Ideal Gas Law — PV = nRT
Formula, Variables, Gas Constants, Assumptions & Solved Problems for Engineering Students
Last Updated: March 2026
📌 Key Takeaways
- PV = nRT — the ideal gas law relates pressure, volume, moles, and temperature of a gas.
- Universal gas constant: R = 8.314 J/(mol·K) = 8.314 kPa·L/(mol·K).
- Mass form: PV = mRsT, where Rs = R/M is the specific gas constant (287 J/kg·K for air).
- Assumptions: No intermolecular forces, negligible molecular volume, perfectly elastic collisions.
- Works well at low pressures and high temperatures. Breaks down near condensation and at very high pressures.
- Foundation for analysing all air-standard thermodynamic cycles (Otto, Diesel, Carnot with ideal gas).
1. The Formula — All Variables Defined
Ideal Gas Law
PV = nRT
| P | = Absolute pressure (Pa or kPa) |
| V | = Volume (m³ or L) |
| n | = Number of moles (mol) |
| R | = Universal gas constant = 8.314 J/(mol·K) |
| T | = Absolute temperature (K) — must be Kelvin, never Celsius |
The ideal gas law is an equation of state — it connects the four macroscopic properties of a gas (P, V, n, T) in a single equation. If any three are known, the fourth can be calculated. It emerges from the kinetic theory of gases, which models gas molecules as tiny particles in constant random motion with no interactions between them.
2. Different Forms of the Ideal Gas Law
| Form | Equation | When to Use |
|---|---|---|
| Molar form | PV = nRT | When you know the number of moles |
| Mass form | PV = mRsT | When you know the mass (engineering applications); Rs = R/M |
| Density form | P = ρRsT | When density is known (fluid mechanics, meteorology) |
| Specific volume form | Pv = RsT | Per unit mass; v = V/m = 1/ρ |
| Two-state form | P&sub1;V&sub1;/T&sub1; = P&sub2;V&sub2;/T&sub2; | When comparing two states of the same gas (constant n) |
3. Universal vs Specific Gas Constant
Universal Gas Constant
R = 8.314 J/(mol·K) = 8.314 kPa·L/(mol·K) = 0.0821 atm·L/(mol·K)
Same for ALL gases. Used with moles (n).
Specific Gas Constant
Rs = R / M
Where M = molar mass (kg/mol). Different for each gas. Used with mass (m).
| Gas | Molar Mass M (kg/mol) | Rs (J/kg·K) |
|---|---|---|
| Air | 0.02897 | 287 |
| Oxygen (O&sub2;) | 0.032 | 260 |
| Nitrogen (N&sub2;) | 0.028 | 297 |
| Hydrogen (H&sub2;) | 0.002 | 4,157 |
| Carbon Dioxide (CO&sub2;) | 0.044 | 189 |
| Helium (He) | 0.004 | 2,077 |
Exam tip: For air in GATE and university problems, use Rs = 287 J/(kg·K) unless otherwise stated. This value is assumed standard for air-standard cycle analysis.
4. Combined and Individual Gas Laws
The ideal gas law is a unification of several historically earlier gas laws:
| Law | Condition | Relationship | Equation |
|---|---|---|---|
| Boyle’s Law | Constant T, n | P inversely proportional to V | P&sub1;V&sub1; = P&sub2;V&sub2; |
| Charles’s Law | Constant P, n | V proportional to T | V&sub1;/T&sub1; = V&sub2;/T&sub2; |
| Gay-Lussac’s Law | Constant V, n | P proportional to T | P&sub1;/T&sub1; = P&sub2;/T&sub2; |
| Avogadro’s Law | Constant T, P | V proportional to n | V&sub1;/n&sub1; = V&sub2;/n&sub2; |
| Combined Gas Law | Constant n | All three combined | P&sub1;V&sub1;/T&sub1; = P&sub2;V&sub2;/T&sub2; |
5. Assumptions — What Makes a Gas “Ideal”
The ideal gas law assumes:
- No intermolecular forces: Gas molecules do not attract or repel each other.
- Negligible molecular volume: The volume of individual molecules is negligible compared to the container volume.
- Perfectly elastic collisions: When molecules collide with each other or the container walls, no kinetic energy is lost.
- Random motion: Molecules move in random directions with a distribution of speeds described by the Maxwell-Boltzmann distribution.
These assumptions are well-satisfied by real gases at low pressures (molecules are far apart) and high temperatures (kinetic energy dominates over intermolecular forces). Most gases behave approximately ideally at atmospheric pressure and room temperature.
6. When It Fails — Real Gases
The ideal gas law becomes inaccurate under these conditions:
| Condition | Why It Fails | Better Model |
|---|---|---|
| High pressure (> 10 atm) | Molecules are close together; molecular volume and intermolecular forces become significant | Van der Waals equation, Redlich-Kwong |
| Low temperature (near boiling point) | Intermolecular attractions dominate; gas approaches condensation | Van der Waals, Peng-Robinson |
| Polar molecules (H&sub2;O, NH&sub3;) | Strong dipole-dipole attractions violate the “no forces” assumption | Real gas tables, equations of state |
| Near the critical point | Gas and liquid become indistinguishable; ideal gas model breaks down completely | Steam tables, compressibility charts |
The compressibility factor Z = PV/(nRT) measures how much a real gas deviates from ideal behaviour. Z = 1 for ideal gas; Z < 1 when attractive forces dominate; Z > 1 when repulsive forces (molecular volume) dominate.
7. Worked Numerical Examples
Example 1: Finding Volume
Problem: 2 moles of an ideal gas are at 300 K and 200 kPa. Find the volume.
Solution
PV = nRT → V = nRT/P
V = (2 × 8.314 × 300) / (200 × 1000) = 4988.4 / 200,000
V = 0.02494 m³ = 24.94 L
Example 2: Two-State Problem (Combined Gas Law)
Problem: A gas at 100 kPa, 27°C occupies 0.5 m³. It is heated to 227°C at constant pressure. Find the new volume.
Solution
T&sub1; = 27 + 273 = 300 K, T&sub2; = 227 + 273 = 500 K
At constant pressure: V&sub1;/T&sub1; = V&sub2;/T&sub2; (Charles’s Law)
V&sub2; = V&sub1; × T&sub2;/T&sub1; = 0.5 × 500/300
V&sub2; = 0.833 m³
Example 3: Finding Density of Air
Problem: Find the density of air at 101.325 kPa and 25°C. Use Rs = 287 J/(kg·K).
Solution
P = ρRsT → ρ = P / (RsT)
T = 25 + 273 = 298 K
ρ = 101,325 / (287 × 298) = 101,325 / 85,526
ρ = 1.184 kg/m³
8. Common Mistakes Students Make
- Using Celsius instead of Kelvin: T in PV = nRT must ALWAYS be absolute temperature (Kelvin). Substituting 25°C instead of 298 K gives a wildly wrong answer.
- Mixing unit systems: If P is in Pa and V in m³, use R = 8.314 J/(mol·K). If P is in kPa and V in litres, use R = 8.314 kPa·L/(mol·K). Mixing Pa with litres or kPa with m³ without conversion causes errors.
- Using gauge pressure instead of absolute: The ideal gas law requires absolute pressure. If a problem states “gauge pressure = 200 kPa,” add atmospheric pressure: Pabs = 200 + 101.325 = 301.325 kPa.
- Confusing R (universal) with Rs (specific): R = 8.314 J/(mol·K) works with moles. Rs = R/M works with mass in kg. Using the wrong R gives answers off by the molar mass factor.
- Applying ideal gas law to steam/water: Steam near saturation is NOT an ideal gas. Use steam tables instead. The ideal gas law only approximates superheated steam at low pressures and high superheat.
9. Frequently Asked Questions
What is the ideal gas law?
The ideal gas law (PV = nRT) is an equation of state that relates the pressure, volume, amount (moles), and temperature of a gas that behaves ideally — meaning its molecules have no intermolecular forces and negligible volume. It unifies Boyle’s, Charles’s, and Avogadro’s laws into a single equation and is the foundation for gas calculations in engineering thermodynamics.
When does the ideal gas law fail?
It fails at high pressures (where molecular volume matters), low temperatures (near condensation), and for polar molecules with strong intermolecular attractions. Under these conditions, the compressibility factor Z deviates significantly from 1, and real gas equations or property tables should be used instead.
What is the difference between the universal gas constant and the specific gas constant?
The universal gas constant R = 8.314 J/(mol·K) is the same for all gases and is used in PV = nRT (with moles). The specific gas constant Rs = R/M depends on the particular gas (since each has a different molar mass M) and is used in PV = mRsT (with mass). For air, Rs = 287 J/(kg·K).