Question

Consider an ideal gas at 30 ∘C and 1.02 atm pressure. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube.

Answer 1

Answer: They are close to each other by 41.03 m^3

Explanation:

From Ideal gas equation, PV = nRT

Where n is negligible

R is gas constant = 8.314 J/mol.k

T = 30 + 273 = 303K

P = 1.02 * 103351.5 = 103351.5 Pascal

Then;

PV = RT

V = P/RT

V = 103351.5/(8.314*303)

V = 41.03m^3

Answer 2

Answer:

See explanation below

Explanation:

This question is incomplete, however, I answered this question on another website, so, I will assume that you need the same answer, cause is the same exercise. If it's not, then please say what's your real question in this exercise.

First of all, you need to know that 1 mole of an ideal gas always occupies 22.4 L. Therefore, using the ideal gas equation, we can solve for the volume that the gas at 1.02 atm occupies, and this will be:

(1 atm x 22.4 L /273 K) x (303 K /1.02 atm ) L = 24.37 L

Converting liter to m³:

24.37 / 1000 = 0.02437 m³

We also know by avogadros number, that 1 mole has  6.02 x 10^23 molecules, and if 1 mole at 1.02 atm has the volume above, then:

0.02437/6.02x10^23 m^3 = 4.05x10^-26 m³

With this result, we can know the length of an edge of the small cube, and we also know that the volume of a cube is a³. We have the volume that the gas occupies, so, we can solve for the length:

l = ∛4x10^-26 m³ = 3.43x10^-9 m

With this, we can compare the distance with the diameter of  a molecule which is 10^-10 m so:

distance = 3.43x10^-9 / 10^-10 = 34.33

This means that is 34 times the diameter of a molecule.