Understanding Water Vapour Density in the Atmosphere

This question asks us to determine the density of water vapour in an urban atmosphere given its pressure and temperature. To solve this, we can use the ideal gas law, which relates pressure, volume, temperature, and the amount of substance (or mass/density).

Applying the Ideal Gas Law to Find Density

The ideal gas law is commonly expressed as:

\(PV = nRT\)

where:

• \(P\) is the pressure of the gas
• \(V\) is the volume of the gas
• \(n\) is the number of moles of the gas
• \(R\) is the ideal gas constant
• \(T\) is the temperature of the gas in Kelvin

We know that the number of moles (\(n\)) is equal to the mass (\(m\)) divided by the molar mass (\(M\)):

\(n = \frac{m}{M}\)

Substituting this into the ideal gas law equation:

\(PV = \left(\frac{m}{M}\right)RT\)

We are looking for density (\(\rho\)), which is defined as mass per unit volume:

\(\rho = \frac{m}{V}\)

We can rearrange the ideal gas equation to solve for \(m/V\):

\(P = \frac{m}{V} \left(\frac{RT}{M}\right)\)

So, the density (\(\rho\)) is:

\(\rho = \frac{PM}{RT}\)

This formula allows us to calculate the density of the water vapour using the given pressure and temperature, along with the molar mass of water and the ideal gas constant.

Converting Units and Gathering Constants

Before plugging the values into the formula, we need to ensure all units are consistent (SI units are standard):

• Pressure (P): Given as 6.0 mb (millibars). We need to convert this to Pascals (Pa).
  1 bar = \(10^5\) Pa
  1 millibar (mb) = \(10^{-3}\) bar = \(10^{-3} \times 10^5\) Pa = \(10^2\) Pa = 100 Pa
  So, \(P = 6.0 \text{ mb} = 6.0 \times 100 \text{ Pa} = 600 \text{ Pa}\)
• Temperature (T): Given as 27°C. We need to convert this to Kelvin (K).
  \(T(\text{K}) = T(\text{°C}) + 273.15\)
  So, \(T = 27 + 273.15 = 300.15 \text{ K}\). For calculation simplicity with common values, sometimes 273 is used, making it 300 K. Let's use 300.15 K first for accuracy, then check with 300 K if needed based on options.
• Molar Mass of Water (M): The chemical formula for water is H\(_2\)O.
  Atomic mass of Hydrogen (H) ≈ 1.008 g/mol
  Atomic mass of Oxygen (O) ≈ 15.999 g/mol
  Molar mass of H\(_2\)O = \(2 \times 1.008 + 15.999 \approx 18.015\) g/mol
  In kilograms per mole: \(M \approx 0.018015\) kg/mol. We can use 0.018 kg/mol for simplicity, which is a common approximation.
• Ideal Gas Constant (R): A standard value is \(R = 8.314 \text{ J/(mol}\cdot\text{K)}\) (or \(8.314 \text{ m}^3\cdot\text{Pa/(mol}\cdot\text{K)}\)).

Calculation of Water Vapour Density

Now we can plug the values into the formula \(\rho = \frac{PM}{RT}\):

Let's use \(P = 600 \text{ Pa}\), \(M = 0.018 \text{ kg/mol}\), \(R = 8.314 \text{ J/(mol}\cdot\text{K)}\), and \(T = 300.15 \text{ K}\).

\(\rho = \frac{600 \text{ Pa} \times 0.018 \text{ kg/mol}}{8.314 \text{ J/(mol}\cdot\text{K)} \times 300.15 \text{ K}}\)

\(\rho = \frac{10.8 \text{ Pa}\cdot\text{kg/mol}}{2496.7521 \text{ J/mol}}\)

Note that J = Pa·m³, so the units work out: \((\text{Pa}\cdot\text{kg/mol}) / (\text{Pa}\cdot\text{m}^3\text{/mol}) = \text{kg/m}^3\).

\(\rho \approx \frac{10.8}{2496.7521} \text{ kg/m}^3\)

\(\rho \approx 0.0043255 \text{ kg/m}^3\)

This can be written in scientific notation as:

\(\rho \approx 4.3255 \times 10^{-3} \text{ kg/m}^3\)

Let's check the calculation using \(T = 300 \text{ K}\) (rounded):

\(\rho = \frac{600 \text{ Pa} \times 0.018 \text{ kg/mol}}{8.314 \text{ J/(mol}\cdot\text{K)} \times 300 \text{ K}}\)

\(\rho = \frac{10.8 \text{ Pa}\cdot\text{kg/mol}}{2494.2 \text{ J/mol}}\)

\(\rho \approx 0.00432996 \text{ kg/m}^3\)

\(\rho \approx 4.330 \times 10^{-3} \text{ kg/m}^3\)

Comparing our results to the given options, the value \(4.330 \times 10^{-3} \text{ kg/m}^3\) is very close to the first option, which is approximately \(4.34 \times 10^{-3} \text{ kg/m}^3\).

Matching with Options

The calculated density is approximately \(4.33 \times 10^{-3} \text{ kg/m}^3\). Let's look at the options:

1. \(\sim 4.34 \times 10^{-3}\text{ kg/m}^3\)
2. \(\sim 1.29 \times 10^{-3}\text{ kg/m}^3\)
3. \(\sim 2.69 \times 10^{-3}\text{ kg/m}^3\)
4. \(\sim 6.31 \times 10^{-3}\text{ kg/m}^3\)

The calculated value of \(4.33 \times 10^{-3} \text{ kg/m}^3\) is closest to the first option, \(4.34 \times 10^{-3} \text{ kg/m}^3\). The minor difference could be due to using slightly different values for R, M, or T in the source calculation, or rounding in the option itself. Given the choices, option 1 is clearly the correct one.

Conclusion

Using the ideal gas law and converting units appropriately, the density of water vapour at 6.0 mb pressure and 27°C temperature is calculated to be approximately \(4.33 \times 10^{-3} \text{ kg/m}^3\), which is best matched by option 1.

Atmospheric Physics Revision Table

Additional Information on Water Vapour and Atmosphere

Water vapour is a crucial component of the Earth's atmosphere, playing a significant role in weather, climate, and the water cycle. Its concentration varies greatly depending on location, temperature, and humidity.

• Partial Pressure: In a mixture of gases like the atmosphere, each gas exerts a partial pressure, which is the pressure it would exert if it alone occupied the volume. The total atmospheric pressure is the sum of the partial pressures of all constituent gases (Dalton's Law of Partial Pressures). The 6.0 mb given in the question is the partial pressure of water vapour.
• Humidity: The amount of water vapour in the air is often described by humidity. Absolute humidity is the density of water vapour (mass of water vapour per unit volume of air), which is what we calculated here. Relative humidity is the ratio of the actual partial pressure of water vapour to the saturation vapour pressure at the same temperature.
• Saturation Vapour Pressure: There is a maximum amount of water vapour that air can hold at a given temperature. This corresponds to the saturation vapour pressure. If the partial pressure of water vapour reaches the saturation vapour pressure, condensation occurs. The saturation vapour pressure increases significantly with temperature.
• Ideal Gas Assumption: The ideal gas law is an approximation. Real gases, including water vapour, deviate from ideal behavior, especially at high pressures and low temperatures. However, for atmospheric conditions at typical temperatures and pressures, the ideal gas approximation is generally quite accurate.

Understanding the density and partial pressure of water vapour is important for fields like meteorology, atmospheric science, and environmental studies.