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Chapter 13: Problem 38

At \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, 1.00 \mathrm{mol}\) of a gas occupies a volume of \(0.0224 \mathrm{m}^{3} .\) (a) What is the number density? (b) Estimate the average distance between the molecules. (c) If the gas is nitrogen \(\left(\mathrm{N}_{2}\right),\) the principal component of air, what is the total mass and mass density?

Short Answer

Expert verified
Answer: The number density of nitrogen gas is approximately 44.64 mol/m³, the average distance between molecules is about 3.30 × 10⁻⁹ m, the total mass of the gas is 28 g or 0.028 kg, and the mass density is approximately 1.25 kg/m³.

Step by step solution

01

Finding number density

To find the number density, we can use the formula for number density: Number density = number of moles/volume We are given that we have 1.00 mol of gas occupying 0.0224 m³ volume. So, we can plug these values into the formula to find the number density: Number density = 1.00 mol/0.0224 m³
02

Calculating number density

Now, we can calculate the number density: Number density ≈ 44.64 mol/m³ In part (a), we found the number density of the gas molecules.
03

Finding average distance between molecules

Now let's find the average distance between the molecules in the gas. We can use the formula for the average distance between molecules which is: Average distance = (volume of one molecule /number density)^(1/3) The volume of one molecule can be found by dividing the total volume by the number of gas molecules, which is equal to the number of moles times Avogadro's number (6.022 × 10²³ particles/mol): Volume of one molecule = (0.0224 m³) / (1.00 mol × 6.022 × 10²³ particles/mol)
04

Calculating average distance between molecules

Now, we can calculate the average distance using the values we have found: Average distance ≈ ((0.0224 m³) / (1.00 mol × 6.022 × 10²³ particles/mol))^(1/3) Average distance ≈ 3.30 × 10⁻⁹ m The estimated average distance between molecules is 3.30 × 10⁻⁹ m in part (b).
05

Calculate the mass of one molecule of \(\mathrm{N}_{2}\) gas

To find the total mass of 1.00 mol of nitrogen gas, we need to know the molecular weight of one molecule of N₂. Nitrogen has an atomic mass of approximately 14 amu, so a molecule of \(\mathrm{N}_{2}\) will have a weight of: Mass of one molecule of N₂ = 2 × 14 amu = 28 amu Now converting this mass to grams per mole: 28 amu × (1 g/mol) / (1 amu) = 28 g/mol So, the mass of 1.00 mol of nitrogen gas will be 28 g.
06

Calculate the mass density of the gas

We found the total mass of the gas in Step 5, so we can now calculate the mass density using the formula: Mass density = mass/volume Mass density = (28 g) / (0.0224 m³) Now, we can convert grams to kilograms: 28 g × (1 kg / 1000 g) = 0.028 kg
07

Calculate mass density using the mass in kilograms

Now, we can calculate the mass density in kg/m³: Mass density = (0.028 kg) / (0.0224 m³) Mass density ≈ 1.25 kg/m³ In part (c), we found the total mass of the nitrogen gas to be 28 g or 0.028 kg, and mass density to be approximately 1.25 kg/m³.

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Most popular questions from this chapter

A bubble with a volume of \(1.00 \mathrm{cm}^{3}\) forms at the bottom of a lake that is \(20.0 \mathrm{m}\) deep. The temperature at the bottom of the lake is \(10.0^{\circ} \mathrm{C} .\) The bubble rises to the surface where the water temperature is \(25.0^{\circ} \mathrm{C}\) Assume that the bubble is small enough that its temperature always matches that of its surroundings. What is the volume of the bubble just before it breaks the surface of the water? Ignore surface tension.
A mass of \(0.532 \mathrm{kg}\) of molecular oxygen is contained in a cylinder at a pressure of \(1.0 \times 10^{5} \mathrm{Pa}\) and a temperature of \(0.0^{\circ} \mathrm{C} .\) What volume does the gas occupy?
Find the molar mass of ammonia (NH \(_{3}\) ).
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The volume of a solid cube with side \(s_{0}\) at temperature \(T_{0}\) is \(V_{0}=s_{0}^{3} .\) Show that if \(\Delta s < s_{0},\) the change in volume \(\Delta V\) due to a change in temperature \(\Delta T\) is given by $$ \frac{\Delta V}{V_{0}}=3 \alpha \Delta T $$ and therefore that \(\beta=3 \alpha .\) (Although we derive this relation for a cube, it applies to a solid of any shape.)
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