The gas laws and the ideal gas equation

Pressure and its units

Pressure is force per unit area. The SI unit is the pascal: $1\ \text{Pa} = 1\ \text{N/m}^2$. A pascal is tiny, so we use bigger units:

• $1\ \text{bar} = 100\ \text{kPa} = 10^5\ \text{Pa}$.
• $1\ \text{atm} = 101.325\ \text{kPa} = 1.01325\ \text{bar}$. This is the average air pressure at sea level.
• $1\ \text{atm} = 760\ \text{mm Hg} = 760\ \text{torr}$. A barometer measures it with a column of mercury $760\ \text{mm}$ high. One mm Hg is $133.322\ \text{Pa}$.

To change $750\ \text{mm Hg}$ to kPa, multiply: $750 \times 0.133322 = 99.99\ \text{kPa}$, about $1.00\ \text{bar}$.

A gas has four variables: pressure $P$, volume $V$, temperature $T$ and amount $n$. The gas laws join them two at a time.

Boyle's law: pressure and volume

Robert Boyle found in 1662 that for a fixed amount of gas at constant temperature, the volume falls when the pressure rises:

So for two states of the same gas,

A gas at $125\ \text{kPa}$ fills $3.2\ \text{L}$. It expands to $8.0\ \text{L}$ at the same temperature. We want $P_2$, so $P_2 = \dfrac{P_1V_1}{V_2} = \dfrac{125\ \text{kPa} \times 3.2\ \text{L}}{8.0\ \text{L}} = 50\ \text{kPa}$. The volume grew $2.5$ times, and the pressure fell to $\tfrac{1}{2.5}$ of its value.

A graph of $P$ against $V$ at one temperature is a curve called an isotherm. A graph of $P$ against $1/V$ is a straight line through the origin. A higher temperature gives a higher curve.

Charles' law: volume and temperature

Jacques Charles found that for a fixed amount of gas at constant pressure, the volume rises steadily with the temperature. Plot the volume against the temperature in $^\circ\text{C}$. The graph is a straight line. Extend it backwards. Every line, whatever the pressure or the gas, meets the axis at the same point, $-273.15\ ^\circ\text{C}$.

That temperature is absolute zero, and it is the zero of the kelvin scale. So $T\ (\text{K}) = t\ (^\circ\text{C}) + 273.15$. In kelvin, the volume is proportional to the temperature:

A gas fills $2.40\ \text{L}$ at $27\ ^\circ\text{C}$. It is heated to $87\ ^\circ\text{C}$ at constant pressure. We change to kelvin: $T_1 = 300.15\ \text{K}$ and $T_2 = 360.15\ \text{K}$. Then $V_2 = 2.40\ \text{L} \times \dfrac{360.15}{300.15} = 2.88\ \text{L}$. A common slip is to write $\dfrac{87}{27}$, which gives $7.7\ \text{L}$. A real gas turns to a liquid before it reaches absolute zero, so the dotted part of the lines is an extension only.

Gay-Lussac's law: pressure and temperature

For a fixed amount of gas in a fixed volume, the pressure is proportional to the temperature in kelvin:

A sealed can at $1.00\ \text{atm}$ and $20\ ^\circ\text{C}$ is thrown into a fire and reaches $120\ ^\circ\text{C}$. The pressure becomes $1.00\ \text{atm} \times \dfrac{393.15}{293.15} = 1.34\ \text{atm}$. This law is also called Amontons' law. Some books give the name Gay-Lussac's law to the law of combining volumes, so say what you mean.

Avogadro's law: volume and amount

Equal volumes of different gases at the same temperature and pressure contain the same number of molecules. In symbols, $V \propto n$ at constant $T$ and $P$. So the molar volume, the volume of one mole, is the same for every ideal gas at the same $T$ and $P$.

The ideal gas equation

Put the four laws together. $V$ is proportional to $n$, to $T$ and to $1/P$, so $V = R\,\dfrac{nT}{P}$, which is usually written

An ideal gas is a gas that follows this equation exactly. Real gases follow it closely at low pressure and high temperature. We study the difference in the next course.

$R$ is the gas constant. Its value depends on the units of $P$ and $V$, and a wrong pair of units is the most common source of wrong answers:

• Pressure in Pa and volume in $\text{m}^3$: $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$.
• Pressure in kPa and volume in L: $R = 8.314\ \text{kPa L K}^{-1}\ \text{mol}^{-1}$.
• Pressure in bar and volume in L: $R = 0.08314\ \text{L bar K}^{-1}\ \text{mol}^{-1}$.
• Pressure in atm and volume in L: $R = 0.08206\ \text{L atm K}^{-1}\ \text{mol}^{-1}$.
• Pressure in mm Hg and volume in L: $R = 62.36\ \text{L mm Hg K}^{-1}\ \text{mol}^{-1}$.

All five are the same constant. A pascal times a cubic metre is a joule. One L bar is $10^5\ \text{Pa} \times 10^{-3}\ \text{m}^3 = 100\ \text{J}$, so $0.08314\ \text{L bar} = 8.314\ \text{J}$. One L atm is $101.325\ \text{J}$, and $0.08206 \times 101.325 = 8.315$.

The temperature in the equation is always in kelvin.

Set $n = 1\ \text{mol}$, $T = 300\ \text{K}$, $V = 24\ \text{L}$. Then set $V = 12\ \text{L}$. The product $PV$ does not change. Now double $T$. The product doubles.

Molar volume and standard conditions

One mole of an ideal gas at $273.15\ \text{K}$ and $1\ \text{bar}$ fills $\dfrac{0.08314 \times 273.15}{1} = 22.71\ \text{L}$. This is the molar volume at the IUPAC standard temperature and pressure (STP) in use since 1982. Older books and the older NCERT use $273.15\ \text{K}$ and $1\ \text{atm}$, which gives $22.41\ \text{L}$, usually printed as $22.4\ \text{L}$. At $298.15\ \text{K}$ and $1\ \text{bar}$, the molar volume is $24.79\ \text{L}$. Always read which standard a question means. This course gives the temperature and pressure in each question.

Density and molar mass

The amount of gas is $n = m/M$, where $m$ is the mass and $M$ the molar mass. Put this in $PV = nRT$ and use the density $d = m/V$:

We want $M$, so $M = \dfrac{dRT}{P}$. Carbon dioxide has $M = 44.01\ \text{g/mol}$. At $1.00\ \text{bar}$ and $300\ \text{K}$ its density is $\dfrac{1.00 \times 44.01}{0.08314 \times 300} = 1.76\ \text{g/L}$. Dry air has $M = 28.96\ \text{g/mol}$, so at the same conditions its density is $1.16\ \text{g/L}$. Carbon dioxide is heavier than air. It sinks and collects at the bottom of a pit.

The density of a gas depends strongly on $T$ and $P$. That is why a hot-air balloon rises: the hot air inside is less dense than the cool air outside.

Practise with new numbers

Try it in Python

The program uses the gas constant that matches the units, finds an amount of gas, and then finds the pressure when the flask is heated.

Common mistakes

• Temperature in $^\circ\text{C}$. The gas laws use kelvin. Doubling $20\ ^\circ\text{C}$ gives $40\ ^\circ\text{C}$, but doubling $293\ \text{K}$ gives $586\ \text{K}$.
• The wrong $R$. Check the units of $P$ and $V$ first. Then take the $R$ that has the same units.
• Mixing mm Hg and atm. Convert before you use a law with two pressures. A ratio of two pressures can use any unit, as long as both are the same.
• Using mass instead of moles. In $PV = nRT$, $n$ is in moles. Divide the mass by the molar mass.
• Using Boyle's law when the temperature changes. Boyle's law needs a constant temperature and a constant amount of gas.

Remember these

• $PV = nRT$, with $T$ in kelvin and $R$ chosen to match the units of $P$ and $V$.
• Boyle: $P_1V_1 = P_2V_2$. Charles: $V_1/T_1 = V_2/T_2$. Gay-Lussac: $P_1/T_1 = P_2/T_2$. Avogadro: $V \propto n$.
• Combined law for a fixed amount: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$.
• Density $d = \dfrac{PM}{RT}$, and the molar mass $M = \dfrac{dRT}{P}$.
• Molar volume: $22.71\ \text{L}$ at $273.15\ \text{K}$ and $1\ \text{bar}$, $22.41\ \text{L}$ at $273.15\ \text{K}$ and $1\ \text{atm}$, $24.79\ \text{L}$ at $298.15\ \text{K}$ and $1\ \text{bar}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A balloon holds $2.00\ \text{L}$ of air at $27\ ^\circ\text{C}$. It is cooled to $-23\ ^\circ\text{C}$ at constant pressure. What is its new volume?

2. Pressure is in bar and volume is in litres. Which value of $R$ do you use, and why?

3. A gas has a molar mass of $32.0\ \text{g/mol}$. What is its density at $273.15\ \text{K}$ and $1.00\ \text{atm}$?

4. Why must the temperature be in kelvin in $PV = nRT$?

5. The gas in a rigid steel cylinder is at $2.0\ \text{bar}$ and $300\ \text{K}$. The temperature goes up to $450\ \text{K}$. What is the pressure?

References

• OpenStax, Chemistry 2e, Chapter 9, sections 9.1 and 9.2 (pressure, and the gas laws).
• IUPAC, Compendium of Chemical Terminology (the Gold Book): the entry on standard pressure, 1 bar since 1982.