(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 \((a)\).

Step by step solution

01

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 \(22.4 \times 10^{3} \mathrm{cm}^{3}\).

02

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, \(L = \left(22.4 \times 10^{3}\right)^{\frac{1}{3}} = 28.2 \mathrm{cm}\).

03

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 \(r = L / d = 28.2 / \left(3.00 \times 10^{-8}\right) = 0.94 \times 10^{9}\)

04

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.

05

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 \(L = \left(18\right)^{\frac{1}{3}} = 2.62 \mathrm{cm}\).

06

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 \(r = L / d = 2.62 / \left(3.00 \times 10^{-8}\right) = 0.87 \times 10^{9}\)

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