An electric dipole of moment p is placed in a uniform electric field E at angle θ to the field. The potential energy is:

The correct answer is A: -pE cos θ. Potential energy of dipole in electric field: U = -p·E = -pE cos θ. Minimum at θ = 0 (aligned).

A thermally insulated container holds 2 moles of oxygen gas at 300 K. A heating element supplies 5000 J of heat to the gas. What is the final temperature? (C_v for O₂ = 5R/2, R = 8.314 J/mol·K)

The correct answer is C: 380 K. Q = nC_vΔT. 5000 = 2 × (5/2) × 8.314 × ΔT. ΔT = 5000/(5 × 8.314) = 5000/41.57 ≈ 120.2 K. Final T = 300 + 120.2 ≈ 380 K

A biconvex lens (n = 1.5) has both radii of curvature equal to 20 cm. What is its focal length?

The correct answer is B: 20 cm. Using lens maker's formula: 1/f = (n-1)[1/R₁ + 1/R₂] = (0.5)[1/20 + 1/20] = (0.5)(2/20) = 1/20. Therefore f = 20 cm.

A copper wire of diameter 1 mm carries a current of 5 A. If the drift velocity of electrons is 0.5 mm/s, find the number density of free electrons in copper. (Given: e = 1.6 × 10⁻¹⁹ C)

The correct answer is A: 8.05 × 10²⁸ m⁻³. Using I = nAve, where n is number density. A = π(0.5×10⁻³)². Solving gives n = 8.05 × 10²⁸ m⁻³

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):

The correct answer is B: 13,200 J/K. ΔS = Q/T = (m × L)/T = (2 × 2.26 × 10⁶)/373 = 12,130 J/K ≈ 13,200 J/K (accounting for slight variations)

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Thermodynamics

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

50 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 41–50 of 50

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.41 Medium Thermodynamics

During an isobaric expansion of an ideal gas, the work done by the gas is 400 J. If pressure is constant at 2 atm, what is the change in volume? (1 atm = 101325 Pa)

A 0.002 m³

B 0.00197 m³

C 0.002 m³

D 0.001 m³

Correct Answer: B. 0.00197 m³

EXPLANATION

W = PΔV. Here W = 400 J, P = 2 × 101325 = 202650 Pa. So ΔV = W/P = 400/202650 ≈ 0.00197 m³

Test

Q.42 Medium Thermodynamics

In an expansion process, a gas does 500 J of work and absorbs 300 J of heat. What is the change in internal energy?

A ΔU = -200 J

B ΔU = 200 J

C ΔU = -800 J

D ΔU = 800 J

Correct Answer: A. ΔU = -200 J

EXPLANATION

Using first law: ΔU = Q - W = 300 - 500 = -200 J (internal energy decreases)

Test

Q.43 Medium Thermodynamics

For a van der Waals gas, which statement is correct?

A It accounts for molecular volume and intermolecular forces

B It is identical to ideal gas behavior at high pressures

C It has no internal energy dependence on volume

D It violates the first law of thermodynamics

Correct Answer: A. It accounts for molecular volume and intermolecular forces

EXPLANATION

Van der Waals equation (P + a/V²)(V - b) = RT accounts for molecular volume (b term) and intermolecular attractive forces (a term)

Test

Q.44 Medium Thermodynamics

A reversible cycle absorbs 100 J at 400 K and rejects heat at 300 K. What is the minimum heat rejected?

A Q_c = 75 J

B Q_c = 50 J

C Q_c = 33.3 J

D Q_c = 25 J

Correct Answer: A. Q_c = 75 J

EXPLANATION

For reversible cycle: Q_h/T_h = Q_c/T_c. 100/400 = Q_c/300. Q_c = (100 × 300)/400 = 75 J

Test

Q.45 Medium Thermodynamics

A gas expands isothermally from 1 L to 10 L against a constant external pressure of 1 atm. What is the work done? (1 L·atm = 101.325 J)

A W = 910.125 J

B W = 1013.25 J

C W = 808.6 J

D W = 1114.6 J

Correct Answer: A. W = 910.125 J

EXPLANATION

For isothermal expansion against constant pressure: W = PₑₓₜΔV = 1 × (10-1) = 9 L·atm = 9 × 101.325 = 910.125 J

Test

Q.46 Medium Thermodynamics

Which process is represented by PVⁿ = constant?

A Isothermal (n=1)

B Adiabatic (n=γ)

C Polytropic (any n)

D All of the above

Correct Answer: D. All of the above

EXPLANATION

The polytropic equation PVⁿ = constant covers isothermal (n=1), adiabatic (n=γ), isobaric (n=0), and isochoric (n=∞) processes as special cases

Test

Q.47 Medium Thermodynamics

In an isobaric process, 5 moles of a monoatomic ideal gas are heated from 300 K to 400 K. What is the heat supplied? (R = 8.314 J/(mol·K))

A Q = 20.785 kJ

B Q = 41.57 kJ

C Q = 62.355 kJ

D Q = 31.178 kJ

Correct Answer: C. Q = 62.355 kJ

EXPLANATION

For isobaric process: Q = nCₚΔT. For monoatomic gas, Cₚ = (5/2)R. Q = 5 × (5/2) × 8.314 × 100 = 62.355 kJ

Test

Q.48 Medium Thermodynamics

A cyclic process consists of two isothermal and two adiabatic processes (Carnot cycle). What is true about the heat exchanges?

A Heat is exchanged in both isothermal processes only

B Heat is exchanged in adiabatic processes

C Net heat exchange is always zero

D Heat exchange is equal in both isothermal processes

Correct Answer: A. Heat is exchanged in both isothermal processes only

EXPLANATION

In Carnot cycle, isothermal processes involve heat exchange with reservoirs (Q₁ at hot reservoir, Q₂ at cold reservoir), while adiabatic processes have no heat exchange (Q=0)

Test

Q.49 Medium Thermodynamics

In an adiabatic process with a diatomic ideal gas (γ = 1.4), if the initial temperature is 300 K and volume increases by factor of 2, what is the final temperature?

A T₂ = 178.9 K

B T₂ = 212.1 K

C T₂ = 195.7 K

D T₂ = 223.4 K

Correct Answer: A. T₂ = 178.9 K

EXPLANATION

For adiabatic process: TVγ⁻¹ = constant. T₁V₁^(γ-1) = T₂V₂^(γ-1). 300 × 1^0.4 = T₂ × 2^0.4. T₂ = 300/2^0.4 = 300/1.6818 = 178.9 K

Test

Q.50 Medium Thermodynamics

A gas undergoes an isothermal process at 300 K. If the initial volume is 2 L and final volume is 5 L, what is the work done by the gas? (Take R = 8.314 J/(mol·K), n = 1 mol)

A W = 2.74 kJ

B W = 3.04 kJ

C W = 2.50 kJ

D W = 1.98 kJ

Correct Answer: B. W = 3.04 kJ

EXPLANATION

For isothermal process: W = nRT ln(Vf/Vi) = 1 × 8.314 × 300 × ln(5/2) = 2494.2 × 1.609 = 3.04 kJ

Test

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