calculations to link mass and the number of entities. For example, if you have a mole of water, this means there are \(6.022 \times 10^{23}\) water molecules. In our problem, this value is used to convert moles of water vapor into the actual number of water molecules in a given volume.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. It represents a measure of a fluid's volatility or tendency to evaporate.
As temperature increases, more of the liquid's molecules can overcome the enthalpy of vaporization and transition to the vapor phase, increasing vapor pressure. In our exercise, the vapor pressure of water at \(25^{\circ} \text{C}\) is given as \(23.8 \text{ mm Hg}\). This indicates how much pressure the vapor of water molecules exerts in a closed container at this specific temperature. Understanding these properties is essential in calculating how many molecules will be present in a vapor phase.

Volume Conversion

Volume conversion is a process often necessary in chemistry, especially when dealing with gases in different units of volume. In the context of our problem, volume conversion helps us express results in units more convenient for solving a problem, such as cubic centimeters.
There are common conversions such as \(1 \text{ L} = 1000 \text{ cm}^3\). Converting volumes ensures that when you calculate using the Ideal Gas Law, the outputs can be as precise and relevant as needed for the problem. In our steps, the volume was initially considered in liters for simplicity but was finally converted to cubic centimeters for the correct expression of molecules per unit volume.

Pressure Units

Pressure units are fundamental in applying the Ideal Gas Law, and understanding their conversion is essential in many scientific problems. The most frequently used units include atmospheres (atm), millimeters of mercury (mm Hg), and pascals (Pa).
The given problem initially provides pressure in mm Hg. For use in the Ideal Gas Law, it was converted into atmospheres using the relationship \(1 \textrm{atm} = 760 \textrm{mm Hg}\). Converting to atm is crucial because the gas constant \(R\), used in the Ideal Gas Law, is typically expressed in terms of atm. Understanding these conversions helps ensure accuracy in calculating gas-related properties.