Question

(a) The van der Waals parameters for helium are \(a=\) \(3.41 \times 10^{-2} \mathrm{~L}^{2} \cdot\) atm \(\cdot \mathrm{mol}{ }^{-2}\) and \(b=2.38 \times 10^{-2} \mathrm{~L}^{-\mathrm{mol}^{-1}}\). Calculate the apparent volume (in \(\mathrm{pm}^{3}\) ) and radius (in pm) of a helium atom as determined from the van der Waals parameters. (b) Estimate the volume of a helium atom on the basis of its atomic radius. (c) How do these quantities compare? Should they be same? Discuss.

Short answer

The apparent volume of a helium atom is significantly larger than the volume calculated based on the van der Waals radius because 'b' accounts for additional space taken by van der Waals forces. They should differ, as they represent different physical concepts.

Step-by-step solution

- Understanding the Parameters

The van der Waals parameters 'a' and 'b' represent corrections for the volume and pressure of a gas, respectively. Parameter 'a' corrects for the attractive forces between gas molecules, while 'b' corrects for the volume occupied by the gas molecules themselves. In this exercise, 'a' is not used. To calculate the apparent volume of a helium atom, we use the parameter 'b', which represents the excluded volume per mole of atoms.

- Calculating the Apparent Volume

The apparent volume ('V_{app}') can be calculated from the parameter 'b' by considering that 'b' represents the volume occupied by one mole of gas molecules (Avogadro's number, NA, of molecules). Since 'b' is given per mole, we can find the volume for a single atom by dividing 'b' by Avogadro's number (NA = 6.022 × 10^23 mol^-1): \[V_{app} = \frac{b}{N_A}\].

- Converting to Appropriate Units

Since we want the volume in picometers cubed (pm^3), we need to convert the value of 'b' from liters per mole to cubic picometers (pm^3) per atom. Note that 1 L = 10^3 cm^3, and 1 cm = 10^10 pm, so 1 L = (10^10 pm)^3 = 10^30 pm^3. \[V_{app} = \frac{2.38 \times 10^{-2} \mathrm{L/mol} \times 10^{30} \mathrm{pm}^3/\mathrm{L}}{6.022 \times 10^{23} \mathrm{mol}^{-1}}\].

- Calculating the Apparent Radius

Assuming the helium atom can be modeled as a sphere, the apparent radius ('r') can be determined from the apparent volume using the formula for the volume of a sphere: \[V = \frac{4}{3}\pi r^3\]. Solving for the radius, we have \[r = \left( \frac{3V_{app}}{4\pi} \right)^{\frac{1}{3}}\].

- Calculate the Volume of a Helium Atom Based on its Radius

To calculate the volume ('V_{atomic}') based on the given atomic radius, we use the formula for the volume of a sphere where 'r' is the atomic radius of a helium atom: \[V_{atomic} = \frac{4}{3}\pi r^3\].

- Comparison and Discussion

To compare the calculated apparent volume (V_{app}) with the volume based on the atomic radius (V_{atomic}), we analyze these values to see if they are in the same order of magnitude. These volumes provide us with different information: V_{app} illustrates the volume actually 'unavailable' to other atoms due to van der Waals forces, while V_{atomic} estimates the volume of the space an individual atom occupies. They are not necessarily expected to be the same since 'b' includes additional space due to intermolecular forces.

Key concepts

Atomic Radius

The atomic radius is a measure of the size of an atom. Conceptually, it can be thought of as the distance from the nucleus to the outer boundary of the surrounding cloud of electrons. However, because the electron cloud does not have a hard edge, the atomic radius is often defined in terms of how close one atom can get to another before the electron clouds repel each other.

In the given exercise, the atomic radius is deduced from the van der Waals parameter 'b', which takes into account the volume occupied by a mole of gas and therefore implicates the space an individual atom takes up. By dividing 'b' by Avogadro's number, the volume for a single atom is obtained, and through calculations assuming the atom is a sphere, the radius is derived. This radius, however, may be larger than the atomic radius measured by other methods (such as covalent, metallic, or ionic radii) due to the inclusion of intermolecular forces that repel neighboring atoms.

Molar Volume

Molar volume is defined as the volume that one mole of a substance occupies at a given temperature and pressure. It is a helpful concept in understanding the properties of gases as well as relating to the size of molecules in a substance. In the context of the van der Waals equation, the parameter 'b' is sometimes referred to as the effective molar volume of the gas molecules, as it represents the volume occupied by one mole of atoms or molecules minus the space where other molecules cannot enter due to repulsion.

The molar volume can inform us about the compactness or the dispersion of the atoms within a substance, and it's directly related to Avogadro's number when it comes to determining the volume occupied by a single atom. It is crucial for understanding gas behaviors under different conditions, which in turn can influence how we interpret atomic and molecular measurements such as those performed in this exercise.

Avogadro's Number

Avogadro's number, denoted by NA, is one of the foundational constants in chemistry and physics. It is defined as the number of constituent particles (usually atoms or molecules) in one mole of a substance. This value is approximately 6.022 × 10^23 mol^-1. Avogadro's number allows chemists and physicists to relate macroscopic quantities of a substance to the number of atoms or molecules and vice versa.

In the exercise at hand, Avogadro's number is utilized to convert the van der Waals 'b' parameter, which relates to a mole of atoms, into a volume for a single atom. This conversion is essential for understanding the scale of atoms and translating between the macroscopic and microscopic views of matter. It provides a link between the tangible (measurable) quantity of a gas and the number of particles it contains.