Home Subjects JEE Physics Thermodynamics

JEE Physics
Thermodynamics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

27 Q 9 Topics Take Mock Test
Advertisement
Difficulty: All Easy Medium Hard 11–20 of 27
Topics in JEE Physics
Q.11 Easy Thermodynamics
An ideal gas undergoes adiabatic compression from state (P₁, V₁, T₁) to (P₂, V₂, T₂). Which relation is correct?
A T₁V₁^(γ-1) = T₂V₂^(γ-1)
B T₁V₁^γ = T₂V₂^γ
C P₁V₁ = P₂V₂
D T₁ = T₂
Correct Answer:  A. T₁V₁^(γ-1) = T₂V₂^(γ-1)
EXPLANATION

For adiabatic process: TV^(γ-1) = constant, therefore T₁V₁^(γ-1) = T₂V₂^(γ-1)

Test
Q.12 Easy Thermodynamics
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?
A P₀V₁ ln(2)
B 2P₀V₁ ln(2)
C P₀V₁ ln(0.5)
D P₀V₁
Correct Answer:  A. P₀V₁ ln(2)
EXPLANATION

For isothermal process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁) = P₀V₁ ln(2)

Test
Q.13 Easy Thermodynamics
A heat pump delivers 5000 J of heat to a house while consuming 1500 J of work. Its coefficient of performance is:
A 3.33
B 2.33
C 4.33
D 1.33
Correct Answer:  A. 3.33
EXPLANATION

For heat pump: COP = Q_h/W = 5000/1500 = 3.33. This indicates the pump delivers 3.33 J of heat for every 1 J of work input.

Test
Q.14 Easy Thermodynamics
What is the efficiency of a Carnot engine operating between 400 K and 300 K?
A 25%
B 33.3%
C 50%
D 75%
Correct Answer:  A. 25%
EXPLANATION

Carnot efficiency: η = 1 − (T_cold/T_hot) = 1 − (300/400) = 1 − 0.75 = 0.25 = 25%

Test
Q.15 Easy Thermodynamics
An ideal gas undergoes an adiabatic process. If the gas is compressed, which of the following is true?
A Temperature decreases, internal energy decreases
B Temperature increases, internal energy increases
C Temperature remains constant, internal energy increases
D Temperature decreases, internal energy remains constant
Correct Answer:  B. Temperature increases, internal energy increases
EXPLANATION

In adiabatic compression (Q = 0), work is done on the gas (W < 0). By first law: ΔU = Q − W = 0 − W > 0. Since ΔU ∝ ΔT for ideal gas, temperature increases.

Test
Q.16 Easy Thermodynamics
A system absorbs 500 J of heat and performs 200 J of work on the surroundings. What is the change in internal energy of the system?
A 300 J
B 700 J
C −300 J
D −700 J
Correct Answer:  A. 300 J
EXPLANATION

By first law of thermodynamics: ΔU = Q − W = 500 − 200 = 300 J

Test
Q.17 Easy Thermodynamics
A carnot engine operates between temperatures 500 K and 300 K. What is its maximum efficiency?
A 40%
B 60%
C 83.3%
D 16.7%
Correct Answer:  A. 40%
EXPLANATION

Carnot efficiency: η = 1 - (Tc/Th) = 1 - (300/500) = 1 - 0.6 = 0.4 = 40%

Test
Q.18 Easy Thermodynamics
In an adiabatic process, if a gas is compressed, which statement is correct?
A Temperature decreases
B Temperature increases
C Temperature remains constant
D Heat is absorbed from surroundings
Correct Answer:  B. Temperature increases
EXPLANATION

In adiabatic compression, no heat exchange occurs (Q=0). Work is done on the gas, so ΔU = W (positive). Since ΔU increases, temperature must increase.

Test
Q.19 Easy Thermodynamics
For one mole of an ideal monatomic gas, the ratio Cp/Cv is:
A 1.40
B 1.67
C 1.33
D 1.50
Correct Answer:  B. 1.67
EXPLANATION

For monatomic gas: Cv = (3/2)R and Cp = (5/2)R. Therefore Cp/Cv = (5/2)/(3/2) = 5/3 ≈ 1.67

Test
Q.20 Easy Thermodynamics
A thermodynamic system undergoes a process where internal energy increases by 150 J while the system does 100 J of work on surroundings. What is the heat absorbed by the system?
A 250 J
B 50 J
C -50 J
D 150 J
Correct Answer:  A. 250 J
EXPLANATION

By first law: ΔU = Q - W. Here ΔU = 150 J, W = 100 J (work done by system). So Q = ΔU + W = 150 + 100 = 250 J

Test
IGET
IGET AI
Online · Exam prep assistant
Hi! 👋 I'm your iget AI assistant.

Ask me anything about exam prep, MCQ solutions, study tips, or strategies! 🎯
UPSC strategy SSC CGL syllabus Improve aptitude NEET Biology tips