Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
Which process results in zero change of entropy for an ideal gas?
Answer: B
For a reversible adiabatic process, dQ = 0, so dS = dQ/T = 0, hence ΔS = 0. This is also called isentropic process.
Q.142Medium
In a throttling process (Joule-Thomson expansion), which of the following remains constant?
Answer: D
Throttling is an irreversible, adiabatic process where enthalpy remains constant (H_initial = H_final). Temperature and pressure both change, and internal energy remains nearly constant only for ideal gases.
Q.143Hard
Two moles of an ideal diatomic gas undergo a process where temperature changes from 400 K to 800 K. If the process is such that PV^(1.4) = constant, then the work done by the gas is approximately (R = 8.314 J/mol·K):
Answer: C
For polytropic process: W = nR(T_i − T_f)/(γ − n) = 2 × 8.314 × (400 − 800)/(1.4 − 1.4). Since n = γ, this is adiabatic: W = nCvΔT = 2 × (25) × 8.314 × (400 − 800) = 5 × 8.314 × (−400) ≈ −16,628 J. But if heating occurs, work done is positive. Actual calculation needs clarification on process direction.
Q.144Hard
A system undergoes a process where both pressure and volume increase. Which thermodynamic quantity must increase?
Answer: A
If both P and V increase for an ideal gas (PV = nRT), then T must increase, hence internal energy U ∝ T increases. Work done by system depends on process path; entropy and heat depend on specific process.
Q.145Hard
Four moles of an ideal gas are heated from 250 K to 350 K at constant volume. Simultaneously, it is allowed to expand at constant temperature. Which process involves greater entropy change?
Answer: B
Isochoric: ΔS = nCv ln(T_f/T_i) = 4 × Cv × ln(250350). Isothermal: ΔS = nR ln(V_f/V_i). For large expansions, isothermal entropy change is typically larger.
Q.146Easy
A heat pump delivers 5000 J of heat to a house while consuming 1500 J of work. Its coefficient of performance is:
Answer: A
For heat pump: COP = Q_h/W = 15005000 = 3.33. This indicates the pump delivers 3.33 J of heat for every 1 J of work input.
Q.147Easy
A diatomic ideal gas expands isothermally from volume V₁ to 2V₁. If the initial pressure is P₀, what is the work done by the gas?
Answer: A
For isothermal process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) = P₀V₁ ln(2)
Q.148Easy
An ideal gas undergoes adiabatic compression from state (P₁, V₁, T₁) to (P₂, V₂, T₂). Which relation is correct?
Answer: A
For adiabatic process: TV^(γ-1) = constant, therefore T₁V₁^(γ-1) = T₂V₂^(γ-1)
Q.149Easy
The internal energy of an ideal gas depends on:
Answer: C
Internal energy U of ideal gas is U = nCᵥT, which depends only on temperature, not on pressure or volume individually
Q.150Easy
A Carnot engine operates between 400 K and 300 K. Its maximum efficiency is:
Two kilograms of water at 100°C is converted to steam at 100°C at 1 atm pressure. The change in entropy is (Latent heat of vaporization = 2.26 × 10⁶ J/kg):
A real gas deviates from ideal behavior. Which condition favors ideal behavior?
Answer: B
At low pressure, molecules are far apart (negligible intermolecular forces), and at high temperature, kinetic energy dominates, making gases behave ideally
Q.156Medium
An open system differs from a closed system in that:
Answer: A
An open system allows both mass and energy exchange with surroundings (e.g., a boiling kettle), while a closed system allows only energy exchange
Q.157Hard
Three moles of ideal gas undergo polytropic process with n = 1.5. If temperature increases from 300 K to 450 K, the work done by gas is:
Answer: B
W = nR(T₂-T₁)/(1-n) = 3 × 8.314 × 150/(1-1.5) = 3,741/(-0.5) = -3,741 J (compression), |W| ≈ 3,372 J accounting for polytropic work formula
Q.158Hard
A heat engine operates between 600 K and 300 K reservoirs. It absorbs 5000 J from hot reservoir. For a Carnot engine operating between same temperatures, maximum work output would be:
