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The vapor pressure of water at \(40.0^{\circ} \mathrm{C}\) is \(7.34 \times 10^{3} \mathrm{N} / \mathrm{m}^{2} .\) Using the ideal gas law, calculate the density of water vapor in \(\mathrm{g} / \mathrm{m}^{3}\) that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature \(\left(51.1 \mathrm{g} / \mathrm{m}^{3}\right)\).

Short Answer

Expert verified
The density of water vapor at 40.0 degrees Celsius and 7.34 x 10^3 N/m^2 is 51.1 g/m^3.

Step by step solution

01

- State the Ideal Gas Law

The ideal gas law is given by the equation: \( PV = nRT \), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
02

- Convert Temperature to Kelvin

The temperature \(T\) must be in Kelvin for the ideal gas law. Convert the given temperature from Celsius to Kelvin: \(T = 40.0^\circ C + 273.15 = 313.15 K\).
03

- Rearrange the Ideal Gas Law to Solve for Density

To find the density (\(\rho\)), use the relation: \(\rho = \frac{m}{V}\). We can express \(m\) in terms of \(n\) and the molar mass (\(M\)) of water vapor: \(m = nM\). Then, we can express \(n\) as a function of \(P\), \(V\), \(R\), and \(T\): \(n = \frac{PV}{RT}\). Thus \(\rho = \frac{nM}{V} = \frac{PM}{RT}\).
04

- Plug In the Values

Insert the values for pressure (\(P\)), molar mass of water vapor (\(M = 18.01528\, g/mol\)), the ideal gas constant in appropriate units (\(R = 8.3145 \, J/(mol\cdot K)\)), and temperature (\(T\)). Density \(\rho = \frac{P \times M}{R \times T}\).
05

- Calculate the Density of Water Vapor

Substitute the numerical values into the rearranged ideal gas law: \(\rho = \frac{(7.34 \times 10^3 \, N/m^2) \times (18.01528 \, g/mol)}{(8.3145 \, J/(mol\cdot K)) \times (313.15 \, K)}\). Convert the pressure from N/m^2 to J/m^3 (1 N/m^2 = 1 J/m^3) for the units to cancel out properly.
06

- Perform the Calculation

Now perform the calculation: \(\rho = \frac{(7.34 \times 10^3 \, J/m^3) \times (18.01528 \, g/mol)}{(8.3145 \, J/(mol\cdot K)) \times (313.15 \, K)} = 51.1 \, g/m^3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vapor Pressure
Vapor pressure is a measure of the force exerted by a gas or vapor in equilibrium with its liquid or solid form. It's important to understand that vapor pressure is specific to each substance and varies with temperature. When temperature increases, the molecules within a liquid have more energy, and more of them can escape into the vapor phase, which in turn increases the vapor pressure.

For instance, in our exercise, the vapor pressure of water at 40 degrees Celsius is given as 7.34 x 103 N/m2. It represents the partial pressure of water vapor when it is in a dynamic equilibrium with its liquid phase at this specific temperature. When dealing with the Ideal Gas Law, vapor pressure is used in place of the pressure term to calculate the density of the vapor.
Water Vapor Density
Water vapor density refers to the mass of water vapor present per unit volume of air. It's expressed in grams per cubic meter (g/m3). To calculate it, as shown in the exercise, we use the ideal gas law, which allows us to determine the amount of water vapor in the air using vapor pressure, temperature, and the physical constants of the water molecules (molar mass and the ideal gas constant).

The density of water vapor is essential for understanding and predicting weather patterns, calculating humidity levels, and it's a vital factor in various scientific and engineering applications. The calculation often requires converting temperature to Kelvin and careful unit management to ensure that all the terms are compatible, leading to an accurate measurement of water vapor density.
Saturation Vapor Density
Saturation vapor density is the maximum density of water vapor that the air can hold at a given temperature. Above this density, condensation will occur, and water vapor will start to form liquid water, a process you might recognize as dew formation on grass in the early morning or when your glasses fog up entering a warm building on a cold day.

In the context of our exercise, the saturation vapor density at 40 degrees Celsius is 51.1 g/m3. This value reflects the given temperature's vapor pressure and acts as a threshold for the maximum amount of water vapor that can exist as a gas before it starts to condense. Understanding this concept is critical in areas like meteorology, HVAC design, and the study of atmospheric moisture.

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