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Chemical reactions · Gases and solutions

Ideal gases

A 12.3 L scuba tank filled to 200 atm for a dive off the Florida Keys, a $\text{CO}_2$ fire extinguisher in your school hallway, 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. You may have met the relationships between pressure, volume, and temperature of gases before (Boyle's, Charles's, and Gay-Lussac's laws); now you will bring them together into one tool, which also links pressure and volume to the amount of gas in moles.

The ideal gas equation of state

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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 kelvins.
  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 it to solve for volume.
  2. b) Calculate the volume occupied by the gas.
Solve
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A small weather balloon contains $2\ \text{mol}$ of helium. At a certain point in its ascent, the sensors record a pressure of $2\ \text{atm}$ and a volume of…

A small weather balloon contains $2\ \text{mol}$ of helium. At a certain point in its ascent, the sensors record a pressure of $2\ \text{atm}$ and a volume of $24.6\ \text{L}$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Calculate the temperature of the helium in kelvins.
  2. b) Express it in degrees Celsius.
Solve
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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 kelvins and the pressure in atmospheres.
  2. b) Calculate the volume of the balloon.
Solve

Standard temperature and pressure 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 at STP 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 STP (0 °C, 1 atm), 1 mol of any gas is 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: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  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 lab, $32\ \text{g}$ 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,…

In a lab, $32\ \text{g}$ 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
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Two identical $11.2\ \text{L}$ containers are both at STP ($1\ \text{atm}$ and $273\ \text{K}$): one contains nitrogen, $\text{N}2$, and the other hydrogen, $\text{H}2$.…

Two identical $11.2\ \text{L}$ containers are both at STP ($1\ \text{atm}$ and $273\ \text{K}$): one contains nitrogen, $\text{N}_2$, and the other hydrogen, $\text{H}_2$. Data: molar masses of $\text{N}_2$ and $\text{H}_2$, $28\ \text{g/mol}$ and $2\ \text{g/mol}$; molar volume at STP, $22.4\ \text{L/mol}$.

  1. a) How many moles of gas are in each container?
  2. b) Calculate the mass of gas in each one.
  3. c) Which of the two containers weighs more? Does it therefore contain more particles? Give reasons for your answers.
Solve

The density of a gas

At a birthday party in Denver, a helium balloon 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

The kinetic theory explains why gases have such low densities: in a gas the particles are very far apart and mostly empty space lies between them, whereas in a liquid they are packed close together.

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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
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A neon-style sign contains a pure noble gas. In the lab, its density is measured: $1.79\ \text{g/L}$ at $2\ \text{atm}$ and $273\ \text{K}$. Data:…

A neon-style sign contains a pure noble gas. In the lab, its density is measured: $1.79\ \text{g/L}$ at $2\ \text{atm}$ and $273\ \text{K}$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Calculate the volume that $1\ \text{mol}$ of the gas would occupy at $2\ \text{atm}$ and $273\ \text{K}$.
  2. b) Use the density to calculate the mass of that volume, i.e., the molar mass of the gas.
  3. c) Identify it among the noble gases: He ($4\ \text{g/mol}$), Ne ($20\ \text{g/mol}$), Ar ($40\ \text{g/mol}$), and Kr ($84\ \text{g/mol}$).
  4. d) Explain, using the ideal gas equation, why increasing the pressure increases the density of a gas even though its molar mass does not change.
Solve

Mixtures, transfers and balloons: the equation in real situations

When a gas changes conditions but the amount stays the same, $n$ and $R$ are unchanged, so $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ (the combined gas law). If $T$ is also constant, $P_1V_1 = P_2V_2$ (Boyle's law). If $P$ is constant, $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$ (Charles's law). If $V$ and $n$ are constant, $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ (Gay-Lussac's law): doubling the temperature in kelvin doubles the pressure. If $V$ and $T$ are constant, $P \propto n$. The kinetic theory explains why: hotter particles move faster and hit the walls harder and more often; more particles means more collisions.

Comparison between the air in a 12.3-liter diving cylinder at 200 atmospheres and the 2460-liter 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. A hot-air balloon over the Albuquerque skies at the October balloon fiesta works the same way.

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A diving cylinder of $12.3\ \text{L}$ is filled with air at $200\ \text{atm}$ and $27\ \celsius$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.…

A diving cylinder of $12.3\ \text{L}$ is filled with air at $200\ \text{atm}$ and $27\ \celsius$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$. The pressure-volume graph shows the compressed air in the cylinder and its expansion at constant temperature to surface pressure.

V (L) p (atm) 12.3 L 200 atm cylinder V = ? 1 atm surface
  1. a) Express the temperature in kelvin and calculate how many moles of air the cylinder contains.
  2. b) Calculate what volume all that air would occupy at surface pressure, $1\ \text{atm}$, and at the same temperature.
  3. c) A diver consumes about $60\ \text{L}$ of air per minute measured at the surface: estimate, in minutes, the theoretical duration of the cylinder.
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 kelvins 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 valve that is closed. The first contains oxygen at $4\ \text{atm}$ and $300\ \text{K}$;…

Two rigid containers, of $10\ \text{L}$ and $30\ \text{L}$, are connected by a valve 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 valve still closed.

10 L O₂ · 4 atm · 300 K closed valve vacuum 30 L
  1. a) Calculate the moles of enclosed oxygen.
  2. b) The valve 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

Dalton's law of partial pressures

In a mixture of gases that do not react, each gas behaves as if it were alone in the container. The total pressure is the sum of the partial pressures:

$$P_{\text{total}} = P_1 + P_2 + P_3 + \cdots$$

The partial pressure of each gas is $P_i = \dfrac{n_i RT}{V}$, so it is proportional to its mole fraction. When a gas is collected over water, the total pressure includes the water vapor: $P_{\text{gas}} = P_{\text{total}} - P_{\text{H}_2\text{O}}$ (the vapor pressure of water is given in a table).

Ideal and real gases

An ideal gas is a useful model, but when do real gases stop obeying it, and how can we tell which assumptions of the kinetic theory fail?

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A weather balloon is filled on the ground with $100\ \text{mol}$ of helium, at $1\ \text{atm}$ and $300\ \text{K}$. It rises to the stratosphere, where the pressure is…

A weather balloon is filled on the ground with $100\ \text{mol}$ of helium, at $1\ \text{atm}$ and $300\ \text{K}$. It rises to the stratosphere, where the pressure is $0.2\ \text{atm}$ and the temperature is $-48\ \celsius$. Data: $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$. The figure shows the balloon at launch and in the stratosphere.

1 atm · 300 K V = ? 0.2 atm · −48 °C V = ?
  1. a) Calculate the volume of the balloon at the moment of launch and express it in $\text{m}^3$.
  2. b) Express the temperature in the stratosphere in kelvin and calculate the volume of the balloon at that altitude.
  3. c) In light of the results, explain why these balloons are launched partially inflated and why they end up bursting at high altitude.
Solve
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At the summit of Mount Everest, the atmosphere is much less dense than at sea level. Data: average molar mass of air, $29\ \text{g/mol}$;…

At the summit of Mount Everest, the atmosphere is much less dense than at sea level. Data: average molar mass of air, $29\ \text{g/mol}$; $R=0.082\ \text{atm}\times\text{L}/(\text{mol}\times\text{K})$.

  1. a) Calculate the density of air at sea level, at $1\ \text{atm}$ and $15\ \celsius$ ($288\ \text{K}$).
  2. b) At the summit, the pressure is $0.33\ \text{atm}$ and the temperature is $-30\ \celsius$ ($243\ \text{K}$): calculate the air density there.
  3. c) Calculate how many grams of air a climber inhales in each breath of $2\ \text{L}$ at each of the two locations.
  4. d) Use the results to explain why many climbers need oxygen tanks at the summit even though the proportion of oxygen in the air is the same at both locations.
Solve

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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