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Chemical reactions · Gases and solutions

Ideal Gases

Before you start, make sure you know: Gases (High School Year 1 (semesters 1 and 2))

A tank of LP gas, a $\text{CO}_2$ fire extinguisher, a hot air balloon over Teotihuacan: all three can be explained with a single equation. In Mexico City (at about 2,240 m above sea level) atmospheric pressure is less than 1 atm, and that changes the volume of a balloon you carry up from the coast. How do we relate pressure, volume, temperature, and the amount of gas? That is what you are going to work out here.

The Ideal Gas Equation of State

The state of a gas enclosed in a cylinder with a piston is described by four variables: the pressure $p$ (how hard the particles hit the walls), the volume $V$ (the space it occupies), the temperature $T$ (how fast the particles move on average), and the amount of substance $n$ (the moles of particles).

Try the simulator: change the temperature, the volume, and the moles, and observe what happens to the pressure.

To convert from degrees Celsius to kelvins: $T(\text{K}) = t(°\text{C}) + 273$. Other pressure units: $1\ \text{atm} = 760\ \text{mmHg} = 101.3\ \text{kPa}$.

Practice★☆☆Not started
In a laboratory, the behavior of a gas enclosed in a cylinder with a piston is studied. The state of the gas is described by four variables: pressure, volume,…

In a laboratory, the behavior of a gas enclosed in a cylinder with a piston is studied. The state of the gas is described by four variables: pressure, volume, temperature and amount of substance. The diagram shows the gas enclosed in the cylinder with its four variables.

p n V T
  1. a) Explain in your own words what each of these variables indicates.
  2. b) State the SI unit for each one.
  3. c) When applying the ideal gas equation, $p\times V=n\times R\times T$, with $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$, in which units should pressure, volume, temperature and amount of substance be expressed?
Solve
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A rigid laboratory container contains a gas at $2\ \text{atm}$ and $27\ \celsius$, and occupies a volume of $24.6\ \text{L}$. Data:…

A rigid laboratory container contains a gas at $2\ \text{atm}$ and $27\ \celsius$, and occupies a volume of $24.6\ \text{L}$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Express the temperature in kelvin.
  2. b) Apply the ideal gas equation, $p\times V=n\times R\times T$, and calculate how many moles of gas are in the container.
Solve
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A cylinder with a piston contains $2\ \text{mol}$ of nitrogen at $300\ \text{K}$, subjected to a pressure of $4\ \text{atm}$. Data:…

A cylinder with a piston contains $2\ \text{mol}$ of nitrogen at $300\ \text{K}$, subjected to a pressure of $4\ \text{atm}$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Write down the ideal gas equation and rearrange for volume.
  2. b) Calculate the volume occupied by the gas.
Solve

Standard Conditions and Molar Volume

In the lab, measure the volume of $1\ \text{mol}$ of helium ($4\ \text{g}$), of nitrogen ($28\ \text{g}$), and of carbon dioxide ($44\ \text{g}$) at $0\ °\text{C}$ and $1\ \text{atm}$: in all three cases you get $22.4\ \text{L}$. Why do such different molecules take up the same amount of space?

The reason: in a gas the particles are very far apart, so the size of each one hardly matters; the volume depends on how many particles there are, not on their mass (Avogadro's law).

Three identical 22.4 L containers at standard conditions with 1 mole of helium (4 g), 1 mole of nitrogen (28 g), and 1 mole of carbon dioxide (44 g): same number of molecules, same volume, different masses. At SC (0 °C and 1 atm), 1 mole of any gas occupies 22.4 L V = 22.4 L V = 22.4 L V = 22.4 L 1 mol of He 1 mol of N₂ 1 mol of CO₂ m = 4 g m = 28 g m = 44 g same number of molecules, same volume
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In an experiment, exactly $1\ \text{mol}$ of an ideal gas is enclosed in a $22.4\ \text{L}$ container kept at $273\ \text{K}$. Data:…

In an experiment, exactly $1\ \text{mol}$ of an ideal gas is enclosed in a $22.4\ \text{L}$ container kept at $273\ \text{K}$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Calculate the pressure of the gas, in atmospheres.
  2. b) Which known value does the result approximate? What are these pressure and temperature conditions called?
Solve
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In chemistry, standard temperature and pressure (STP) are used as reference conditions. a) Which values of pressure and temperature define standard conditions (STP)?…

In chemistry, standard temperature and pressure (STP) are used as reference conditions.

  1. a) Which values of pressure and temperature define standard conditions (STP)? Express the temperature in kelvin as well.
  2. b) Use the ideal gas equation, $p\times V=n\times R\times T$ with $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$, to calculate the volume occupied by $1\ \text{mol}$ of any ideal gas at standard temperature and pressure.
  3. c) What is this reference volume called?
Solve
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In a laboratory, $32\ \text{g}$ of oxygen, $\text{O}2$, is weighed. Data: molar mass of $\text{O}2$, $32\ \text{g/mol}$; molar volume at standard temperature and…

In a laboratory, $32\ \text{g}$ of oxygen, $\text{O}_2$, is weighed. Data: molar mass of $\text{O}_2$, $32\ \text{g/mol}$; molar volume at standard temperature and pressure, $22.4\ \text{L/mol}$.

  1. a) How many moles of oxygen is this?
  2. b) What volume would this gas occupy at standard temperature and pressure?
Solve

The Density of a Gas

A helium balloon rises to the ceiling; one inflated with $\text{CO}_2$ from a cartridge falls. How does the mass of the molecules decide whether a gas floats or sinks in air?

The same gas with the same particles: at high temperature it fills a larger container and is less dense; at low temperature it fills a smaller container and is denser. Larger volume High T: less dense Smaller volume Low T: denser Same number of particles (5) in both
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A $5\ \text{L}$ cylinder contains $10\ \text{g}$ of nitrogen gas. The figure compares this cylinder with the same volume of water using the particle model. a) Calculate…

A $5\ \text{L}$ cylinder contains $10\ \text{g}$ of nitrogen gas. The figure compares this cylinder with the same volume of water using the particle model.

N₂ gas 5 L · 10 g d = ? g/L water 5 L d = 1000 g/L
  1. a) Calculate the density of the gas in $\text{g/L}$ (mass divided by volume).
  2. b) The density of water is $1000\ \text{g/L}$: how many times denser is water than this gas?
  3. c) Explain, using the particle model, why gases have such low densities compared to liquids.
Solve
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$\text{CO}2$ is denser than air, which explains why it can accumulate at the bottom of wells and poorly ventilated cellars. Data: molar mass of $\text{CO}2$,…

$\text{CO}_2$ is denser than air, which explains why it can accumulate at the bottom of wells and poorly ventilated cellars. Data: molar mass of $\text{CO}_2$, $44\ \text{g/mol}$; $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$; density of air at STP, $1.29\ \text{g/L}$.

  1. a) Calculate the mass of $1\ \text{mol}$ of $\text{CO}_2$ and the volume it occupies at STP ($1\ \text{atm}$ and $273\ \text{K}$).
  2. b) Calculate the density of $\text{CO}_2$ at STP, to one decimal place.
  3. c) Compare it with that of air and explain why $\text{CO}_2$ accumulates in low areas.
Solve
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A cylinder containing an unknown gas is received in a laboratory. Measuring its mass and volume gives a density of $1.43\ \text{g/L}$ at STP. Data: molar volume at STP,…

A cylinder containing an unknown gas is received in a laboratory. Measuring its mass and volume gives a density of $1.43\ \text{g/L}$ at STP. Data: molar volume at STP, $22.4\ \text{L/mol}$.

  1. a) Calculate the mass of $22.4\ \text{L}$ of the gas, i.e. its molar mass.
  2. b) Which gas could it be: $\text{H}_2$ ($2\ \text{g/mol}$), $\text{O}_2$ ($32\ \text{g/mol}$) or $\text{CO}_2$ ($44\ \text{g/mol}$)? Justify your answer.
Solve

Tanks, Balloons, and Containers: The Equation in Real Situations

Almost all problems are solved the same way: identify what you are given, convert to the units of $R$, and solve for the unknown. If the gas changes conditions, first calculate the moles (which are conserved if no gas escapes) and then apply the equation to the new state.

Comparison between the air in a 12.3-liter diving tank at 200 atmospheres and the 2460-liter volume it occupies at 1 atmosphere, about 200 times larger. 12.3 L 200 atm expands at 1 atm same T = 300 K 2,460 L at atmospheric pressure ≈ 200 times more volume

A hot air balloon works with this same equation: it is open at the bottom, so the pressure inside is always atmospheric. If you heat the air, its density $\rho = pM/(RT)$ drops and becomes lower than that of the cold air outside: the balloon floats.

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A balloon contains $0.25\ \text{mol}$ of helium at $27\ \celsius$ and the barometer measures an external pressure of $380\ \text{mmHg}$. Data:…

A balloon contains $0.25\ \text{mol}$ of helium at $27\ \celsius$ and the barometer measures an external pressure of $380\ \text{mmHg}$. Data: $1\ \text{atm}=760\ \text{mmHg}$; $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Express the temperature in kelvin and the pressure in atmospheres.
  2. b) Calculate the volume of the balloon.
Solve
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A factory boiler contains $4\ \text{mol}$ of water vapour in a volume of $16.4\ \text{L}$ at a pressure of $10\ \text{atm}$. Data:…

A factory boiler contains $4\ \text{mol}$ of water vapour in a volume of $16.4\ \text{L}$ at a pressure of $10\ \text{atm}$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Calculate the temperature of the vapour in kelvin.
  2. b) Express it in degrees Celsius.
  3. c) Explain why it is essential to use kelvin and not degrees Celsius in the ideal gas equation: what would happen to the result if you used the temperature in $\celsius$?
Solve
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A hot-air balloon is open at the bottom, so that the pressure of the air inside is always atmospheric pressure, $1\ \text{atm}$. The outside air is at $27\ \celsius$ and…

A hot-air balloon is open at the bottom, so that the pressure of the air inside is always atmospheric pressure, $1\ \text{atm}$. The outside air is at $27\ \celsius$ and the burner heats the inside air up to $127\ \celsius$. Data: average molar mass of air, $29\ \text{g/mol}$; $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$. The diagram represents the balloon and the inside and outside air.

p = 1 atm (inside and outside) 127 °C d = ? 27 °C d = ?
  1. a) Express both temperatures in kelvin and calculate the density of the outside air.
  2. b) Calculate the density of the hot air inside.
  3. c) Compare the mass of $1000\ \text{L}$ ($1\ \text{m}^3$) of air inside and outside, and use this difference to explain why the balloon rises.
Solve
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Two rigid containers, of $10\ \text{L}$ and $30\ \text{L}$, are connected by a tap that is closed. The first contains oxygen at $4\ \text{atm}$ and $300\ \text{K}$; the…

Two rigid containers, of $10\ \text{L}$ and $30\ \text{L}$, are connected by a tap that is closed. The first contains oxygen at $4\ \text{atm}$ and $300\ \text{K}$; the second is empty. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$. The diagram shows the setup with the tap still closed.

10 L O₂ · 4 atm · 300 K closed tap vacuum 30 L
  1. a) Calculate the moles of enclosed oxygen.
  2. b) The tap is opened and the gas spreads throughout the entire available volume, without the temperature changing. Calculate the final pressure.
  3. c) Explain, using the kinetic model, why the pressure drops even though not a single particle has left the system.
Solve

Back to the beginning: in the extinguisher, the $\text{CO}_2$ is stored compressed; when you open the valve it expands to $1\ \text{atm}$ and takes up hundreds of times its volume. And since its density (about $1.97\ \text{g/L}$) is greater than that of air, it blankets the fire and displaces the oxygen.

Summary

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