Ideal Gas Law PV = nRT — Units, R Values and Solved Problems
PV = nRT is one equation that replaces Boyle's law, Charles's law, Gay-Lussac's law and Avogadro's law all at once. Almost nobody loses marks on the physics of it. Marks are lost on units — degrees Celsius left in the temperature slot, millilitres left in the volume slot, or the wrong value of R. This guide gives you the equation, every common R value, and four solved problems that show the unit discipline in action.
The equation
Rearranged: P = nRT/V · V = nRT/P · n = PV/RT · T = PV/nR
What each symbol means
| Symbol | Meaning | Usual unit |
|---|---|---|
| P | Pressure of the gas | atm, Pa, bar or mmHg — must match R |
| V | Volume of the container | L or m³ — must match R |
| n | Amount of gas = mass ÷ molar mass | mol |
| R | Universal gas constant | depends on the units chosen |
| T | Absolute temperature | kelvin, always (K = °C + 273.15) |
Every R value, and when to use it
R is one physical constant expressed in different units. Pick the row whose pressure and volume units match your data — that is the entire decision.
| Value of R | Units | Use it when |
|---|---|---|
| 0.0821 | L atm K⁻¹ mol⁻¹ | P in atm, V in litres — the default in Indian chemistry papers |
| 8.314 | J K⁻¹ mol⁻¹ (= Pa m³ K⁻¹ mol⁻¹) | P in pascals, V in m³; also for energy work (ΔG, Nernst, Arrhenius) |
| 8.314 | L kPa K⁻¹ mol⁻¹ | P in kPa, V in litres (1 kPa·L = 1 J) |
| 0.08314 | L bar K⁻¹ mol⁻¹ | P in bar, V in litres |
| 62.36 | L Torr K⁻¹ mol⁻¹ | P in mmHg/torr, V in litres |
| 1.987 | cal K⁻¹ mol⁻¹ | Older thermodynamics problems quoted in calories |
Conversions worth memorising: 1 atm = 101325 Pa = 101.325 kPa = 1.01325 bar = 760 mmHg.
Worked example 1 — Find the pressure
2.00 mol of a gas occupies 10.0 L at 300 K. Find the pressure in atm.
Units are litres and we want atm, so R = 0.0821 L atm K⁻¹ mol⁻¹.
P = nRT / V = (2.00 × 0.0821 × 300) ÷ 10.0
0.0821 × 300 = 24.63
24.63 × 2.00 = 49.26
P = 49.26 ÷ 10.0 = 4.93 atm
Worked example 2 — Molar volume at STP (both definitions)
This is where "22.4 litres" comes from — and why the modern answer is 22.7.
Find the volume of 1 mol of an ideal gas at 273.15 K.
(a) At 1 bar (100000 Pa) — current IUPAC STP:
V = RT/P = (8.314 × 273.15) ÷ 100000 = 2270.98 ÷ 100000 = 0.0227098 m³ =
22.71 L
(b) At 1 atm (101325 Pa) — the older definition still printed in many books:
V = 2270.98 ÷ 101325 = 0.0224140 m³ = 22.414 L
So 22.4 L mol⁻¹ is correct at 1 atm and 22.7 L mol⁻¹ at 1 bar. Use whichever your syllabus or the question specifies, and say which you used.
Worked example 3 — Molar mass from gas density
Substituting n = m/M into PV = nRT gives one of the most useful derived forms:
A gas has a density of 1.25 g L⁻¹ at 1.00 atm and 273 K. Identify it.
M = (1.25 × 0.0821 × 273) ÷ 1.00
0.0821 × 273 = 22.4133
M = 1.25 × 22.4133 = 28.0 g mol⁻¹
28 g mol⁻¹ matches N₂ (2 × 14.007 = 28.014) or CO (12.011 + 15.999 = 28.010). Density alone cannot separate them — you would need a chemical test.
Worked example 4 — Mixed units, done carefully
Find the moles of gas in 250 mL at 27 °C and 700 mmHg.
Convert everything first — this is the habit that saves marks:
P = 700 ÷ 760 = 0.9211 atm
V = 250 mL = 0.250 L
T = 27 + 273 = 300 K
n = PV / RT = (0.9211 × 0.250) ÷ (0.0821 × 300)
Numerator = 0.23026
Denominator = 24.63
n = 0.23026 ÷ 24.63 = 9.35 × 10⁻³ mol
Alternative route with no conversion: keep P in mmHg and use R = 62.36 L Torr K⁻¹ mol⁻¹.
n = (700 × 0.250) ÷ (62.36 × 300) = 175 ÷ 18708 = 9.35 × 10⁻³ mol ✔ identical.
When gases are not ideal
PV = nRT assumes gas molecules have zero volume and do not attract one another. Both assumptions fail at high pressure (molecular volume is no longer negligible compared with the container) and at low temperature (attractions matter because molecules move slowly). Real gases behave most ideally at low pressure and high temperature. When accuracy matters, the van der Waals equation adds a correction term for each effect:
For Class 11–12 and most entrance-exam numericals, the ideal form is what is expected unless the question hands you a and b.
Common mistakes that cost marks
- Leaving temperature in Celsius. At 27 °C, using 27 instead of 300 makes the answer eleven times wrong. Convert first, every single time.
- Leaving volume in millilitres while using R = 0.0821, which expects litres. Divide by 1000.
- Mixing R with the wrong pressure unit. 700 mmHg with R = 0.0821 gives an answer 760 times too large.
- Using 22.4 L mol⁻¹ at every temperature and pressure. It is the molar volume at STP only; at 300 K and 2 atm it is nowhere near.
- Forgetting to convert mass to moles. n is moles, not grams — divide by the molar mass.
- Using gauge pressure instead of absolute pressure. P in PV = nRT is always absolute.
Where PV = nRT appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | States of matter, gas laws, molar volume, Dalton's law of partial pressures |
| CBSE/ICSE Class 12 | Volume of gas evolved in a reaction, solutions and vapour pressure |
| JEE / NEET | Molar mass from density, gas mixtures, non-ideal behaviour and Z factor |
| IIT-JAM / CUET-PG | Kinetic theory, van der Waals constants, critical conditions |
| GATE / CSIR-NET | Physical chemistry thermodynamics, reaction engineering, gas-phase kinetics |
Let the calculator handle the units. The Ideal Gas Law tool solves PV = nRT for whichever quantity you leave blank and accepts atm, kPa, bar or mmHg with L or mL, so a unit slip cannot silently ruin your answer.
Open the Ideal Gas Law (PV = nRT) Calculator →Gas laws are pure marks if the unit discipline is trained early. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India, with home tuition available in Delhi-NCR — see abcchemistry.in.