PV = nRT, where n is number of moles, R = 8.314 J/mol/K. Equivalently PV = NkT (k = Boltzmann constant). Combines Boyle's, Charles's, and Avogadro's laws.
-- NCERT, p. 2Equation of State Perfect Gas
8 MCQs1 revision card9-step worked example
Source: NCERT Kinetic TheoryPYQ coverage: NEET 2020, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The ideal gas equation PV = nRT is the single equation of state for a perfect gas, and NEET tests it with a predictable pattern: give you two or three state variables, change one, ask for the missing one. The trap is almost always temperature conversion — using °C where the formula demands kelvin.
The equation and its two forms. NCERT Class 11 Physics Chapter 12, page 2 states the ideal gas law as:
PV = nRT (molar form) = NkT (molecular form)
Here P is pressure (Pa), V is volume (m³), n is moles, R = 8.314 J mol⁻¹ K⁻¹ is the universal gas constant, N is the total number of molecules, k = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant, and T is absolute temperature in kelvin.
When it applies. The equation holds for an ideal gas — low pressure, high temperature (far from liquefaction). Real gases deviate at high pressures and low temperatures.
The temperature-conversion trap. When a problem states temperature as 27 °C, you must convert: T = 27 + 273 = 300 K. Using 27 directly in PV = nRT gives an answer off by a factor of roughly 11. This is a high-frequency distractor in NEET options — the wrong answer from forgetting conversion often appears as a choice.
Combined gas form for process problems. For a fixed amount of gas undergoing a change from state 1 to state 2:
P₁V₁/T₁ = P₂V₂/T₂
This is just PV = nRT applied to two states with n constant. Watch for problems that hold one variable fixed (isobaric: P constant; isothermal: T constant; isochoric: V constant) — the equation simplifies accordingly.
Connecting R and k. Since R = Nₐk (where Nₐ is Avogadro's number), the molar and molecular forms are equivalent. NEET occasionally tests whether you can switch between the two.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The equation of state of a perfect gas is PV = nRT. What does R represent, and what is its SI value?
MCQ 2Easy RecallPractice
The molecular form of the ideal gas equation is PV = NkT. The relationship between R and k is:
MCQ 3Easy RecallPractice
Under which condition does a real gas behave approximately as an ideal gas?
MCQ 4Direct ApplicationPractice
Two moles of an ideal gas occupy a volume of 0.050 m³ at a pressure of 1.0 × 10⁵ Pa. What is the temperature of the gas? (R = 8.314 J mol⁻¹ K⁻¹)
MCQ 5Direct ApplicationPractice
An ideal gas at 27 °C and 1.0 × 10⁵ Pa is heated at constant volume until the pressure doubles. What is the final temperature?
MCQ 6Direct ApplicationPractice
An ideal gas at temperature T and pressure P occupies volume V. If the temperature is raised to 3T at constant pressure, the new volume is:
MCQ 7CalculationPractice
A container holds an ideal gas at 2.0 × 10⁵ Pa and 400 K. The gas is compressed to half its volume while being cooled to 200 K. What is the final pressure?
MCQ 8CalculationPractice
A fixed mass of ideal gas at 1.0 × 10⁵ Pa and 27 °C occupies 0.030 m³. It is first heated at constant pressure to 127 °C, then compressed at constant temperature until its volume returns to 0.030 m³. What is the final pressure?
Quick recall before you leave
Worked Example
- 1
Given
- P = 3.0 × 10⁵ Pa - V₁ = 2.0 × 10⁻³ m³ - T₁ = 27 °C = 300 K - V₂ = 4.0 × 10⁻³ m³ - Process: constant pressure - R = 8.314 J mol⁻¹ K⁻¹
- 2
Required
(a) Number of moles n. (b) Final temperature T₂ in °C.
- 3
Concept
PV = nRT connects P, V, T, and n. For part (a), use the initial state. For part (b), constant-pressure heating means V₁/T₁ = V₂/T₂.
- 4
Formula
(a) n = PV₁/(RT₁) (b) T₂ = T₁ × (V₂/V₁)
- 5
Substitution
(a) n = (3.0 × 10⁵ × 2.0 × 10⁻³) / (8.314 × 300) (b) T₂ = 300 × (4.0 × 10⁻³ / 2.0 × 10⁻³)
- 6
Calculation
(a) Numerator = 6.0 × 10² = 600. Denominator = 2494.2. n = 600/2494.2 ≈ 0.241 mol. Note on exact constants: R = 8.314 J mol⁻¹ K⁻¹ is a defined constant. The given values (3.0 × 10⁵, 2.0 × 10⁻³, 300 K) each have 2 significant figures, so the answer is reported to 2 significant figures. (b) T₂ = 300 × 2 = 600 K. In Celsius: 600 − 273 = 327 °C. The factor 2 (volume ratio) and 273 (K-to-°C offset) are exact and do not limit significant figures.
- 7
Final answer
(a) n ≈ 0.24 mol (2 significant figures, limited by the given pressure and volume). (b) T₂ = 600 K = 327 °C.
- 8
Common trap
The high-frequency distractor here is forgetting to convert 27 °C to 300 K. Using T₁ = 27 in PV = nRT gives n = 600/(8.314 × 27) ≈ 2.67 mol — more than 10× too large. NEET options regularly include this wrong answer.
- 9
Similar NEET-style question
An ideal gas at 1.0 × 10⁵ Pa and 127 °C has a volume of 5.0 × 10⁻³ m³. It is cooled at constant pressure to 27 °C. Find the final volume and the number of moles. (Answer: V₂ = 3.75 × 10⁻³ m³, n ≈ 0.150 mol.) ---
Before solving, remember these
Formulas
5 formulas — click to collapse
Average translational KE per molecule
Microscopic interpretation of temperature: T is direct measure of average translational kinetic energy.
| Symbol | Quantity | SI Unit |
|---|---|---|
| k | Boltzmann constant | J/K |
| T | absolute temperature | K |
Valid when
- Translational degrees of freedom only
- Ideal gas
Cv from degrees of freedom
Each quadratic DoF contributes (1/2)R to molar Cv. Mono: f=3, Cv=3R/2; di-rigid: f=5, Cv=5R/2; poly-rigid: f=6, Cv=3R.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Cv | molar specific heat | J/mol/K |
| f | degrees of freedom | - |
| R | gas constant | J/mol/K |
Valid when
- Equipartition holds (temperature high enough)
- Quadratic energy modes
Ideal gas equation
Fundamental equation of state of ideal gas relating pressure, volume, temperature.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| V | volume | m^3 |
| n | moles | mol |
| R | 8.314 | J/mol/K |
| N | molecule count | - |
| k | Boltzmann 1.38e-23 | J/K |
| T | temp | K |
Valid when
- Gas obeys ideal gas approximation (low pressure, high temperature relative to phase transitions)
Mean free path of gas molecule
Average distance between successive molecular collisions.
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | mean free path | m |
| n | number density | 1/m^3 |
| d | molecular diameter | m |
Valid when
- Hard-sphere model
- Equilibrium gas
RMS speed of gas molecules
Root-mean-square molecular speed; depends on T and molar mass M (or molecular mass m).
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | gas constant | J/mol/K |
| T | temp | K |
| M | molar mass | kg/mol |
| k | Boltzmann | J/K |
| m | molecular mass | kg |
Valid when
- Ideal gas
- Maxwell-Boltzmann distribution
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student treats v_rms ∝ T instead of √T. Doubling T does NOT double v_rms; it multiplies by √2.
When it triggers
Question asks for new v_rms after T change.
How to avoid
v_rms = √(3RT/M). v_rms ∝ √T. To double v_rms, T must quadruple.
Root cause: formula misuse
Correction
v_rms = √(3RT/M), so v_rms ∝ √T. To double v_rms, T must quadruple (factor of 4). Common error: assume doubling T doubles v_rms.
Past Year Questions
6 questions from NEET 2020, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
10.156 kg
20.125 kg
30.144 kg
40.116 kg
NTA Answer: Option 4(final)
NEET 2024Revised key
1P3 > P2 > P1
2P1 > P3 > P2
3P2 > P1 > P3
4P1 > P2 > P3
NTA Answer: Option 4(revised_final)
NEET 2023
13295°C
23097 K
3223 K
4669°C
NTA Answer: Option 1(final)
NEET 2022
15.6 m3
25.6 × 106 m3
35.6 × 103 m3
45.6 × 10–3 m3
NTA Answer: Option 1(final)
NEET 2020
12 k B T 1 1
22 k B T 3 3
32 k B T 5 5
42 k B T
NTA Answer: Option 3(final)
NEET 2020
12 n 2 2 d 2 1 1
22 n d 1 1
32 n d2 1 1
42 n 2 d 2
NTA Answer: Option 3(final)
How NEET usually asks this
4 recurring patterns from past papers — click to collapse
Average translational KE per molecule = (3/2)kT (monoatomic). Total via DoF.
Direct ApplicationEasy
Common distractors
uses 3 2 RT not 3 2 kT
Confuses per-mole RT with per-molecule kT
PV = nRT. Given two of (P, V, T, n) and changes, find missing.
Multi StepEasy
Common distractors
forgets temperature conversion
Mixes °C with K
Mean free path scaling with n, d. lambda = 1/(sqrt(2)*pi*n*d^2).
InterpretationEasy
Common distractors
ignores d squared
Treats d linearly
Find new T given v_rms scales by factor k. v_rms ∝ sqrt(T), so T_new = T*k^2.
Multi StepMedium
Common distractors
uses linear scaling
Treats v_rms ∝ T not sqrt(T)
Sources
NCERT refs: Class 11 Physics Chapter 12, p.2
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