Question

Estimate the van der Waals constant \(b\) for \(\mathrm{H}_{2} \mathrm{O}\) knowing that one kilogram of water has a volume of \(0.001 \mathrm{~m}^{3}\). The molar mass of water is \(18 \mathrm{~g} / \mathrm{mol}\).

Short answer

The estimated van der Waals constant \(b\) for water is approximately \(1.8 \times 10^{-5} \mathrm{~m}^{3/mol}\).

Step-by-step solution

Determine the number of moles in 1 kilogram of water

First, convert the mass of water from kilograms to grams (since the molar mass is given in grams). Since \(1 \mathrm{~kg} = 1000 \mathrm{~g}\), there are \(1000 \mathrm{~g}\) of water in one kilogram. The number of moles of water can be found using the molar mass \(M\), by applying the following formula: \[N = \frac{m}{M}\]where:\(m\) = mass of the substance = 1000 g\(N\) = number of moles\(M\) = molar mass of the substance = 18 g/mol

Calculate the number of moles

Substitute the given values into the formula:\[N = \frac{1000 \mathrm{~g}}{18 \mathrm{~g/mol}} = 55.56 \mathrm{~mol}\]This means there are approximately 55.56 moles of water in one kilogram.

Calculate the van der Waals constant

The volume occupied by one mole of a real gas at zero pressure according to the van der Waals equation is the van der Waals constant \(b\). As the total volume of the water is given, \(b\) can be calculated by dividing the volume by the number of moles:\[b = \frac{V_{total}}{N}\]where:\(V_{total}\) = total volume = 0.001 m^3\(N\) = number of moles = 55.56 mol

Final Calculation

Substitute the given values into the formula:\[b = \frac{0.001 \mathrm{~m}^{3}}{55.56 \mathrm{~mol}} = 1.8 \times 10^{-5} \mathrm{~m}^{3/mol}\]So, the estimated van der Waals constant \(b\) for water is approximately \(1.8 \times 10^{-5} \mathrm{~m}^{3/mol}\).

Key concepts

Molar Mass

Molar mass is a fundamental concept vital to understanding both chemistry and physics. It refers to the mass of a single mole of a substance, expressed in grams per mole (g/mol). This measurement allows us to transition from atomic scale to a macroscopic scale easily recognizable in laboratory and industrial settings. For water, the molar mass is measured at 18 g/mol. This tells you how much 6.022 x 10^23 molecules of water—that's a mole—actually weighs.

This conversion is incredibly useful when dealing with equations like the van der Waals equation or when attempting stoichiometric calculations. It connects the mass of a substance, which we can measure easily, to the number of moles, which helps us understand the number of molecules involved in reactions or behaviors under different conditions.

Number of Moles

The concept of the number of moles allows us to count particles when they're too small to see. When calculating the number of moles, remember to convert mass to grams if necessary since molar mass is usually given in grams per mole (g/mol).

For instance, in the exercise above, you needed to convert the mass of water from kilograms to grams to match the molar mass unit. Using the formula \[N = \frac{m}{M}\]where \(m\) is the mass and \(M\) is the molar mass, allows you to find the number of moles. By substituting the values, you arrive at approximately 55.56 moles for 1000 grams of water.

This figure is essential when exploring volume and pressure in gases or moving on to other calculations such as volume calculation in van der Waals equation, where having the right number of moles ensures accuracy.

Volume Calculation

Volume calculation becomes crucial, especially in the context of gases. For real gases, deviations from ideal behavior need special consideration, which is where equations and calculations like van der Waals come in. The van der Waals equation accounts for the volume occupied by gas molecules and the attractive forces between them, differing from the ideal gas law.

In our exercise, the goal was to estimate the van der Waals constant \(b\), which involves the volume calculation using \[b = \frac{V_{total}}{N}\]where \(V_{total}\) is the total volume and \(N\) the number of moles.

Dividing the total volume of 0.001 m^3 by the number of moles (55.56 mol) gives you the value of \(b\), approximating the space a mole of water molecules occupies under zero pressure. This not only helps us understand the spatial distribution of water molecules but also plays a role in exploring gas behaviors further.

Real Gas

Understanding real gases is important when working with gases that deviate significantly from ideal conditions. The van der Waals equation is one such method that helps calculate and predict the behavior of real gases by incorporating two correction factors: the volume occupied by gas molecules and the intermolecular forces between them.

Real gases differ from ideal gases as they exist in everyday life, where pressure and temperature vary, preventing gases from always behaving predictably. The van der Waals equation: \[\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT\]in which \(a\) and \(b\) are constants specific to each gas, provides more accurate predictions by accounting for these deviations.

In the application presented in the exercise, determining the constant \(b\) allows us to better understand and anticipate the molecular volume, improving calculations for practical designs and operations involving gases.