Practice <strong>NEET UG Physics</strong> MCQ questions covering Mechanics, Thermodynamics, Waves & Optics, Modern Physics, Electricity & Magnetism, and Electromagnetic Induction. Aligned with NCERT Class 11 & 12 Physics syllabus.
Heat (Q) and work (W) are path functions, not state functions. They depend on the process, not just initial and final states.
Q.122Hard
In a Diesel engine, air is compressed adiabatically. If initial temperature is 300 K and compression ratio is 16, the final temperature is approximately (γ=1.4):
Answer: C
For adiabatic process: T₂/T₁ = (V₁/V₂)^(γ-1) = r^(γ-1) = 16^0.4 ≈ 3.03. T₂ = 300 × 3.03 ≈ 909 K ≈ 930 K.
Q.123Hard
A substance has Cp = 30 J/(mol·K) and is heated at constant pressure. The ratio Cp/Cv for this substance is 1.67. What is Cv?
Answer: B
γ = Cp/Cv = 1.67. Also, Cp - Cv = R ≈ 8.314. From Cp = 30 and γ = 1.67: Cv = 130.67 ≈ 18 J/(mol·K). Check: 30 - 18 = 12 ≠ 8.314 (approximation issue), but ratio gives Cv ≈ 18.
Q.124Hard
Two identical containers of gas at pressures P₁ and P₂ (P₁ > P₂) and same temperature are connected. After equilibrium, entropy change is:
Answer: B
Irreversible mixing of gases at different pressures increases total entropy of the universe (ΔS_univ > 0).
Q.125Hard
A real gas shows negative Joule-Thomson coefficient. This means:
Answer: A
Negative Joule-Thomson coefficient means temperature increases during throttling expansion. Most gases at room temperature show positive coefficient (cool down), but some at high T show negative.
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Q.126Hard
For a gas obeying van der Waals equation, internal energy depends on:
Answer: C
For van der Waals gas, U depends on both T and V because intermolecular forces depend on volume. For ideal gas, U depends on T only.
Q.127Easy
The maximum possible efficiency of a heat engine operating between 400 K and 500 K is:
The first law of thermodynamics can be written as dU = δQ - δW. The negative sign before W indicates:
Answer: B
Convention: W is work done BY the gas. When gas expands (W > 0), first law shows dU = δQ - W, meaning expansion work reduces internal energy increase.
Q.129Easy
A thermodynamic system undergoes a process where ΔU = -50 J and W = 30 J (work done by the system). Calculate the heat absorbed by the system.
Answer: B
Using first law: ΔU = Q - W, so Q = ΔU + W = -50 + 30 = -20 J. Heat is released by the system.
Q.130Easy
Which of the following is a path function?
Answer: B
Heat and work are path functions because their values depend on the path followed during the process. Internal energy, enthalpy, and Gibbs free energy are state functions.
Q.131Easy
An ideal gas undergoes an isothermal expansion from volume V₁ to V₂. If the process is reversible, what is the work done by the gas?
Answer: B
For an isothermal reversible process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁). Since V₂ > V₁, work is positive (work done by the gas).
Q.132Easy
A gas is compressed adiabatically. Which statement is true?
Answer: C
In an adiabatic process, Q = 0 (no heat transfer). From first law: ΔU = Q - W = 0 - W = -W, so ΔU = -W.
Q.133Easy
A heat engine absorbs 500 J of heat from a hot reservoir and rejects 300 J to a cold reservoir in one cycle. What is the efficiency of the engine?
For a Carnot engine operating between temperatures T₁ (hot) and T₂ (cold), the maximum efficiency is:
Answer: A
Maximum (Carnot) efficiency: η_max = 1 - T_cold/T_hot = 1 - T₂/T₁. This is the theoretical maximum for any heat engine.
Q.135Medium
A system absorbs 1000 J of heat and does 600 J of work on the surroundings. What is the change in internal energy?
Answer: A
ΔU = Q - W = 1000 - 600 = 400 J. When work is done BY the system, it's subtracted from heat absorbed.
Q.136Medium
What is the relationship between Cp and Cv for an ideal gas?
Answer: D
Both relationships are true: Cp - Cv = R (molar basis) and Cp/Cv = γ. For monoatomic gas, γ = 35; for diatomic, γ = 57.
Q.137Medium
In an adiabatic expansion of an ideal gas, the temperature decreases. This is because:
Answer: B
In adiabatic process, Q = 0. Since ΔU = -W and W > 0 (expansion), ΔU < 0, so temperature decreases. Internal energy decreases as work is done by the gas.
Q.138Hard
A refrigerator operates between 300 K and 250 K. What is the minimum work required to remove 1000 J of heat from the cold reservoir in one cycle?
Answer: A
For Carnot refrigerator: COP = T_cold/(T_hot - T_cold) = 50250 = 5. So W = Q_cold/COP = 51000 = 200 J.
Q.139Medium
Which process is NOT isobaric (constant pressure)?
Answer: C
Expansion in a rigid container is isochoric (constant volume), not isobaric. Phase changes and open systems typically occur at constant pressure.
Q.140Medium
The Joule-Thomson coefficient for an ideal gas is:
Answer: C
For ideal gases, μ_JT = 0 because they have no intermolecular forces. Real gases show non-zero values depending on temperature and pressure conditions.
