NCERT Solutions Class 11 Physics Thermodynamics – Free PDF Download
*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics will give you the self-confidence to face the exam without fear. In this chapter, students shall study the laws which govern thermal energy. The process where work is converted into heat and vice versa is explained in brief in this chapter. To help students understand the basic concepts of this chapter, we at BYJU’S, have designed the chapter-wise NCERT Solutions according to the updated CBSE Syllabus 2023-24 in PDF format, which can be accessed at any time and at any place.
Thermodynamics in this chapter pertains to the concepts of temperature and heat and their conversion to other forms of energy. This chapter also contains numericals, and their regular practice is important to perform well in the exam. The solutions contain important concepts highlighted, formulas and definitions in simple language to help students revise the chapter a few days before the exam. Get the free PDF of NCERT Solutions for Class 11 Physics from the link given below.
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics
Access the Answers to NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics
1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g?
Solution:
Given
Water is flowing at a rate of 3.0 litre/min
The geyser heats the water, raising the temperature from 270 C to 770 C.
Initial temperature, T1 = 270 C
Final temperature, T2 = 770 C
Rise in temperature, T = T2 – T1
= 77 – 27
= 500 C
Heat of combustion = 4 x 104 J / g
Specific heat of water, C = 4.2 J / g 0C
Mass of flowing water, m = 3.0 litre / min
= 3000 g / min
Total heat used, Q = mcT
= 3000 x 4.2 x 50
On calculation, we get,
= 6.3 x 105 J / min
Rate of consumption = 6.3 x 105 / (4 x 104)
We get,
= 15.75 g/min
Therefore, rate of consumption is 15.75 g/min
2. What amount of heat must be supplied to 2.0 × 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)
Solution:
Given,
Mass of nitrogen = 2 x 10-2 kg
= 20 g
Rise in temperature = ΔT
= 450 C
Heat required = Q =?
Q = nCT
We know,
C = 7R / 2 (diatomic molecule)
C = 7 x 8.3 / 2
n (no. of moles) = w / m
where,
w = 20 g
m = 28 u
n = 20 / 28
n = 1/ 1.4 moles
Let the temperature be 45 K
Q = 10 / 14 x 7 / 2 x 8.3 x 45
We get,
Q = 933.75 J
3. Explain why
(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact, do not necessarily settle to the mean temperature (T1 + T2)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Solution:
(i). When two bodies having different temperatures, say T1 and T2 are brought in thermal contact with each other, there is a flow of heat from the body at the higher temperature to the body at the lower temperature till both the bodies reach an equilibrium position, i.e., both the bodies are having equal temperature. The equilibrium temperature is only equal to the mean temperature when the thermal capacities of both the bodies are equal.
(ii).The coolant used in a chemical or nuclear plant should always have a high specific heat. Higher is the specific heat of the coolant, higher is its capacity to absorb heat and release heat. Therefore, a liquid with a high specific heat value is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
(iii). When the driver is driving a vehicle then due to the motion of air molecules the air temperature inside the tyre increases. And according to the Charles’ law, the temperature is directly proportional to pressure.
Therefore, when the temperature inside a tyre increases, then there is also an increase of air pressure.
(iv). The relative humidity in a harbour town is more than that of the relative humidity in a desert town. Humidity is a measure of water vapor in the atmosphere and the specific heat of water vapor is very high. Therefore, the climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution:
The cylinder is completely insulated from its surroundings.
Therefore, no heat is exchanged between the system (cylinder) and its surroundings.
Thus, the process is adiabatic
Initial pressure inside the cylinder = P1
Final pressure inside the cylinder = P2
Initial volume inside the cylinder = V1
Final volume inside the cylinder = V2
Ratio of specific heat, γ = Cp / Cv = 1.4
For an adiabatic process, we have:
P1V1γ = P2V2γ
The final volume is compressed to half of its initial volume
Hence,
V2 = V1 / 2
P1V1γ = P2(V1 / 2)γ
P2 / P1 = V1γ / (V1 / 2)γ
= 21.4
We get,
= 2.639
Therefore, the pressure increases by a factor of 2.639
5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Solution:
Given
The work done (W) on the system while the gas changes from state A to state B is 22.3 J
This is an adiabatic process.
Therefore, change in heat is zero.
So,
ΔQ = 0
ΔW = – 22.3 (Since the work is done on the system)
From first law of thermodynamics, we have:
ΔQ = ΔU + ΔW
Where,
ΔU = Change in the internal energy of the gas
Hence,
ΔU = ΔQ – ΔW
= 0 – (-22.3 J)
We get,
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:
ΔQ = 9.35 cal
= 9.35 x 4.19
On calculation, we get,
= 39.1765 J
Heat absorbed, ΔQ = ΔU + ΔW
Thus,
ΔW = ΔQ – ΔU
= 39.1765 – 22.3
We get,
= 16.8765 J
Hence, 16.88 J of work is done by the system
6. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
Solution:
(a). Now, as soon as the stop cock is opened the gas will start flowing from cylinder P to cylinder Q which is completely evacuated, and thus the volume of the gas will be doubled because both the cylinders have equal capacity. And since the pressure is inversely proportional to volume, hence the pressure will get decreased to half of the original value.
Since the initial pressure of the gas in cylinder P is 1 atm, therefore, the pressure in each of the cylinder will now be 0.5 atm.
(b). Here, in this case, the internal energy of the gas will not change i.e. ΔU = 0. It is because the internal energy can change only when the work is done by the system or on the system. Since in this case, no work is done by the gas or on the gas.
Therefore, the internal energy of the gas will not change.
c) There will be no change in the temperature of the gas. It is because during the expansion of gas there is no work being done by the gas.
Therefore, there will be no change in the temperature of the gas in this process.
d). The above case is the clear case of free expansion and free expansion is rapid and it cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non – equilibrium states, they do not lie on the Pressure-Volume – Temperature surface of the system
7. A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Given
Work done by the steam engine per minute, W = 5.4 x 108 J
Heat supplied from the boiler, H = 3.6 x 109 J
Efficiency of the engine = Output Energy / Input Energy
Hence,
η = W / H
= 5.4 x 108 / (3.6 x 109)
On simplification, we get,
= 0.15
Therefore, the percentage efficiency of the engine is 15%
Amount of heat wasted = 3.6 x 109 – 5.4 x 108
We get,
= 30.6 x 108
= 3.06 x 109 J
Hence, the amount of heat wasted per minute is 3.06 x 109 J
8. An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:
According to law of conservation of energy
Total energy = work done + internal energy
ΔQ = ΔW + ΔU
Here,
Rate at which heat is supplied ΔQ = 100 W
Rate at which work is done ΔW = 75 Js-1
Rate of change of internal energy is ΔU
ΔU = ΔQ – ΔW
ΔU = 100 – 75
We get,
ΔU = 25 J s-1
Hence,
The internal energy of the system is increasing at a rate of 25 W
9. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F
Solution:
Total work done by the gas from D to E to F = Area of ∆DEF
Area of ∆DEF = (1/2) x DE x EF
Where,
DF = Change in pressure
= 600 N/m2 – 300 N/m2
We get,
= 300 N/ m2
FE = Change in volume
= 5.0 m3 – 2.0 m3
We get,
= 3.0 m3
Area of ∆DEF = (1/ 2) x 300 x 3
On further calculation, we get,
= 450 J
Hence, the total work done by the gas from D to E to F is 450 J
10. A refrigerator is to maintain eatables kept inside at 90 C. If room temperature is 360 C, calculate the coefficient of performance.
Solution:
Temperature inside the refrigerator, T1 = 90 C
= 273 + 9
= 282 K
Room temperature, T2 = 360 C
= 273 + 36
= 309 K
Coefficient of performance = (T1) / (T2 – T1)
On substituting, we get,
= 282 / (309 – 282)
We get,
= 10.44
Hence, the coefficient of performance of the given refrigerator is 10.44
| Also Access |
| NCERT Exemplar for Class 11 Physics Chapter 12 |
| CBSE Notes for Class 11 Physics Chapter 12 |
Thermodynamics is a hot topic for the experts who prepare question papers. In order to get more marks in Class 11 Physics, it is crucial to study these solutions as one can expect many questions from this resource being asked often in entrance exams and other competitive exams.
Thermodynamics is one of the most scoring sections in Physics NCERT Solutions for Class 11. Students must study this Chapter in-depth to excel in the exam. Some key points of Thermodynamics are given below.
- Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system does not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. The temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed.
- Heat capacity, in general, depends on the process of the system that goes through when the heat is supplied.
- In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium.
- In isothermal quasi-static processes, heat is absorbed or given out by the system even though, at every stage, the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir.
Subtopics of Class 11 Physics Chapter 12 Thermodynamics
- Introduction
- Thermal equilibrium
- Zeroth law of Thermodynamics
- Heat, internal energy and work
- First law of thermodynamics
- Specific heat capacity
- Thermodynamic state variables and the equation of state
- Thermodynamic processes
- Heat engines
- Refrigerators and heat pumps
- The second law of thermodynamics
- Reversible and irreversible processes
- Carnot Engine
In India, CBSE is one of the most widely accepted educational boards in the country. The syllabus covered by this board is also helpful in attempting other competitive exams such as JEE and NEET. NCERT Solutions is one of the best study materials to prepare physics for Class 11. The NCERT Solutions for Class 11 Physics Chapter 12 is given here so that students can prepare effectively for their exams.
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Disclaimer –
Dropped Topics –
12.9 Heat Engines
12.10 Refrigerators and Heat Pumps
Exercises 12.7 and 12.10
Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 12
How will using the NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics help you with your exam preparation?
What will I learn from Chapter 12 of NCERT Solutions for Class 11 Physics?
1. Introduction
2. Thermal equilibrium
3. Zeroth law of Thermodynamics
4. Heat, internal energy and work
5. First law of thermodynamics
6. Specific heat capacity
7. Thermodynamic state variables and the equation of state
8. Thermodynamic processes
9. Heat engines
10. Refrigerators and heat pumps
11. The second law of thermodynamics
12. Reversible and irreversible processes
13. Carnot Engine
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