NEET Physics - Kinetic Theory of Gases
Pressure of an Ideal Gas – Complete Notes, Revision, Important Questions & Downloads
Pressure of an Ideal Gas is developed from the single TOC subtopic Derivation of Pressure, where molecular collisions with a wall are converted into macroscopic pressure using momentum transfer per unit time and per unit area. The derivation sequence in the source page runs through collision frequency, change in momentum, force on one wall, and symmetry of velocity components to obtain P = (1/3)rho(v_rms)^2. NEET tests this topic through one-step numericals, assertion-reason items on isotropy (vx^2 = vy^2 = vz^2), and conceptual traps that ask why pressure is same in all directions for random motion. The section also links pressure with translational kinetic energy density through P = (2/3)E, which is a frequent bridge to kinetic interpretation questions.
2 SubtopicsDerivation + FormulaPressure-KE Link
Expected QuestionsQ
1
Typically one question appears from derivation logic, formula application, or pressure-energy relation in NEET-style MCQs.
Time Required⏱
60-90 min
One focused session to lock the derivation steps and one practice block for formula conversion and unit-safe numericals.
Difficulty⚡
Medium
The formulas are short, but marks are lost when students confuse mean square speed with average speed or ignore component-wise symmetry.
NRI USA Curriculum GapUS
Medium
US high-school courses often state ideal-gas relations qualitatively, while NEET expects quick derivation-level reasoning from molecular collision to pressure expression.
2Subtopics
8+Practice Questions
4Free Downloads
60-90 minPrep Time
NEET Weightage - Pressure of an Ideal Gas
Kinetic Theory of Gases (Chapter 13)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 0 | 0 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 1 | 4 | |
| 6-Year Trend (2019-2024) | 3-4 | 12-16 |
Common framing starts from momentum change on wall collision and asks for force or pressure after dividing by collision time and wall area.
Another frequent pattern tests symmetry of random motion and expects vx^2 = vy^2 = vz^2 = v_rms^2/3 before substituting into pressure expression.
Pressure-energy conversion P = (2/3)E is often used in mixed conceptual-numerical questions where kinetic energy density is given directly.
Another frequent pattern tests symmetry of random motion and expects vx^2 = vy^2 = vz^2 = v_rms^2/3 before substituting into pressure expression.
Pressure-energy conversion P = (2/3)E is often used in mixed conceptual-numerical questions where kinetic energy density is given directly.
0.7
Avg Questions / Year
12-16
Total Marks (6 yrs)
Direct
Pattern
Medium
Difficulty
Exam Strategy for Pressure of an Ideal Gas
1
Lock the collision-to-pressure chain in order Memorize the exact sequence Delta p -> Delta t -> F -> P. In questions, verify whether momentum change is for molecule or wall before using sign and magnitude.
2
Check component direction before substitution Use only x-component for collisions with one wall and convert to full speed using v_rms^2 = 3vx^2 only after applying isotropy. This avoids direct misuse of v instead of vx.
3
Differentiate mean square speed from average speed Pressure derivation uses mean square speed and v_rms, not arithmetic average speed. If a question gives v_av, convert or reject the option path that substitutes it directly.
4
Use P = (2/3)E for energy-density data When kinetic energy per unit volume is given, do not go back to molecule counting. Apply P = (2/3)E directly and keep SI units consistent for quick scoring.
Download Study Notes - Pressure of an Ideal Gas
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete notes on Derivation of Pressure including collision frequency, momentum transfer, isotropy relation, and final pressure formulas.
Download PDFFormula Sheet
Quick sheet for P = (1/3)rho(v_rms)^2, P = (2/3)E, and component relation vx^2 = v_rms^2/3 with unit reminders.
Download PDFMCQ Practice
Practice set focused on momentum-transfer numericals, isotropy-based derivation steps, and pressure-kinetic-energy conversion.
Download PDFPYQ
NEET-style practice aligned to pressure derivation and ideal-gas kinetic interpretation patterns from recent years.
Download PDFSubtopics - Pressure of an Ideal Gas
2-Column Table| Column A | Column B |
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Revision Cards - Pressure of an Ideal Gas
Concept → Trap → Example1) Derivation of Pressure
Core DerivationFor random molecular motion, pressure comes from wall collision momentum transfer and gives P = (1/3)(mN/V)v_rms^2 = (1/3)rho(v_rms)^2, with vx^2 = vy^2 = vz^2 = v_rms^2/3.
- Start with one molecule: Delta p = 2mvx and Delta t = 2L/vx, then force from one molecule is mvx^2/L.
- For all molecules on one wall, sum mean-square x-components and divide by area to reach pressure expression.
- Trap: replacing mean square speed with average speed or forgetting that isotropy relation is for squared components, not for velocities themselves.
Example (NEET-style)If rho = 1.2 kg m^-3 and v_rms = 500 m s^-1, then P = (1/3)rho(v_rms)^2 = (1/3)(1.2)(250000) = 100000 Pa. The same state gives E = (1/2)rho(v_rms)^2 = 150000 J m^-3 and verifies P = (2/3)E.
US Curriculum Gaps
Where NEET handling of this topic is usually stricter than typical US high-school treatmentDerivation depth vs formula recall
Many US high-school Physics courses use pressure relations without requiring derivation from collision frequency and momentum transfer, while NEET asks derivation-logic checkpoints.
- NEET may ask which step converts single-molecule force into macroscopic pressure.
- Students are expected to interpret Delta p and Delta t correctly for wall collisions.
- Option elimination often depends on identifying wrong use of component velocity.
Isotropy and mean-square speed handling
US introductory tracks often emphasize ideal gas law numericals, but NEET expects explicit use of vx^2 = v_rms^2/3 and the pressure-energy relation in fast MCQs.
- Questions mix molecular and macroscopic forms in one item.
- Students must keep clear distinction between v_rms and average speed.
- Pressure from kinetic energy density, P = (2/3)E, is used as a direct scoring shortcut.
NEET-style Practice Questions - Pressure of an Ideal Gas
4 Questions1A molecule of mass m collides elastically with a wall with x-component of velocity vx and rebounds with -vx. If time between two successive hits on the same wall is 2L/vx, the force by this molecule on that wall is:Derivation of Pressure
mvx/L
mvx^2/L
2mvx^2/L
mL/vx^2
Correct option is B. In one collision, momentum transferred to the wall is +2mvx in magnitude. The collision interval for the same wall is Delta t = 2L/vx. Therefore force from one molecule is F = Delta p/Delta t = (2mvx)/(2L/vx) = mvx^2/L. Option A misses one factor of vx. Option C keeps an extra factor 2 after simplification. Option D is dimensionally incorrect for force. This is a standard NEET derivation checkpoint where students often confuse momentum change with impulse sign or skip the return-trip time term.
2For an ideal gas with density rho and rms speed v_rms, which relation gives pressure correctly?Derivation of Pressure
P = rho(v_rms)^2
P = (1/2)rho(v_rms)^2
P = (1/3)rho(v_rms)^2
P = (3/2)rho(v_rms)^2
Correct option is C because isotropic random motion gives equal mean-square components in three perpendicular directions, so only one-third of total mean-square speed contributes to pressure on any selected wall. Hence P = (1/3)rho(v_rms)^2. Option A ignores directional averaging. Option B incorrectly mixes with kinetic energy density expression E = (1/2)rho(v_rms)^2. Option D is the inverse factor and appears when students mistakenly move 3 to numerator. NEET often builds such options to test whether the 1/3 factor is conceptually understood or merely memorized.
3If translational kinetic energy density of an ideal gas is E = 1.5 x 10^5 J m^-3, then pressure is:Derivation of Pressure
1.0 x 10^5 Pa
1.5 x 10^5 Pa
2.25 x 10^5 Pa
0.5 x 10^5 Pa
Correct option is A by direct use of P = (2/3)E. Substituting E = 1.5 x 10^5 J m^-3 gives P = (2/3)(1.5 x 10^5) = 1.0 x 10^5 Pa. Option B wrongly assumes P = E. Option C multiplies by 3/2 instead of 2/3. Option D halves E without reason. In NEET, this conversion is a quick score if you instantly recognize that energy is per unit volume and therefore pressure has the same SI unit scale (N m^-2).
4In the pressure derivation, why is vx^2 averaged instead of vx when summing molecular contributions?Derivation of Pressure
Because vx is always positive for all molecules.
Because force expression contains vx^2 after Delta p/Delta t and molecules move in both +/-x directions.
Because vx^2 equals v_rms for every molecule.
Because averaging vx^2 removes dependence on molecular mass.
Correct option is B. The single-molecule force on a wall comes out proportional to vx^2, not vx, after dividing momentum transfer by return-collision time. Also, molecular motion is random in both positive and negative x directions, so linear velocity averages can cancel, while squared components remain positive and capture net pressure effect. Option A is false because vx can be positive or negative depending on direction. Option C is false since v_rms is a root of total mean-square speed, not identical to each component square. Option D is false because mass remains explicitly present in derivation before converting to density form.
Practice Questions - Pressure of an Ideal Gas
Click "Reveal Answer" after attempting1For an ideal gas, rho = 0.8 kg m^-3 and v_rms = 600 m s^-1. Find pressure.
4.8 x 10^4 Pa
9.6 x 10^4 Pa
1.2 x 10^5 Pa
2.88 x 10^5 Pa
👁 Reveal Answer
Correct option: B (9.6 x 10^4 Pa). Use P = (1/3)rho(v_rms)^2 = (1/3)(0.8)(600^2) = (1/3)(0.8)(360000) = 96000 Pa. Option A misses the factor 1/3 handling during multiplication. Option C comes from using 1/2 instead of 1/3. Option D ignores division by 3 completely. This checks correct use of pressure formula and SI-unit consistency.
2An ideal gas has pressure 1.2 x 10^5 Pa. What is its translational kinetic energy density E?
0.8 x 10^5 J m^-3
1.2 x 10^5 J m^-3
1.8 x 10^5 J m^-3
2.4 x 10^5 J m^-3
👁 Reveal Answer
Correct option: C (1.8 x 10^5 J m^-3). From P = (2/3)E, we get E = (3/2)P = (3/2)(1.2 x 10^5) = 1.8 x 10^5 J m^-3. Option B assumes E = P directly. Option D multiplies by 2. Option A inverts the relation. This type appears in NEET when pressure is given and energy density is asked in one step.
3If mass and temperature of a gas are constant but volume is halved, pressure becomes:
P/2
P
2P
4P
👁 Reveal Answer
Correct option: C (2P). From the source relation P proportional to (mN)T/V, with fixed mass and temperature pressure varies inversely with volume. So halving volume doubles collision frequency with walls and pressure doubles. Option A is opposite trend. Option B ignores volume effect. Option D overestimates by applying square relation that does not belong here.
4A student uses average speed v_av in P = (1/3)rho v^2 to calculate pressure. What is the conceptual error?
No error, v_av and v_rms are always equal
Pressure relation requires mean square speed, so v_rms must be used
Pressure depends only on number of molecules and not on speed
v_av should be cubed instead of squared
👁 Reveal Answer
Correct option: B. The derivation gives pressure in terms of mean square speed and therefore v_rms, not arithmetic average speed. Using v_av directly changes numerical value and breaks the isotropy-based derivation step. Option A is false because v_rms > v_av for the same gas. Option C is false because molecular speed enters through momentum transfer. Option D invents a power with no physical basis.
Pressure of an Ideal Gas Revision Checklist
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Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ - Pressure of an Ideal Gas
Notes · Downloads · Revision · Important QuestionsWhy does the pressure formula have the factor 1/3?
Pressure on a given wall comes from the velocity component perpendicular to that wall. In random isotropic motion, the mean square speed is equally shared among x, y, and z directions, so mean(vx^2) = v_rms^2/3. Substituting this directional average into the derivation produces the 1/3 factor. If this symmetry step is skipped, students often get P = rho(v_rms)^2, which is incorrect by a factor of 3.
Why do we use v_rms and not average speed in pressure derivation?
The derivation involves momentum transfer rate, and after dividing Delta p by Delta t the force depends on vx^2. Summing molecular contributions therefore requires mean square speed, not arithmetic mean speed. v_rms is defined from mean square speed, so it naturally appears in pressure relation. Using average speed gives a numerically different value and has no direct derivation path from collision mechanics.
Is pressure inside the gas only at the wall or everywhere in the gas?
Pressure is a state property of the gas at every point in equilibrium, not only at the wall. The wall-collision derivation is used because wall force is measurable, but the same microscopic random motion supports pressure transmission throughout the gas volume. In NEET conceptual items, statements that pressure exists only at boundaries should be marked incorrect.
How is P = (2/3)E obtained from kinetic theory?
From kinetic theory, P = (1/3)rho(v_rms)^2. Translational kinetic energy per unit volume is E = (1/2)rho(v_rms)^2. Eliminating rho(v_rms)^2 between these two relations gives P = (2/3)E. This relation is highly useful in objective questions where one quantity is given directly and the other is asked without additional gas-law data.
What does collision frequency mean in this context?
Collision frequency here means the number of collisions per second made by molecules with a wall. For one molecule bouncing between opposite walls separated by L, the interval between two hits on the same wall is 2L/vx, so frequency is vx/2L. This term controls how quickly momentum is transferred to the wall and therefore directly affects force and pressure.
How does pressure change when temperature increases at fixed mass and volume?
At fixed mass and volume, pressure rises with temperature because molecular mean square speed increases with temperature. Faster molecules collide with walls more frequently and with larger momentum change per collision, so net pressure increases. In kinetic form, P proportional to v_rms^2 and v_rms^2 proportional to T, so P proportional to T under these constraints.
What is the most common NEET trap in this topic?
A common trap is mixing scalar speed relations with directional component relations, especially replacing vx^2 = v_rms^2/3 by vx = v_rms/3 or using v_av in place of v_rms. Another trap is missing the factor 2 in momentum change during elastic reversal and then carrying that error through force and pressure expressions. Careful tracking of each algebraic step avoids these mistakes.
Can this derivation be applied to real gases directly?
The derivation assumes ideal-gas conditions: negligible molecular volume and no intermolecular force except during brief elastic collisions. Real gases approximate this behavior best at low pressure and high temperature; otherwise corrections are needed. For NEET-level problems in this topic, use ideal-gas assumptions unless the question explicitly introduces non-ideal behavior or real-gas corrections.
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