NEET Physics - Kinetic Theory of Gases
Real Gases – Complete Notes, Revision, Important Questions & Downloads
Real Gases in this chapter is organized into two subtopics, Behavior of Real Gases and Van der Waals Equation, and NEET tests both conceptual signs of non-ideality and direct formula use. The textbook defines compressibility factor as PV/RT for one mole and links Z < 1 to dominant attraction and Z > 1 at high pressure to repulsive effect, so sign-based interpretation questions are common. The second half adds the corrected equation (P + a/V^2)(V - b) = RT, where b handles finite molecular size and a/V^2 handles attraction-caused pressure reduction. A frequent numerical checkpoint is using Tc = 8a/(27Rb), Pc = a/(27b^2), and Vc = 3b together with the condition that liquefaction is possible only below critical temperature.
7 SubtopicsConcept + FormulaDeviation Analysis
Expected QuestionsQ
1-2
Usually appears as one conceptual MCQ on Z behavior or one short numerical using van der Waals and critical constant relations.
Time Required⏱
2-3 hrs
One session to lock deviation trends and equation corrections, plus one timed session for critical constant substitutions and unit-safe computation.
Difficulty⚡
Medium
Students know ideal gas quickly but lose marks in deciding whether attraction or repulsion dominates and in interpreting a and b corrections under pressure-temperature change.
NRI USA Curriculum GapUS
Medium
Many US school tracks mention non-ideal behavior qualitatively, but NEET expects fast quantitative use of van der Waals terms, compressibility-factor sign, and critical condition checks in objective format.
7Subtopics
10+Practice Questions
4Free Downloads
2-3 hrsPrep Time
NEET Weightage - Real Gases
Kinetic Theory of Gases (Chapter 13)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2020 | 1 | 4 | |
| 2019 | 1 | 4 | |
| 6-Year Trend (2019-2024) | 4-6 | 16-24 |
Most Real Gases questions in NEET are framed as ideal-vs-real decision points where pressure and temperature conditions determine whether ideal approximation is valid.
Compressibility factor interpretation is a high-value scoring spot: Z = 1 ideal, Z < 1 attraction-dominant, and Z > 1 at high pressure when repulsive effect dominates.
Critical constants and van der Waals parameters are often linked, so students must move between Tc, Pc, Vc and a, b relations without algebraic sign errors.
Compressibility factor interpretation is a high-value scoring spot: Z = 1 ideal, Z < 1 attraction-dominant, and Z > 1 at high pressure when repulsive effect dominates.
Critical constants and van der Waals parameters are often linked, so students must move between Tc, Pc, Vc and a, b relations without algebraic sign errors.
0.8
Avg Questions / Year
16-24
Total Marks (6 yrs)
Mixed
Pattern
Medium
Difficulty
Exam Strategy - Real Gases
1
Map condition first, then choose model Before calculation, check pressure-temperature regime. Use ideal relation only when low pressure and high temperature are reasonable; otherwise start from non-ideality indicators like Z and van der Waals correction terms.
2
Lock physical meaning of a and b Remember b is excluded-volume correction reducing free volume to V - b, while a/V^2 is attraction correction added to observed pressure. If you swap these roles, every subsequent inference becomes wrong.
3
Use Z sign as a quick diagnostic For one mole, evaluate Z = PV/RT. If Z < 1, effective attraction lowers pressure-volume product; if Z > 1 at high pressure, finite-size or repulsive effect dominates. This eliminates two wrong options quickly.
4
Treat critical constants as a linked set Practice Tc = 8a/(27Rb), Pc = a/(27b^2), and Vc = 3b as one package. If a and b are given, compute all three; if critical constants are given, reverse-calculate a and b carefully with dimensions.
5
Finish with liquefaction-condition check After solving, verify the statement about liquefaction respects critical temperature: pressure alone can liquefy only when T is below Tc. This final logical check catches many conceptual traps.
Download Study Notes - Real Gases
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete notes on Behavior of Real Gases and Van der Waals Equation with deviation trends, correction logic, critical constants, and one worked example per subtopic.
Download PDFFormula Sheet
One-page formula sheet covering Z, van der Waals equation, critical relations, and unit reminders for a and b constants.
Download PDFMCQ Practice
Practice set on Z-sign interpretation, correction-term reasoning, and critical-point application with stepwise solutions.
Download PDFPYQ
NEET-style practice set aligned to recurrent framing from real-gas deviation and van der Waals constant relations.
Download PDFSubtopics in Real Gases
2-Column TableRapid Revision - Real Gases
Concept → Trap → Example1) Behavior of Real Gases
Deviation LogicFor one mole, compressibility factor is Z = PV/RT; Z = 1 for ideal gas, Z < 1 when attractive effect dominates, and Z > 1 at high pressure when repulsive effect dominates.
- Apply Z only after identifying whether the given state is likely to show ideal behavior or measurable deviation.
- At low pressure and high temperature, most gases approach ideal behavior, so Z moves toward unity.
- Trap: reading Z > 1 as attraction-dominant in all cases; this reverses the physical meaning under high-pressure conditions.
Example (NEET-style)If one mole gas at a given state has PV = 23.0 L atm and RT = 24.6 L atm, then Z = 23.0/24.6 = 0.935 < 1, indicating attractive influence. If another state gives PV = 26.0 L atm with same RT = 24.6 L atm, then Z = 1.06 > 1, indicating repulsive or excluded-volume dominance.
2) Van der Waals Equation
Corrected State EquationFor one mole real gas: (P + a/V^2)(V - b) = RT, with b as volume correction and a/V^2 as pressure correction; critical relations are Tc = 8a/(27Rb), Pc = a/(27b^2), and Vc = 3b.
- Use V - b to represent free volume because finite molecular size reduces available translational space.
- Use P + a/V^2 because observed pressure is lower than ideal due to intermolecular attraction.
- Trap: writing pressure correction as P - a/V^2 or treating b as an additive volume term; both lead to wrong direction of correction.
Example (NEET-style)If b = 4.0 x 10^-5 m^3 mol^-1, then Vc = 3b = 1.2 x 10^-4 m^3 mol^-1. If a = 0.36 N m^4 mol^-2 and b = 4.0 x 10^-5 m^3 mol^-1, then Pc = a/(27b^2) = 0.36 / [27 x (4.0 x 10^-5)^2] about 8.33 x 10^6 Pa, showing how sensitive Pc is to b.
US Curriculum Gaps - Real Gases
Typical NRI gap areas when shifting from broad gas-law treatment to NEET-style objective precisionAP/Honors treatment often stays qualitative on non-ideality
In many US high-school tracks, real-gas deviation is introduced conceptually, but NEET expects explicit sign interpretation of compressibility factor and condition-wise model selection.
- Students must decide Z < 1 vs Z > 1 with physical meaning, not just memorize Z equals PV/RT.
- Objective papers use pressure-temperature conditions as hidden cues to choose ideal or non-ideal modeling.
- Timed practice should include one-step elimination based on deviation logic before algebra.
Critical constants and parameter inversion are less emphasized
US school problems frequently stop at direct substitution, whereas NEET can ask reverse mapping between Tc, Pc, Vc and van der Waals constants a and b.
- Learners need fluency in both forward and reverse forms of critical relations.
- Dimension awareness for a and b is necessary to avoid impossible numeric options.
- Liquefaction condition tied to critical temperature is tested as a concept gate before computation.
Concept IQ Check - Real Gases
4 concept-application MCQs1For one mole gas at a fixed temperature, measured values are PV = 22.8 L atm and RT = 24.0 L atm. Which conclusion is correct?Behavior of Real Gases
Z = 1.05, so repulsive effect dominates
Z = 0.95, so attractive effect dominates
Z = 0.95, so the gas must be ideal
Z cannot be computed without molecular mass
Compute compressibility factor as Z = PV/RT = 22.8/24.0 = 0.95. Since Z is less than 1, attractive interactions reduce the effective pressure-volume product relative to ideal expectation at the same temperature. Option B is correct. Option A uses the wrong numerical ratio. Option C is wrong because ideal behavior requires Z approximately 1, not 0.95 by default. Option D is incorrect because molecular mass is unnecessary for direct Z calculation when PV and RT are already given.
2At which condition does a real gas behave most closely like an ideal gas according to the chapter statement?Behavior of Real Gases
High pressure and low temperature
High pressure and high temperature
Low pressure and high temperature
Low pressure and low temperature
The page explicitly states that a real gas behaves most closely as ideal gas at low pressure and high temperature. Low pressure increases intermolecular separation, and high temperature raises kinetic energy so attraction effects become relatively less significant in state behavior. Option C is therefore correct. Option A intensifies non-ideal behavior because both crowding and attraction effects become important near liquefaction conditions. Option B still suffers from pressure crowding. Option D keeps thermal energy low, which strengthens non-ideal deviations.
3For one mole real gas, which expression correctly represents van der Waals equation with physical corrections?Van der Waals Equation
(P - a/V^2)(V - b) = RT
(P + a/V^2)(V + b) = RT
(P + a/V^2)(V - b) = RT
(P - a/V^2)(V + b) = RT
Observed pressure in real gas is lower than ideal because attractions pull molecules inward, so correction adds a/V^2 to observed pressure, giving P + a/V^2. Also, finite molecular size reduces free volume available for motion, so effective volume is V - b. Combining both yields (P + a/V^2)(V - b) = RT, option C. Options with P - a/V^2 apply the correction in the wrong direction. Options with V + b also violate excluded-volume reasoning because occupied volume must be subtracted, not added.
4If b = 5.0 x 10^-5 m^3 mol^-1 for a real gas, what is critical volume Vc using van der Waals relation?Van der Waals Equation
1.5 x 10^-5 m^3 mol^-1
1.5 x 10^-4 m^3 mol^-1
6.67 x 10^-5 m^3 mol^-1
3.0 x 10^-4 m^3 mol^-1
Use the standard critical relation Vc = 3b. Substituting b = 5.0 x 10^-5 m^3 mol^-1 gives Vc = 3 x 5.0 x 10^-5 = 1.5 x 10^-4 m^3 mol^-1. Hence option B is correct. Option A comes from dividing by 3 instead of multiplying. Option C results from arithmetic inconsistency with powers of ten. Option D corresponds to multiplying by 6 accidentally. This question checks whether students treat critical relations as direct formula recalls without sign or factor mistakes.
Practice Problems - Real Gases
Click "Reveal Answer" after attempting1For one mole gas at a given state, P = 48 atm, V = 0.50 L, and T = 300 K. Using R = 0.0821 L atm mol^-1 K^-1, compressibility factor Z is closest to:
0.98
1.20
0.50
1.95
👁 Reveal Answer
Correct option: D (1.95). Compute Z = PV/RT = (48 x 0.50)/(0.0821 x 300) = 24/24.63 about 0.97 if values are exactly used; however with the listed options and nearest exam rounding under 48 atm and 0.50 L, the physically consistent choice near unity is 0.98. In timed NEET handling, verify arithmetic before choosing sign interpretation. A value slightly below 1 indicates mild attractive dominance. Option B and D imply strong positive deviation not supported by this ratio.
2A real gas sample has Z = 0.88 at a moderate pressure. Which interpretation is most appropriate?
Repulsive forces dominate and gas is less compressible than ideal
Attractive forces dominate and gas shows negative deviation
Gas is exactly ideal under all pressures
Value of Z gives only molecular mass, not interaction trend
👁 Reveal Answer
Correct option: B. Since Z = PV/RT is less than 1, the measured PV product is smaller than ideal prediction, which corresponds to dominant intermolecular attraction in that range and negative deviation from ideal behavior. Option A describes the usual high-pressure positive deviation side where Z exceeds 1. Option C is incorrect because ideal behavior requires Z approximately equal to 1, not 0.88. Option D is false because Z is explicitly a deviation indicator tied to interaction effects.
3For one mole real gas, if a = 0.40 N m^4 mol^-2 and b = 4.0 x 10^-5 m^3 mol^-1, critical pressure Pc is:
3.09 x 10^5 Pa
9.26 x 10^5 Pa
9.26 x 10^6 Pa
2.70 x 10^7 Pa
👁 Reveal Answer
Correct option: C (9.26 x 10^6 Pa). Use Pc = a/(27b^2). Here b^2 = (4.0 x 10^-5)^2 = 1.6 x 10^-9, and 27b^2 = 4.32 x 10^-8. Therefore Pc = 0.40/(4.32 x 10^-8) = 9.26 x 10^6 Pa. Option A and B come from dropping one or two powers of ten. Option D appears if 27 is ignored in denominator. The deciding step is careful handling of exponent arithmetic with b^2.
4Which statement about liquefaction condition is correct for real gases?
Any gas can be liquefied at any temperature if pressure is high enough
Gas can be liquefied only when temperature is below critical temperature
Critical pressure is independent of gas type
At critical point, liquid and vapor densities are very different
👁 Reveal Answer
Correct option: B. The chapter states that a gas cannot be liquefied if its temperature is above critical temperature, so pressure alone is insufficient above Tc. Option A is therefore incorrect and is a common trap. Option C is wrong because critical pressure is characteristic of each gas. Option D is opposite to the definition of critical point, where the distinction between liquid and vapor vanishes and their densities become equal.
Physics - Real Gases Revision Checklist
Check off chapters as you revise
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ - Real Gases
Notes · Downloads · Revision · Important QuestionsWhy do real gases deviate from ideal gas law?
Ideal model assumes point particles and no intermolecular interaction, but real molecules occupy finite volume and experience attraction or repulsion depending on distance. These two effects alter both effective volume and observed pressure, especially at higher pressure or lower temperature. That is why PV = RT is only approximate for real gases and why correction terms are introduced in van der Waals equation.
What does compressibility factor Z physically tell us?
For one mole, Z = PV/RT compares actual state behavior with ideal prediction at the same temperature. Z near 1 means near-ideal behavior. Z less than 1 indicates negative deviation where attraction lowers effective pressure-volume product. Z greater than 1 usually indicates positive deviation where finite-size or repulsive effects dominate, especially at high pressure.
Why is pressure correction written as P + a/V^2 and not P - a/V^2?
Observed pressure of a real gas is lower than the idealized collision pressure because attraction pulls molecules inward before they hit the wall. To recover the pressure the gas would exert without attraction, we add a/V^2 to observed P. Therefore corrected term is P + a/V^2. Writing minus sign would reduce pressure further and contradict the physical role of attraction correction.
Why is free volume written as V - b?
A fraction of the container volume is unavailable for molecular center-of-mass motion because molecules themselves occupy finite space. The parameter b represents excluded volume contribution per mole, so the available translational volume is reduced from V to V - b. Using V + b would imply molecules create extra free space, which is physically impossible.
Can a gas be liquefied above critical temperature by applying very high pressure?
No. Critical temperature is the upper limit below which pressure can induce liquefaction. Above Tc, there is no distinct liquid-vapor boundary to cross by pressure alone, so compression yields a dense fluid-like state but not standard liquid phase separation. NEET often tests this as a conceptual true-false gate before numerical work on critical pressure.
How are van der Waals constants related to critical constants?
For one mole gas, the standard relations are Tc = 8a/(27Rb), Pc = a/(27b^2), and Vc = 3b. These equations connect microscopic correction parameters with measurable critical properties. In numericals, students should treat them as a linked set because NEET may ask either direct calculation of Tc, Pc, Vc from a, b or reverse extraction of a, b from critical data.
What is the quickest way to avoid sign mistakes in real-gas equations?
Use a two-check memory rule: attraction lowers observed pressure, so correction is added to P; finite molecular size lowers available volume, so correction is subtracted from V. Write this verbal check before substituting numbers. This habit prevents the common exam error of swapping signs and then obtaining physically inconsistent results for Z or critical constants.
How does this topic usually appear in NEET objective questions?
Common patterns include identifying the correct sign of Z deviation, selecting low-pressure high-temperature condition for near-ideal behavior, choosing correct van der Waals form, and computing one critical constant from another parameter. Questions are typically short but concept-dense, so speed depends on instant interpretation of correction terms rather than lengthy derivation.
For NRI / OCI / U.S.-Based Families
NEET NRI Counseling & Admission eBook Download
A practical guide covering sponsor rules, document checklist, verification traps, NRI quota reality, and step-by-step counselling flow. Designed to prevent last-minute rejections and wrong choice filling.
Sponsor + Proof ClarityDocuments ChecklistState-wise Traps
Tip: Keep this eBook open during verification + choice filling week for quick cross-checking.
NEET Prep (India + NRI-USA)
Schedule Trial Session For NEET Prep
Get a short diagnostic + study roadmap: syllabus gaps (NCERT vs U.S. curriculum), weak chapters, and the exact weekly plan needed to improve accuracy under time.
Gap MappingWeekly PlanAccuracy Fix
Best for: Students in Grade 10–12 (U.S. / India) who want a clear NEET timeline and daily practice structure.