What is a gas, in simple words?
A gas is a huge crowd of very small particles called molecules. They fly in straight lines, hit each other and hit the walls. They are far apart, so a gas is mostly empty space.
We describe a gas with four numbers:
- P, pressure: push on each square metre of wall. Unit pascal (Pa = N/m²).
- V, volume: space the gas fills. Unit m³ (1 L = 10⁻³ m³).
- T, temperature in kelvin: T(K) = t(°C) + 273.15.
- n, amount in moles.
The three gas laws
Boyle's law (T fixed)
P ∝ 1/V, so PV = constant. Halve the volume, the pressure doubles. Reason: the same molecules hit a smaller wall area more often.
Charles's law (P fixed)
V ∝ T, so V/T = constant. T must be in kelvin. Hot molecules move faster and need more room to keep the same pressure.
Gay-Lussac's (pressure) law (V fixed)
P ∝ T. Heat a closed can and its pressure rises.
Avogadro's law
At the same P and T, equal volumes of all gases have the same number of molecules. So V ∝ n.
The perfect gas equation PV = nRT
Join the laws: V ∝ nT/P. Put in a constant R and you get
PV = nRT
R = 8.314 J mol⁻¹ K⁻¹ is the universal gas constant. It is the same for every gas.
Another form uses the number of molecules N = n NA: PV = N kB T, where kB = R/NA = 1.38 × 10⁻²³ J/K is the Boltzmann constant.
Using density ρ and molar mass M: n = m/M, so P = ρRT/M.
A gas that follows PV = nRT exactly at all P and T is a perfect (ideal) gas. Real gases come close at low pressure and high temperature, when molecules are far apart and fast.
Avogadro's number and the mole
One mole of anything has NA = 6.022 × 10²³ particles. This is Avogadro's number.
- Number of moles n = N/NA = (mass in g)/(molar mass in g).
- At STP (273 K, 1 atm) one mole of any ideal gas fills about 22.4 L.
- So 22.4 L of hydrogen and 22.4 L of oxygen at STP hold the same 6.022 × 10²³ molecules, though oxygen is 16 times heavier.
Work done in compressing a gas
When a piston of area A moves in by a small distance Δx against gas pressure P, the force is F = PA and the work done on the gas is F Δx = P A Δx = P ΔV. Add up all the small pieces:
W = ∫ P dV = area under the P–V graph.
At constant pressure
W = P (V₁ − V₂).
At constant temperature (isothermal)
Here P = nRT/V changes as V changes. Adding up gives
W on gas = nRT ln(V₁/V₂) = 2.303 nRT log₁₀(V₁/V₂)
Sign rule used in NCERT style: work done by the gas is positive when it expands; when you compress it, the gas does negative work and you do positive work on it.
Try it: the 3D and at home
- In the 3D: set T = 300 K, V = 25 L. Predict P when you halve V. Then move the slider and check. Next, double T and see P double.
- At home: close the tip of a plastic syringe (no needle) with your finger and push the plunger. Feel it push back harder as the air gets squeezed. Put a slightly blown balloon in the fridge for 20 minutes: it shrinks (Charles's law).
Board exam focus
- Always turn °C into K and litres into m³ before using PV = nRT.
- Know both forms: PV = nRT and PV = N kB T; kB = R/NA.
- Isothermal work: W = 2.303 nRT log(V₁/V₂). Remember the area idea.
- Draw P–V graphs: Boyle is a hyperbola; PV vs P is a flat line for an ideal gas.
Key formulas and definitions
- PV = nRT
- PV = N k_B T, k_B = R/N_A = 1.38 × 10⁻²³ J/K
- R = 8.314 J mol⁻¹ K⁻¹
- N_A = 6.022 × 10²³ mol⁻¹
- P = ρRT/M
- Boyle: P₁V₁ = P₂V₂ (T fixed)
- Charles: V₁/T₁ = V₂/T₂ (P fixed)
- P₁V₁/T₁ = P₂V₂/T₂
- W = ∫P dV; isothermal W = nRT ln(V₁/V₂)
Worked examples
1. A gas occupies 6 L at 2 atm. At the same temperature it is squeezed to 3 L. Find the new pressure.
Step 1: T is fixed, so use Boyle: P₁V₁ = P₂V₂. Step 2: 2 × 6 = P₂ × 3. Step 3: P₂ = 12/3. Answer: P₂ = 4 atm.
2. A balloon holds 2.0 L of air at 27 °C. At the same pressure it is warmed to 177 °C. Find its new volume.
Step 1: change to kelvin: T₁ = 300 K, T₂ = 450 K. Step 2: Charles: V₂ = V₁ × T₂/T₁ = 2.0 × 450/300. Answer: V₂ = 3.0 L.
3. How many molecules are in 8 g of oxygen gas (O₂, M = 32 g/mol)?
Step 1: n = 8/32 = 0.25 mol. Step 2: N = n N_A = 0.25 × 6.022 × 10²³. Answer: N ≈ 1.51 × 10²³ molecules.
4. Find the pressure of 2 mol of an ideal gas in a 0.05 m³ vessel at 300 K. (R = 8.31 J/mol K)
Step 1: P = nRT/V. Step 2: P = 2 × 8.31 × 300 / 0.05 = 4986 / 0.05. Answer: P ≈ 9.97 × 10⁴ Pa (about 1 atm).
5. A tyre has air at 27 °C and 2.0 × 10⁵ Pa. After a drive the air is at 57 °C, volume unchanged. Find the new pressure.
Step 1: V fixed, so P/T is constant. Step 2: T₁ = 300 K, T₂ = 330 K. Step 3: P₂ = 2.0 × 10⁵ × 330/300. Answer: P₂ = 2.2 × 10⁵ Pa.
6. Find the density of nitrogen (M = 28 g/mol) at 1.0 × 10⁵ Pa and 280 K.
Step 1: ρ = PM/(RT). Step 2: M = 0.028 kg/mol. Step 3: ρ = 1.0 × 10⁵ × 0.028 / (8.31 × 280) = 2800 / 2326.8. Answer: ρ ≈ 1.20 kg/m³.
7. 1 mol of an ideal gas at 300 K is compressed slowly at constant temperature from 20 L to 10 L. Find the work done on the gas. (ln 2 = 0.693)
Step 1: isothermal, so W on gas = nRT ln(V₁/V₂). Step 2: = 1 × 8.314 × 300 × ln(20/10). Step 3: = 2494.2 × 0.693. Answer: W ≈ 1.73 × 10³ J done on the gas (the gas does −1.73 kJ).
8. A gas at a constant pressure of 1.0 × 10⁵ Pa is compressed from 3.0 L to 1.0 L. Find the work done on it.
Step 1: W = P (V₁ − V₂). Step 2: ΔV = 2.0 L = 2.0 × 10⁻³ m³. Step 3: W = 1.0 × 10⁵ × 2.0 × 10⁻³. Answer: W = 200 J on the gas.
Common mistakes
- Using °C instead of kelvin in PV = nRT or Charles's law. Always add 273.
- Putting litres straight into SI formulas. 1 L = 10⁻³ m³.
- Mixing up R (per mole) and k_B (per molecule): PV = nRT but PV = N k_B T.
- Forgetting the sign of work: compressing a gas means work is done on it, and the gas's own work is negative.