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Chapter 14: Problem 18

What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of nitrogen gas (molecular mass \(=28 \mathrm{u}\) ) at a pressure of 2.0 atmospheres and a temperature of \(310 \mathrm{~K} ?\)

Short Answer

Expert verified
The density of nitrogen gas is approximately 3.7 kg/m鲁.

Step by step solution

01

Use the Ideal Gas Law

The Ideal Gas Law is given by the formula \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
02

Calculate the Number of Moles per Volume Unit

We can rearrange the Ideal Gas Law to determine \( \frac{n}{V} \) (moles per volume unit): \( \frac{n}{V} = \frac{P}{RT} \).First, convert the pressure from atmospheres to pascals: \( P = 2.0 \text{ atm} = 2.0 \times 101325 \text{ Pa} = 202650 \text{ Pa} \).Substitute values into the rearranged formula:\[ \frac{n}{V} = \frac{202650}{8.314 \times 310} \approx 79.56 \text{ moles/m}^3 \].
03

Convert Molecular Mass to Kilograms

The molecular mass of nitrogen is given as 28 u. To find the mass in kg, convert this by using 1 u = \( 1.66 \times 10^{-27} \text{ kg} \).Thus, the molecular mass in kg is \( 28 \times 1.66 \times 10^{-27} = 4.648 \times 10^{-26} \text{ kg} \).
04

Calculate Density from Moles and Molecular Mass

Density \( \rho \) is mass per unit volume. Using \( \rho = \frac{m}{V} \), and knowing \( \frac{n}{V} \) from Step 2, calculate:\[ \rho = \frac{79.56 \text{ moles/m}^3 \times 4.648 \times 10^{-26} \text{ kg/mole}}{1} \approx 3.7 \text{ kg/m}^3 \].

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation used to describe the behavior of gases. It relates several key variables: pressure (\( P \)), volume (\( V \)), the number of moles (\( n \)), temperature (\( T \)), and a constant known as the ideal gas constant (\( R \)). The law is concisely expressed as \( PV = nRT \). This formula is extremely useful for calculating any one of these variables when the others are known.
Let's break down each component:
  • Pressure (\( P \)): The force exerted by the gas particles against the walls of its container. It's often measured in atmospheres (atm) but can be converted to pascals (Pa) for standard calculations.
  • Volume (\( V \)): The space that the gas occupies, usually in cubic meters (\( m^3 \)).
  • Number of moles (\( n \)): The amount of substance, measured in moles, which describes how many molecules are present.
  • Temperature (\( T \)): Measured in Kelvin (K), this is a measure of the thermal energy of the gas.
  • Ideal Gas Constant (\( R \)): A constant that is \( 8.314 \, \text{J/(mol} \, \text{K)} \), which serves as a bridge between these units.
By understanding this equation, we can solve for the density of a gas, given its temperature and pressure, which is a common application in physics and chemistry.
Molecular Mass
Molecular mass is the mass of a single molecule of a substance, expressed in atomic mass units (\( u \)), where 1 \( u \) is defined as \( 1/12 \) of the mass of a carbon-12 atom.
To apply this in calculations like determining the density of a gas, it needs to be converted to kilograms. This is because the SI unit of mass is kilograms. The conversion from \( u \) to kilograms uses the factor \( 1 \, u = 1.66 \times 10^{-27} \, \text{kg} \).For example, the molecular mass of nitrogen is given as 28 \( u \). To convert it to kilograms:
  • Multiply by 1.66 \( \times 10^{-27} \) kg.
  • This gives us the mass of one molecule of nitrogen as \( 4.648 \times 10^{-26} \) kg.
Knowing the molecular mass allows us to calculate the density of the gas when combined with the number of moles per volume, as it gives us the mass of each mole in terms of the standard measurement of mass.
Pressure Conversion
Pressure conversion is crucial when working with the Ideal Gas Law, especially because pressure can be measured in various units. One common unit is atmospheres (atm), but in scientific calculations, the SI unit is the pascal (Pa). Knowing how to convert between these units allows for accurate problem-solving.
To convert from atmospheres to pascals, use the fact that:
  • 1 atm = 101325 Pa.
In our exercise, we had a pressure of 2.0 atm. Convert this to pascals by multiplying:
  • 2.0 atm \( \times \) 101325 Pa/atm = 202650 Pa.
This conversion is necessary as many gas law calculations require using pascals to maintain consistency with other SI units, such as the volume in cubic meters and the ideal gas constant in \( \text{J/(mol} \, \text{K)} \). By using these conversions, understanding and solving gas law problems becomes much more manageable.

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Most popular questions from this chapter

Initially, the translational rms speed of a molecule of an ideal gas is \(463 \mathrm{~m} / \mathrm{s}\). The pressure and volume of this gas are kept constant, while the number of molecules is doubled. What is the final translational rms speed of the molecules?

A spherical balloon is made from a material whose mass is \(3.00 \mathrm{~kg} .\) The thickness of the material is negligible compared to the 1.50 -m radius of the balloon. The balloon is filled with helium (He) at a temperature of \(305 \mathrm{~K}\) and just floats in air, neither rising nor falling. The density of the surrounding air is \(1.19 \mathrm{~kg} / \mathrm{m}^{3}\). Find the absolute pressure of the helium gas.

If the translational rms speed of the water vapor molecules \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) in air is \(648 \mathrm{~m} / \mathrm{s}\), what is the translational rms speed of the carbon dioxide molecules \(\left(\mathrm{CO}_{2}\right)\) in the same air? Both gases are at the same temperature.

The pressure of sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) is \(2.12 \times 10^{4} \mathrm{~Pa}\). There are 421 moles of this gas in a volume of \(50.0 \mathrm{~m}^{3}\). Find the translational rms speed of the sulfur molecules.

On the sunlit surface of Venus, the atmospheric pressure is \(9.0 \times 10^{6} \mathrm{~Pa}\), and the temperature is \(740 \mathrm{~K}\). On the earth's surface the atmospheric pressure is \(1.0 \times 10^{5} \mathrm{~Pa}\), while the surface temperature can reach \(320 \mathrm{~K}\). These data imply that Venus has a "thicker" atmosphere at its surface than does the earth, which means that the number of molecules per unit volume \((N / V)\) is greater on the surface of Venus than on the earth. Find the ratio \((N / V)_{\text {Venus }} /(N / V)_{\text {Earth }}\).
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