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To derive the ideal-gas equation, we assume that the volume of a gas's atoms/molecules can be neglected. Given that the atomic radius of argon \(0.097 \mathrm{nm},\) and knowing that a sphere has a volume of \(4 \pi r^{3} / 3,\) calculate the fraction of space that Ar atoms occupy in a sample of argon at STP.

Short Answer

Expert verified
Argon atoms occupy approximately 0.1% of the space in a sample at STP.

Step by step solution

01

Find the Volume of One Argon Atom

The first step involves calculating the volume of a single argon atom using the formula for the volume of a sphere: \[ V_{atom} = \frac{4}{3} \pi r^3 \] where \( r = 0.097 \) nm is the radius of the argon atom. Convert the radius to meters (1 nm = \(10^{-9}\) m) before substituting it into the equation. This gives:\[ V_{atom} = \frac{4}{3} \pi (0.097 \times 10^{-9})^3 \approx 3.82 \times 10^{-29} \ m^3 \]
02

Determine the Molar Volume of Argon at STP

STP (Standard Temperature and Pressure) conditions specify a temperature of 273.15 K and a pressure of 1 atm. Under these conditions, 1 mole of any ideal gas occupies 22.4 liters. Convert this volume to cubic meters: \[ 22.4 \ liters = 22.4 \times 10^{-3} \ m^3 \] Thus, the molar volume of argon at STP is \(22.4 \times 10^{-3} \ m^3\).
03

Calculate the Number of Argon Atoms in a Mole

One mole of a substance contains Avogadro's number of particles, which is approximately:\[ N_A = 6.022 \times 10^{23} \text{ atoms/mole} \]
04

Find the Total Volume Occupied by Argon Atoms

Multiply the volume of one argon atom by Avogadro's number to find the total volume occupied by all the argon atoms in one mole:\[ V_{total} = V_{atom} \times N_A = 3.82 \times 10^{-29} \ m^3 \times 6.022 \times 10^{23} \approx 2.30 \times 10^{-5} \ m^3 \]
05

Calculate the Fraction of Space Occupied by Argon Atoms

To find the fraction of space occupied by argon atoms, divide the total volume of the argon atoms by the molar volume:\[ \text{Fraction} = \frac{V_{total}}{V_{molar}} = \frac{2.30 \times 10^{-5} \ m^3}{22.4 \times 10^{-3} \ m^3} \approx 0.0010 \] This fraction is approximately 0.1%, indicating that the argon atoms themselves occupy a very small portion of the total space.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Argon Atom
An argon (Ar) atom is a single particle of the noble gas argon. This element is quite prevalent, making up a small but significant fraction of Earth's atmosphere.
What makes the argon atom interesting, especially in the context of the Ideal Gas Law, is its atomic structure. It has a full outer shell of electrons, which makes it very stable and inert.
When examining the size of an argon atom, its radius is about 0.097 nm. We often describe an atom’s shape as being roughly spherical, which allows us to calculate its volume using the formula for the volume of a sphere:
  • Volume, \( V_{atom} = \frac{4}{3} \pi r^3 \)
Understanding this equation is vital because it helps us comprehend how these small atomic units contribute to the behavior of gases as described by the Ideal Gas Law.
Molar Volume
Molar volume refers to the volume occupied by one mole of a substance at a given temperature and pressure.
At Standard Temperature and Pressure (STP), which is 273.15 K and 1 atm, the molar volume of an ideal gas is universally noted as 22.4 liters.
Converting liters to cubic meters, a unit often used in scientific calculations, gives us:
  • 22.4 liters equals 22.4 \( \times 10^{-3} \) m³

This measure is essential when dealing with gases, as it provides a reference point for comparing how different gases behave under identical conditions. For instance, even though gases like argon and oxygen are different in terms of mass and structure, one mole of each will still take up the same molar volume when measured at STP. Thus, molar volume serves as a baseline measurement that illustrates the collective behavior of gas molecules.
Avogadro's Number
Avogadro's Number is a fundamental constant in chemistry, representing the number of atoms, molecules, or particles in one mole of a substance.
This number is approximately \( 6.022 \times 10^{23} \).
This insight is critical because it allows us to convert between the bulk properties of substances measured in moles to the microscopic scale measured as individual atoms or molecules.
For example, if you know the volume occupied by an individual argon atom, Avogadro's number helps estimate how much space all the atoms in a mole of argon would occupy. It connects the macroscopic and microscopic worlds, aiding chemists and physicists in understanding material properties. It's this seamless transition between scales that makes Avogadro's number so vital in the realm of gas laws.
STP Conditions
Standard Temperature and Pressure (STP) conditions are a set point for the measurement of gases.
In scientific contexts, STP provides a basis for comparison and consistency.
According to the International Union of Pure and Applied Chemistry (IUPAC), STP is set at 273.15 Kelvin for temperature and 1 atmosphere (atm) for pressure.
These conditions are optimal because they reflect an easily reproducible set of circumstances that allow scientists to experiment and compare findings across different labs and studies.
  • STP ensures uniformity when studying how gases expand, contract, and occupy space.
  • This is particularly relevant when dealing with ideal gases, where volume directly relates to temperature and pressure.
Without STP, it would be challenging to establish consistent standards for evaluating gaseous behavior. This makes it a cornerstone in the studies involving the Ideal Gas Law, providing predictable and reliable bases for calculations and comparisons.

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Most popular questions from this chapter

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. \((\mathbf{e})\) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s) $$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(152.0 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

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