Question

(a) Consider \(1.00 \mathrm{~mol}\) of an ideal gas at \(285 \mathrm{~K}\) and \(1.00 \mathrm{~atm}\) pressure. Imagine that the molecules are for the most part evenly spaced at the centers of identical cubes. Using Avogadro's constant and taking the diameter of a molecule to be \(3.00 \times 10^{-8} \mathrm{~cm}\), find the length of an edge of such a cube and calculate the ratio of this length to the diameter of a molecule. The edge length is an estimate of the distance between molecules in the gas. (b) Now consider a mole of water having a volume of \(18 \mathrm{~cm}^{3}\). Again imagine the molecules to be evenly spaced at the centers of identical cubes and repeat the calculation in \((a)\).

Short answer

The ratio of the edge length to the diameter of the molecule for the gas is \(0.94 \times 10^{9}\). The ratio for the water molecules is \(0.87 \times 10^{9}\).

Step-by-step solution

Calculation of the volume of the gas

As per Avogadro's law, one mole of any gas at standard temperature and pressure conditions occupies a volume of \(22.4L = 22.4 \times 10^{3} \mathrm{cm}^{3}\). Thus, the volume \(V\) of the gas is \(22.4 \times 10^{3} \mathrm{cm}^{3}\).

Calculation of edge length for gas molecules

Assume that the molecules of the gas occupy the centre of cubes having each edge length equal to \(L\). Since the volume of a cube equals edge length cubed (\(L^3\)), we can express \(L\) as the cube root of \(V\). Using the volume calculated in Step 1, \(L = \left(22.4 \times 10^{3}\right)^{\frac{1}{3}} = 28.2 \mathrm{cm}\).

Calculation of ratio of edge length to diameter of gas molecule

The diameter of the molecule is given as \(3.00 \times 10^{-8} \mathrm{cm}\). The ratio of edge length to the diameter of the molecule is therefore \(r = L / d = 28.2 / \left(3.00 \times 10^{-8}\right) = 0.94 \times 10^{9}\)

Calculation of the volume of a mole of water

In contrast to the gas, a mole of water occupies a volume of \(18 \mathrm{cm}^{3}\). This is the volume \(V\) of the water mole.

Calculation of edge length for water molecules

Following the same logic as in Step 2, the edge length \(L\) of the identical cubes for the water molecules is calculated as \(L = \left(18\right)^{\frac{1}{3}} = 2.62 \mathrm{cm}\).

Calculation of ratio of edge length to diameter for water molecules

The diameter of the molecule remains the same. The ratio of edge length to the diameter of the molecule is therefore \(r = L / d = 2.62 / \left(3.00 \times 10^{-8}\right) = 0.87 \times 10^{9}\)

Key concepts

Avogadro's Constant

In the world of chemistry, Avogadro's Constant plays a critical role in bridging the microscopic and macroscopic scales. Avogadro's Constant, denoted as NA, is approximately equal to \(6.022 \times 10^{23}\) molecules per mole. This number is significant because it allows us to count atoms or molecules in a sample by referring to macroscopic quantities like moles.

When considering ideal gas calculations, Avogadro’s Constant assists in understanding how molecules are arranged within a given volume. For example, in the case of our ideal gas with a known volume and the number of moles, the constant helps determine the spacing between molecules. By calculating the volume that one mole of a gas occupies at standard temperature and pressure (STP), which is approximately 22.4 liters, we can infer the average distance between these molecules.

Molecular Spacing

Molecular spacing refers to the average distance between molecules within a given substance. In the context of gases, these molecules are far apart compared to liquids and solids. This spacing can be visualized as molecules being at the center of cubes, with each edge representing the distance between neighboring molecules. This is particularly useful in our ideal gas scenario, where we want to find the edge length \(L\) of such a cube.

For example, when dealing with one mole of an ideal gas at standard conditions, the typical molecular spacing can be estimated from the cube root of the volume of the gas, \(22.4 \times 10^3\) cm³. This leads to an edge length \(L\) of 28.2 cm, indicating how far apart the molecules are on average within the gas.

In contrast, when considering a liquid like water, the spacing is much smaller due to the denser packing of molecules. For one mole of water, occupying only 18 cm³, the edge length becomes 2.62 cm. This smaller edge length reflects the closer packing of molecules in liquid water compared to the gas.

Volume Calculations

Volume calculations are essential to understanding the physical behavior of gases and liquids. For gases, knowing the volume occupied by a certain number of moles at given conditions helps in understanding the nature of gas molecules. The Ideal Gas Law, represented as \(PV = nRT\), sets out the relationship between pressure \(P\), volume \(V\), temperature \(T\), and moles \(n\) of a gas, where \(R\) is the gas constant.

This law provides a means to calculate the volume a gas occupies under different conditions. At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters, which is valuable for calculating molecular spacing.

In our textbook problem, we see this principle in action: the gas volume is calculated and used to determine how far apart gas molecules are. For water, the calculation uses the liquid's volume, which is much less than that of a gas at STP, to show the denser molecular packing in the liquid state.

Edge Length Ratio

The edge length ratio is a useful concept in understanding the relationship between the size of a molecule and the average spacing between them. This ratio is calculated by dividing the edge length of the cube containing the molecule by the diameter of the molecule itself, given as \(3.00 \times 10^{-8}\) cm for our problem.

In the case of the ideal gas from our example, this ratio, calculated as \(28.2/3.00 \times 10^{-8}\), results in a very large number \(0.94 \times 10^9\). This large ratio implies that the molecules are quite far apart relative to their size, typical in a gas.

For water, repeating the calculation, we use the edge length of the cube that a water molecule occupies, \(2.62\), resulting in a smaller ratio \(0.87 \times 10^9\). Although still large, this indicates significantly less spacing relative to the size of the molecule compared to the gas scenario, again highlighting the denser molecular arrangement in liquids like water.