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Thermodynamics — NEET & JEE MCQs with answers
Thermodynamics studies the laws governing thermal energy and the interconversion of heat and work. It introduces the concepts of thermal equilibrium, internal energy, and the three foundational laws — Zeroth, First, and Second — that constrain all thermodynamic processes. The chapter also analyses ideal thermodynamic processes (isothermal, adiabatic, isochoric, isobaric, cyclic) and the Carnot engine as the theoretical maximum-efficiency heat engine.
Topics in this chapter
- 11.1 Introduction
- 11.2 Thermal Equilibrium
- 11.3 Zeroth Law of Thermodynamics
- 11.4 Heat, Internal Energy and Work
- 11.5 First Law of Thermodynamics
- 11.6 Specific Heat Capacity
- 11.7 Thermodynamic State Variables and Equation of State
- 11.8 Thermodynamic Processes
- 11.9 Second Law of Thermodynamics
- 11.10 Reversible and Irreversible Processes
- 11.11 Carnot Engine
Free practice questions
The efficiency of a Carnot engine is 30% when the sink is at 350 K. The source temperature is
- A. 500 K
- B. 245 K
- C. 455 K
- D. 105 K
Show answer
Answer: A — 0.30 = 1 − 350/T1 ⇒ 350/T1 = 0.70 ⇒ T1 = 350/0.70 = 500 K.
NCERT Class 11 Physics · Ch 11 · 11.11 Carnot Engine
A refrigerator removes $Q_2 = 360\,\text{J}$ from the cold body and rejects $Q_1 = 480\,\text{J}$ to the room. Its coefficient of performance $\alpha = Q_2/W$ is
- A. 0.75
- B. 1.33
- C. 3
- D. 4
Show answer
Answer: C — $W = Q_1 - Q_2 = 480 - 360 = 120\,\text{J}$. $\alpha = Q_2/W = 360/120 = 3$.
NCERT Class 11 Physics · Ch 11 · 11.9 Second Law of Thermodynamics
Through which type of wall can two systems eventually reach thermal equilibrium with each other?
- A. A diathermic (conducting) wall
- B. A perfectly rigid wall regardless of conductivity
- C. Any wall that blocks heat
- D. An adiabatic (insulating) wall
Show answer
Answer: A — Only a diathermic (conducting) wall allows heat to flow, so equilibrium between the two systems is attained in due course; an adiabatic wall blocks heat and keeps them independent.
NCERT Class 11 Physics · Ch 11 · Thermal equilibrium
Consider internal energy $U$ and heat $Q$. Based on NCERT section 11.7, which statement correctly distinguishes them?
- A. $U$ is a state variable, whereas $Q$ is not a state variable
- B. Both $U$ and $Q$ are state variables
- C. Neither $U$ nor $Q$ is a state variable
- D. $U$ is not a state variable, whereas $Q$ is
Show answer
Answer: A — NCERT lists internal energy $U$ as a state variable (extensive), while the footnote states that heat $Q$ is not a state variable. $U$ depends only on the present state; $Q$ is not a property of the state.
NCERT Class 11 Physics Ch 11.7
A gas absorbs 1500 J of heat. Of this, 40% goes into doing external work. The increase in internal energy of the gas is:
- A. 900 J
- B. 1500 J
- C. 2100 J
- D. 600 J
Show answer
Answer: A — ΔW = 0.40 × 1500 = 600 J. ΔU = ΔQ − ΔW = 1500 − 600 = 900 J.
NCERT Class 11 Physics · Ch 11 · 11.5 First Law of Thermodynamics
For an ideal gas undergoing an isochoric process, the change in temperature for a given amount of heat is determined by the specific heat at:
- A. Constant volume
- B. Constant temperature
- C. Constant pressure
- D. Constant internal energy
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Answer: A — In an isochoric process, the change in temperature for a given amount of heat is determined by the specific heat of the gas at constant volume (NCERT 11.8.4).
NCERT Class 11 Physics · Ch 11 · 11.8 Thermodynamic Processes
Thermodynamics is concerned with:
- A. the forces and torques on the body
- B. the motion of the body's centre of mass
- C. the internal macroscopic state of the body
- D. the trajectory of a projectile
Show answer
Answer: C — NCERT: 'It is concerned with the internal macroscopic state of the body.'
NCERT Class 11 Physics · Ch 11 · Introduction
A gas is taken from state 1 to state 2 by two different paths. Which quantity is necessarily the same for both paths?
- A. The change in internal energy ΔU
- B. The work done
- C. Both the heat supplied and the work done
- D. The heat supplied
Show answer
Answer: A — Since U is a state variable, the change in internal energy ΔU depends only on the initial and final states and not on the path. Heat and work, in general, depend on the path.
NCERT Class 11 Physics · Ch 11 · 11.4 Heat, Internal Energy and Work
From the previous data, the same gas (2 moles, $C_v = 20\ \text{J mol}^{-1}\text{K}^{-1}$) is heated at constant pressure through $15\ \text{K}$. The heat needed is (take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$):
- A. $849.4\ \text{J}$
- B. $249.4\ \text{J}$
- C. $600\ \text{J}$
- D. $1200\ \text{J}$
Show answer
Answer: A — $C_p = C_v + R = 20 + 8.314 = 28.314\ \text{J mol}^{-1}\text{K}^{-1}$. $\Delta Q = \mu\,C_p\,\Delta T = 2 \times 28.314 \times 15 = 849.4\ \text{J}$.
NCERT Class 11 Physics · Ch 11 · 11.6 Specific Heat Capacity
When a moving body comes to a stop, losing its mechanical energy as heat to the floor and the body, this is cited by NCERT as an example of:
- A. a dissipative effect causing irreversibility
- B. an equilibrium state of the system
- C. a process consistent with reversibility
- D. a quasi-static reversible process
Show answer
Answer: A — NCERT gives this as an example of a dissipative effect (the second cause of irreversibility): 'a moving body coming to a stop and losing its mechanical energy as heat to the floor and the body.'
NCERT Class 11 Physics · Ch 11 · Reversible and Irreversible Processes
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