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Chemical reactions · Gases and solutions

Ideal gases

A 12,3 L diving cylinder stored at 200 atm, a $\text{CO}_2$ fire extinguisher, a hot-air balloon that rises because the hot air inside it is less dense than the air outside: all three cases are solved with a single equation, the same for any gas in any conditions. In 4th ESO you studied Boyle's, Charles's and Gay-Lussac's laws separately; now you are going to bring them together into one tool, which can also link pressure and volume to the real amount of substance, in moles.

Before you start, make sure you know: Gases (4th ESO)
Before you start, make sure you know: The mole and molar mass (4th ESO)

The ideal gas equation of state

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In a laboratory, the behaviour 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 behaviour 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, $p\cdot V=n\cdot R\cdot T$, with $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$, 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: $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$.

  1. a) Express the temperature in kelvin.
  2. b) Apply the ideal gas equation, $p\cdot V=n\cdot R\cdot T$, 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: $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$.

  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

If you measure in the laboratory the volume of $1\ \text{mol}$ of helium ($4\ \text{g}$), $1\ \text{mol}$ of nitrogen ($28\ \text{g}$) and $1\ \text{mol}$ of carbon dioxide ($44\ \text{g}$) at $0\ ^\circ\text{C}$ and $1\ \text{atm}$, you get exactly $22{,}4\ \text{L}$ in all three cases: why do gases whose molecules have such different masses and sizes occupy exactly the same space in those conditions?

Three identical cubic containers of 22,4 L in standard conditions containing 1 mol of helium (4 g), 1 mol of oxygen O₂ (32 g) and 1 mol of carbon dioxide CO₂ (44 g). All three have the same number of molecules and occupy the same volume, although their masses are very different. At SC (0 °C and 1 atm), 1 mol 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 O₂ 1 mol of CO₂ m = 4 g m = 32 g m = 44 g Atoms: He O C same number of molecules (Avogadro's law) → 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: $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$.

  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, $p\cdot V=n\cdot R\cdot T$ with $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$, 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 balloon filled with helium rises quickly to the ceiling, whereas if you inflate it with carbon dioxide from a cartridge it falls to the floor immediately, even though both are at the same pressure and temperature: how does the mass of each molecule determine whether a gas floats or sinks in air?

Comparison of the same gas with the same particles: at high temperature it occupies a larger container and is less dense; at low temperature it occupies a smaller container and is denser. Larger volume High T: less dense Smaller volume Low T: denser Same number of particles (5) in both containers
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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 $\text{CO}_2$, $44\ \text{g/mol}$; $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$; 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

Mixtures, transfers and balloons: the equation in real situations

Comparison between the air in a 12,3-litre diving cylinder at 200 atmospheres and the 2460-litre volume it occupies when expanded to 1 atmosphere, about 200 times greater. 12,3 L 200 atm expands at 1 atm same T = 300 K 2460 L at atmospheric pressure ≈ 200 times more volume

A hot-air balloon rises because of the same equation: if the air inside is heated, its density falls (at the same pressure, which is always atmospheric pressure) and ends up below the density of the cold air outside, just as an object floats in a liquid that is denser than it is.

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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: $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$.

  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, $29\ \text{g/mol}$; $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$. 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: $R=0,082\ \text{atm}\cdot\text{L}/(\text{mol}\cdot\text{K})$. 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

Partial pressures and collecting gases

When you collect a gas given off in a reaction by bubbling it into an inverted measuring cylinder full of water, the pressure gauge shows a higher pressure than expected because there is also evaporated water vapour inside: how can we tell which part of the measured pressure really corresponds to the gas we are interested in?

The ideal approximation works best at low pressure and far from condensation. The conditions must be stated: at $0\ \celsius$ and $1\ \text{atm}$ the molar volume is approximately $22,4\ \text{L/mol}$; at $0\ \celsius$ and $1\ \text{bar}$ it is approximately $22,7\ \text{L/mol}$. Do not use $22,4$ without checking the conditions.

The diagram links the main ideas:

Inverted measuring cylinder over a trough of water containing dry gas and water vapour spread throughout its volume; on the right, the total pressure as the sum of the pressure of the dry gas and that of the water vapour. Water Dry gas Water vapour Water vapour Dry gas Sum of pressures

Back to the start: In the extinguisher, the $\text{CO}_2$ is stored compressed; when the valve is opened it expands towards atmospheric pressure ($1\ \text{atm}$), occupying hundreds of times its volume and smothering the fire by displacing the oxygen, thanks to its density (about $1{,}96\ \text{g/L}$) being greater than that of air.

Summary

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