States of Matter • Topic 1 of 3

Gas Laws & Ideal Gas Equation

Matter exists in three familiar physical states — solid, liquid and gas — and the same substance can move between them by changing temperature or pressure. What decides the state is the tug-of-war between the intermolecular forces that pull molecules together and the thermal energy that keeps them moving apart.

Intermolecular forces (van der Waals forces) are much weaker than chemical bonds. Three types matter here:

London dispersion forces — momentary dipoles in every atom or molecule; they grow with molecular size and explain why larger non-polar molecules have higher boiling points.
Dipole–dipole forces — act between permanent dipoles in polar molecules such as $\text{HCl}$.
Hydrogen bonding — a specially strong dipole attraction when H is bonded to the small, highly electronegative atoms N, O or F (as in water and ammonia).

Gases have the weakest grip of these forces relative to their thermal energy, so they fill any container. Their behaviour is summarised by simple gas laws, each relating two variables while the others stay fixed.

Boyle’s law (constant $T$, fixed $n$): pressure is inversely proportional to volume, $P\propto\frac{1}{V}$, so $P_1V_1=P_2V_2$.

Charles’ law (constant $P$): volume is directly proportional to absolute temperature, $\frac{V}{T}=\text{constant}$, so $\frac{V_1}{T_1}=\frac{V_2}{T_2}$.

Gay-Lussac’s law (constant $V$): pressure is directly proportional to absolute temperature, $\frac{P}{T}=\text{constant}$.

Avogadro’s law: at the same $T$ and $P$, equal volumes contain equal numbers of molecules, so $V\propto n$. One mole of any gas occupies $22.4\ \text{L}$ at STP.

Combining all four gives the combined gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ and the ideal gas equation $PV=nRT$, where $R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}=0.0821\ \text{L}\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1}$.

For a mixture of non-reacting gases, Dalton’s law of partial pressures states $P_{total}=p_1+p_2+\dots$, and each partial pressure equals the total pressure times that gas’s mole fraction: $p_i=x_i\,P_{total}$. A key reminder: every temperature in these laws must be in kelvin, $T(\text{K})=t(^\circ\text{C})+273.15$.

Solved Examples

Worked Example

A gas occupies 2.0 L at 760 mmHg. What volume will it occupy at 380 mmHg, temperature constant?

Solution

Step 1: Boyle’s law (constant $T$): $P_1V_1=P_2V_2$.
Step 2: Substitute: $760\times2.0=380\times V_2$.
Step 3: Solve: $V_2=\frac{1520}{380}=4.0\ \text{L}$.

Answer: $V_2=4.0\ \text{L}$ (halving the pressure doubles the volume).

Worked Example

A gas has a volume of 300 mL at $27^\circ\text{C}$. Find its volume at $127^\circ\text{C}$ at constant pressure.

Solution

Step 1: Convert to kelvin: $T_1=300\ \text{K}$, $T_2=400\ \text{K}$.
Step 2: Charles’ law: $\frac{V_1}{T_1}=\frac{V_2}{T_2}$.
Step 3: $V_2=300\times\frac{400}{300}=400\ \text{mL}$.

Answer: $V_2=400\ \text{mL}$.

Worked Example

The pressure of a fixed-volume gas is 1.5 atm at $27^\circ\text{C}$. Find the pressure at $327^\circ\text{C}$.

Solution

Step 1: $T_1=300\ \text{K}$, $T_2=600\ \text{K}$.
Step 2: Gay-Lussac’s law (constant $V$): $\frac{P_1}{T_1}=\frac{P_2}{T_2}$.
Step 3: $P_2=1.5\times\frac{600}{300}=3.0\ \text{atm}$.

Answer: $P_2=3.0\ \text{atm}$.

Worked Example

How many moles of an ideal gas occupy 4.92 L at 2.0 atm and 300 K? ($R=0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$)

Solution

Step 1: Ideal gas equation: $PV=nRT\Rightarrow n=\frac{PV}{RT}$.
Step 2: Substitute: $n=\frac{2.0\times4.92}{0.0821\times300}$.
Step 3: $n=\frac{9.84}{24.63}\approx0.40\ \text{mol}$.

Answer: $n\approx0.40\ \text{mol}$.

Worked Example

A 2.0 L flask at 300 K contains 0.20 mol $\text{N}_2$ and 0.30 mol $\text{O}_2$. Find the partial pressure of $\text{O}_2$. ($R=0.0821$)

Solution

Step 1: Each gas obeys $p_iV=n_iRT$, so $p_{\text{O}_2}=\frac{n_{\text{O}_2}RT}{V}$.
Step 2: $p_{\text{O}_2}=\frac{0.30\times0.0821\times300}{2.0}$.
Step 3: $p_{\text{O}_2}=\frac{7.389}{2.0}\approx3.69\ \text{atm}$.

Answer: $p_{\text{O}_2}\approx3.69\ \text{atm}$.

Worked Example

A gas at 1 atm and 273 K has a density of $1.25\ \text{g/L}$. Find its molar mass. ($R=0.0821$)

Solution

Step 1: From $PV=nRT$ and $n=\frac{m}{M}$, density form is $M=\frac{dRT}{P}$.
Step 2: Substitute: $M=\frac{1.25\times0.0821\times273}{1}$.
Step 3: $M\approx28.0\ \text{g/mol}$ (i.e. $\text{N}_2$).

Answer: $M\approx28\ \text{g/mol}$.

Key Points

Intermolecular forces (London dispersion, dipole–dipole, hydrogen bonding) compete with thermal energy to decide the physical state; gases have the weakest relative grip.
Boyle’s law: $P\propto\frac{1}{V}$ at constant $T$; Charles’ law: $\frac{V}{T}=\text{constant}$ at constant $P$; Gay-Lussac’s law: $\frac{P}{T}=\text{constant}$ at constant $V$.
Avogadro’s law: $V\propto n$ at fixed $T,P$; one mole of any gas occupies $22.4\ \text{L}$ at STP.
Ideal gas equation: $PV=nRT$ with $R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}=0.0821\ \text{L}\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1}$.
Dalton’s law: $P_{total}=p_1+p_2+\dots$ and $p_i=x_i\,P_{total}$; always use absolute temperature in kelvin.

Quick Check — 4 MCQs

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Q1.At constant temperature, if the volume of a fixed mass of gas is reduced to one-third, its pressure becomes:

Explanation: Boyle’s law: $P\propto\frac{1}{V}$, so reducing $V$ to $\frac{1}{3}$ triples $P$.

Q2.Which temperature scale must be used in the gas laws?

Explanation: Gas-law proportionalities hold only for absolute (kelvin) temperature, where $0\ \text{K}$ is true zero.

Q3.Equal volumes of two gases at the same $T$ and $P$ contain equal numbers of molecules. This is:

Explanation: Avogadro’s law links volume directly to the number of molecules (moles).

Q4.In a gas mixture, the partial pressure of a component equals:

Explanation: By Dalton’s law, $p_i=x_i P_{total}$, where $x_i$ is the mole fraction.

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