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Ever curious about how heat moves, why engines function, or why a hot cup of tea gets cold after some time? Thermodynamics is the answer to all these interesting phenomena!
In NCERT Exemplar Class 11 Physics Solutions Chapter 12, students will learn about concepts such as heat, work, internal energy, and thermodynamic laws. This chapter is the gateway to understanding practical applications, ranging from refrigerators to automobile engines. Get started to learn these crucial concepts with step-by-step solutions!
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NCERT Exemplar Class 11 Physics Solutions Chapter 12 deals with significant topics of heat, energy, and work. This chapter discusses how heat is transferred to energy and how it influences matter. Students will study the laws of thermodynamics and their uses in real life. These solutions make it easy to solve exam questions and understand significant formulas for better preparation.
Question:1
An ideal gas undergoes four different processes from the same initial state (Figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic.
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
The answer is the option (c) 4 is the isobaric process, 1 is isochoric. Out of 3 and 2, 3 has a smaller slope hence is isothermal. Remaining process 2 is adiabatic.Question:2
If an average person jogs, he produces
(a) 0.025 kg
(b) 2.25 kg
(c) 0.05 kg
(d) 0.20 kg
Answer:
The answer is the option (a) 0.025 kg
Question:3
Consider P-V diagram for an ideal gas shown in Figure.
Out of the following diagrams (Figure), which represents the T-P diagram?
(a) (iv)
(b) (ii)
(c) (iii)
(d) (i)
Answer:
The answer is the option (c)Question:4
Answer:
The answer is the option (b). Since the direction of arrows is anticlockwise, so work done is negative equal to the area of the loopQuestion:5
Answer:
The answer is the option (a). Let us consider a P-V diagram for container A and B. Compression of gas is involved in both the cases. For the isothermal process (gas A) during
For adiabatic process
Question:6
Answer:
The answer is the option (b) Let the equilibrium temperature of the system=TQuestion:7
Which of the processes described below are irreversible?
(a) The increase in temperature of an iron rod by hammering it.
(b) A gas in a small container at a temperature T1 is brought in contact with a big reservoir at a higher temperature T2 which increases the temperature of the gas.
(c) A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless piston.
(d) An ideal gas is enclosed in a piston-cylinder arrangement with adiabatic walls. A weight W is added to the piston, resulting in compression of the gas.
Answer:
The answer is the option (a), (b), (d)Question:8
An ideal gas undergoes isothermal process from some initial state I to final state f. Choose the correct alternatives
a) dU = 0
b) dQ = 0
c) dQ = dU
d) dQ = dW
Answer:
The answer is the option (a) and (d).
Since the process is isothermal
For an ideal gas, dU=Change in internal energy=nCvdT
dT=0 ;Thus dU=0
dQ=dU+dW
dQ=dW
Hence, a and d are correct.
Question:9
Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
(a) Change in internal energy is same in IV and III cases, but not in I and II.
(b) Change in internal energy is same in all the four cases.
(c) Work done is maximum in case I
(d) Work done is minimum in case II.
Answer:
The answer is the option (b) and (c). dU is independent of the path followed in a P-V diagram. It depends only on the initial and final position. Work done is the area enclosed with the V-axis.Question:10
Consider a cycle followed by an engine
1 to 2 is isothermal
2 to 3 is adiabatic
3 to 1 is adiabatic
Such a process does not exist because
a) heat is completely converted to mechanical energy in such a process, which is not possible
b) mechanical energy is completely converted to heat in this process, which is not possible
c) curves representing two adiabatic processes don’t intersect
d) curves representing an adiabatic process and an isothermal process don’t intersect
Answer:
The answer is the option (a) and (c).
The given process is cyclic, which starts at 1 and ends at 1.
So, dU=0
And dQ= dU+dW
dQ=dW
The heat energy supplied to the system gets converted to mechanical work which isn't possible by the second law of thermodynamics.
Question:11
Consider a heat engine as shown on the figure.
If W > 0, then possibilities are:
a)
b)
c)
d)
Answer:
The answer is the option (a) and (c)Question:12
Can a system be heated, and its temperature remains constant?
Answer:
A system can be heated, but with a constant temperature when it does work against the surrounding to compensate for the heat supplied.Question:13
Answer:
For path 1 Q1=1000J
Question:14
If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Answer:
When the refrigerator door is kept open, the amount of heat absorbed from inside the refrigerator and the work is done on it by electricity both will be rejected by the refrigerator in the room. This will make the room hotter.Question:15
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Answer:
In adiabatic compression, the temperature of gas increases while no heat is added to the system.
dQ=0
dQ=dU+dW
dU=-dW
So in compression work done on the system is negative. Thus, dU is positive, increasing the temperature of the system. As the internal energy of gas increases, its temperature decreases.
Question:16
Air pressure in a car tyre increases during driving. Explain.
Answer:
During driving, a reaction force is applied due to the force on tyres. The temperature of gas increases due to this reaction force causing the gas inside the tyre to expand while the volume inside the tyre remains constant. In accordance with the Charles Law, the temperature of the car tyre increases as well as the pressureQuestion:17
Answer:
The efficiency of Carnot's engineQuestion:18
Answer:
The energy produced by 1 kg fat=7000 k calQuestion:19
Answer:
Air is transferred into tyre adiabatically.Question:20
Answer:
Here refrigerator acts as a Carnot's engine in reverse order with an efficiencyQuestion:21
Answer:
Question:22
The initial state of a certain gas is
a) The expansion takes place at constant temperature
b) The expansion takes place at constant pressure
Plot the P-V diagram for each case. In which of the two cases is the work done by the gas more?
Answer:
The expansion from Vi to Vf at a constant temperature in (a) is an isothermal expansion.i.e.Question:23
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure.
a) find the work done when the gas is taken from state 1 to state 2
b) what is the ratio of temperature
c) given the internal energy for one mole of gas at temperature T is
Answer:
LetQuestion:24
A cycle followed by an engine is shown in the figure.
A to B: volume constant
B to C: adiabatic
C to D: volume constant
D to A: adiabatic
a) in which part of the cycle heat is supplied to the engine from outside?
b) in which part of the cycle heat is being given to the surrounding by the engine?
c) what is the work done by the engine in one cycle in terms of
d) what is the efficiency of the engine?
Answer:
(a) Heat is supplied to the engine in part AB where dV=0Question:25
A cycle followed by an engine is shown in the figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle considering
AB: constant volume
BC: constant pressure
CD: adiabatic
DA: constant pressure
Answer:
Ans.Question:26
Answer:
Slope of graph atQuestion:27
Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached. A spring is attached to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from V0 to V1.
a) what is the initial pressure of the system?
b) what is the final pressure of the system?
c) using the first law of thermodynamics, write down the relation between Q, Pa, V, V0, and k.
Answer:
1. Thermal Equilibrium - When two bodies in contact stop exchanging heat, they are in thermal equilibrium.
2. Zeroth Law of Thermodynamics - If
3. First Law of Thermodynamics - Energy cannot be created or destroyed, only transferred as heat or work.
4. Specific Heat Capacity - The heat required to raise the temperature of 1 kg of a substance by
5. Thermodynamic Processes -
6. Second Law of Thermodynamics - Heat flows from hot to cold objects naturally; the efficiency of a heat engine is always less than
7. Carnot Engine - An ideal engine with maximum efficiency.
Chapter 1 | Units and Measurement |
Chapter 2 | Motion in a straight line |
Chapter 3 | Motion in a Plane |
Chapter 4 | Laws of Motion |
Chapter 5 | Work, Energy and Power |
Chapter 6 | System of Particles and Rotational motion |
Chapter 7 | Gravitation |
Chapter 8 | Mechanical Properties of Solids |
Chapter 9 | Mechanical Properties of Fluids |
Chapter 10 | Thermal Properties of Matter |
Chapter 11 | Thermodynamics |
Chapter 12 | Kinetic Theory |
Chapter 13 | Oscillations |
Chapter 14 | Waves |
Yes, this chapter is important for NEET and the medical entrance exam. The physics papers tend to have 1-2 questions from this chapter every year.
Our teams of experienced teachers who have industry and teaching knowledge for years have solved these questions keeping in mind the student’s level of understanding.
Yes, these class 11 physics NCERT exemplar solutions chapter 12 are designed in a way so that the students can score the maximum marks as per the marking scheme.
The second law explains why heat moves from hot to cold, why engines aren’t 100% efficient, and why systems tend toward disorder over time.
Entropy measures disorder. It increases in natural processes (like melting ice) but remains constant in ideal reversible processes.
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