Home
Coaching Programs
NEET Physics Coaching Delhi JEE Physics Coaching Delhi CBSE Class 11 Physics CBSE Class 12 Physics Online Physics Classes Physics Doubt Solving
Academic Calendar
Online Live Class – XI Online Live Class – XII Online Live Class – Dropper Batch
Locations Near You
Physics Coaching East Delhi Physics Coaching South Delhi Physics Coaching Noida Physics Coaching Gurgaon Physics Coaching Ghaziabad Physics Coaching Indirapuram Physics Coaching Greater Noida
Study Material
Class 11
Units & Measurements Motion in 1D Motion in 2D Laws of Motion Work, Energy & Power Rotational Motion Gravitation Thermal Properties Thermodynamics Oscillations & SHM Waves
Class 12
Electric Charges Capacitors Current Electricity Moving Charges EMI Alternating Current EM Waves Ray Optics Wave Optics Dual Nature Nuclei Semiconductors
Expert Strategy Guides
Improve Physics Numericals Common JEE Mistakes Score 90 in CBSE Physics NEET Prep Strategy Exam Time Management Problem Solving Framework Derivations Step-by-Step Why Students Struggle How Toppers Study Best Way to Revise
Resources & Reference
📐 Physics Formulas & Concepts ⚠️ Common Mistakes & Corrections
Blog & Articles
Physics Doubts Solving Guide Best Way to Study Physics for NEET How to Score 90 in Class 12 Physics Physics Formula Sheet Class 12
Book a Diagnostic Session
📞 Call Now 🎯 Get Your Physics Assessment

First Law & Work

Most-tested category in all exams
First Law
Change in Internal Energy
ΔU = Q − W
ΔU = change in internal energy, Q = heat absorbed, W = work done by system
CBSENEETJEE MJEE A
Work Done
Isobaric Work
W = PΔV = P(V₂ − V₁)
Work done by gas in isobaric (constant pressure) process
CBSENEETJEE M
Work Done
Isothermal Work
W = nRT ln(V₂/V₁)
Work done in isothermal process. Since PV = nRT and T is constant.
NEETJEE MJEE A
Work Done
Isochoric Work
W = 0 (V = constant)
No work is done at constant volume. All heat goes to changing internal energy.
CBSENEETJEE M
Work Done
Adiabatic Work
W = −ΔU = nCv(T₁ − T₂)
No heat exchange. Work comes entirely from internal energy change.
NEETJEE MJEE A
Work Done
Cyclic Process Work
W_net = Area of PV loop
Net work = area enclosed by cycle. Positive (clockwise). ΔU = 0 for full cycle.
NEETJEE MJEE A
💨

Ideal Gas & Kinetic Theory

Foundation for all process calculations
Ideal Gas Law
Ideal Gas Equation
PV = nRT
P = pressure (Pa), V = volume (m³), n = moles, R = 8.314 J/mol·K, T = temperature (K)
CBSENEETJEE MJEE A
Ideal Gas Law
Gas Law (Boltzmann)
PV = Nk_BT
N = number of molecules, k_B = 1.38 × 10⁻²³ J/K (Boltzmann constant)
NEETJEE MJEE A
Internal Energy
Internal Energy (Ideal Gas)
U = f/2 · nRT
f = degrees of freedom. Monoatomic: f=3, Diatomic: f=5, Polyatomic: f=6
CBSENEETJEE MJEE A
Heat Capacity
Mayer's Relation
Cp − Cv = R
Difference between molar heat capacities at constant pressure and volume equals R (gas constant)
CBSENEETJEE MJEE A
Heat Capacity
Ratio of Heat Capacities
γ = Cp/Cv = (f+2)/f
γ = 5/3 (monoatomic), 7/5 (diatomic), 4/3 (polyatomic)
CBSENEETJEE MJEE A
Heat Transfer
Heat at Constant Volume
Q = nCvΔT (isochoric)
At constant volume, W = 0, so all heat goes to internal energy change
CBSENEETJEE M
🔄

Thermodynamic Processes

Process-specific equations tested heavily in JEE
Adiabatic
Adiabatic Process Laws
PVᵞ = const
Also: TV^(γ-1) = const and T^γP^(1-γ) = const. No heat exchange (Q = 0)
NEETJEE MJEE A
Adiabatic
Adiabatic T-V Relation
TV^(γ−1) = constant
Temperature-volume relation in adiabatic process. Derived from PVᵞ = const and PV = nRT
NEETJEE MJEE A
Adiabatic
Adiabatic T-P Relation
T^γ P^(1−γ) = constant
Temperature-pressure relation for adiabatic process
JEE MJEE A
Isothermal
Boyle's Law
PV = constant (T = const)
P₁V₁ = P₂V₂. ΔU = 0 for ideal gas. Work done = heat absorbed.
CBSENEETJEE M
Slope
PV Diagram Slopes
|Slope_adi| = γ × |Slope_iso|
Adiabatic curve is steeper than isothermal. Critical graph-based question fact.
NEETJEE MJEE A
Polytropic
Polytropic Process
PVⁿ = constant
General process: n=0 isobaric, n=1 isothermal, n=γ adiabatic, n=∞ isochoric
JEE A
🏎️

Carnot Engine & Efficiency

Always in NEET and JEE Main
Efficiency
Carnot Efficiency
η = 1 − T_C/T_H
T_C and T_H must be in Kelvin. Maximum possible efficiency for given temperatures.
CBSENEETJEE MJEE A
Efficiency
Heat Engine Efficiency
η = W/Q₁ = 1 − Q₂/Q₁
W = Q₁ − Q₂ (net work done), Q₁ = heat absorbed from hot source
CBSENEETJEE M
Refrigerator
COP of Refrigerator
COP = Q₂/W = T_C/(T_H−T_C)
Coefficient of Performance. Higher COP = more efficient refrigerator.
CBSENEETJEE M
Entropy
Entropy Change
ΔS = Q_rev / T
For reversible process. Entropy is a state function. ΔS = 0 for reversible adiabatic.
JEE MJEE A
Efficiency
Efficiency Improvement
Δη ≈ ΔT/T_H
Increasing T_H by ΔT increases efficiency more than decreasing T_C by same ΔT if T_H > T_C
JEE A
Heat Engine
Work-Heat Relation
Q₁/T_H = Q₂/T_C (Carnot)
For Carnot cycle (reversible engine). Heat ratios equal temperature ratios.
NEETJEE MJEE A
📏

Dimensional Analysis & SI Units

CBSE derivation marks + JEE verification
QuantitySI UnitDimensionsNotes
Internal Energy (U)J (Joule)[ML²T⁻²]Same as work and heat
Heat (Q)J (Joule)[ML²T⁻²]Not a state function
Work (W)J (Joule)[ML²T⁻²]W = ∫PdV
Entropy (S)J/K[ML²T⁻²Θ⁻¹]State function
Temperature (T)K (Kelvin)[Θ]Always in Kelvin for formulas
Pressure (P)Pa = N/m²[ML⁻¹T⁻²]1 atm = 101325 Pa
Specific Heat (c)J/kg·K[L²T⁻²Θ⁻¹]Per unit mass
Molar Heat Cap. (C)J/mol·K[ML²T⁻²Θ⁻¹mol⁻¹]Per mole
Gas Constant (R)J/mol·K[ML²T⁻²Θ⁻¹mol⁻¹]R = 8.314 J/mol·K
Boltzmann Const. (k_B)J/K[ML²T⁻²Θ⁻¹]k_B = 1.38×10⁻²³ J/K
Efficiency (η)Dimensionless[1]0 ≤ η < 1
COPDimensionless[1]Can be > 1
🔬

Exam Insight — Dimensional Analysis

In JEE, dimensional analysis is often used to verify formula correctness in multi-step problems. Key check: R = Cp − Cv. Both have units J/mol·K = [ML²T⁻²Θ⁻¹mol⁻¹]. The ratio γ = Cp/Cv is dimensionless — it cancels all units. If a formula gives you γ with units, something is wrong.

🔢

Key Constants & Values

Memorize these — they appear repeatedly
Gas Constant
R = 8.314 J/mol·K
≈ 2 cal/mol·K
Boltzmann Constant
k_B = 1.38×10⁻²³ J/K
R = N_A × k_B
Avogadro Number
N_A = 6.022×10²³
molecules per mole
1 Calorie
1 cal = 4.186 J
Mechanical equivalent of heat
← Core Concepts 🔄 Thermodynamic Processes →