Class 11 Physics Chapter 12 Important Questions Thermodynamics

Q. 1. Which law of thermodynamics gives the concept of temperature? Which law gives the concept of internal energy? Ans. Zeroth law of thermodynamics

It is important for the students that all the concepts should be very clear for better marks in future. Here, we are providing important conceptual questions and answers for class 11 physics chapter 12 Thermodynamics. In this lesson, students will learn about Thermodynamics. This will not only help the students to know the important questions but will also help them during revision.

Conceptual Questions for Class 11 Physics Chapter 12 Thermodynamics

Q. 1. Which law of thermodynamics gives the concept of temperature? Which law gives the concept of internal energy?
Ans. Zeroth law of thermodynamics gives the concept of temperature and first law of thermodynamics gives the concept of internal energy.

Q. 2. A piece of metal is hammered? What happens to its internal energy, and why?
Ans. When a piece of metal is hammered, a part of work done on the metal piece during hammering is converted into heat and temperature of metal piece rises. So, the internal energy of metal piece increases.

Q. 3. Out of isochoric or isobaric or isothermal or adiabatic processes, for which process the equation of state PV = µRT true?
Ans. The equation of state PV = µRT is true for all the thermodynamic processes mentioned here provided that the gas behaves as an ideal gas.

Q. 4. What is the value of specific heat capacity of a gas under (a) isothermal, and (b) adiabatic conditions.
Ans. (a) Under isothermal condition, specific heat capacity of a gas is infinite.

(b) Under adiabatic condition, specific heat capacity of a gas is zero.

Q. 5. What are the conditions for complete equilibrium of a system?
Ans. The conditions are as follows:

(a) Temperature of each and every part of the system must be exactly same.

(b) There should be no net unbalanced force or torque on a part or whole of the system.

(c) There should be no change in the system with time due to chemical reactions, i.e., chemical composition of the system must remain same throughout.

Q. 6. Can mechanical energy be converted continuously and completely into heat? Is the reverse also true?
Ans. We can continuously and completely convert mechanical energy into heat. But we cannot convert heat into mechanical work continuously and completely. It is on account of the restriction imposed by the second law of thermodynamics.

Q. 7. Give one example each of an isochoric process and an isobaric process.
Ans. Cooling or heating of a gas enclosed in a chamber is an isochoric process. Expansion of a gas enclosed in a cylinder fitted with a piston capable of motion is an isobaric process.

Q. 8. What is the shape of P-V indicator diagram for (a) an isochoric process, and (b) an isobaric process?
Ans. P-V indicator diagram for an:

(a) isochoric process is a straight line parallel to pressure axis (or perpendicular to the volume axis), and

(b) isobaric process is a straight line parallel to volume axis.

Read also: Kinetic Theory Class 11 Physics Notes Chapter 13

Q. 9. Can you give examples of approximate reversible processes?
Ans. Isothermal and adiabatic expansion or compression of a gas under ideal condition may be considered as reversible processes. Again, process of melting of a solid or boiling of a liquid may be considered as a reversible process provided that no heat transfer is taking place to or from the surroundings.

Q. 10. Milk is poured into a cup of tea and is mixed with a spoon. Is this a reversible process? Why?
Ans. The process is an irreversible process. When milk poured into a cup of tea is mixed by stirring with a spoon, work is done on the system and hence, internal energy of the system slightly increases. If, by some process, we want to separate milk from the tea, we cannot recover the work done during mixing process. Rather we shall require more work to be done. Thus, it is clear that the process is irreversible.

Q. 11. What factors cause reduction in the efficiency of a heat engine from its ideal value?
Ans. Friction, turbulence, finite values of the temperatures of hot and cold reservoirs and conduction of heat from the system to the surroundings are the important factors which reduce the efficiency of an engine from its ideal value.

Q. 12. What is the most important conclusion drawn from the second thermodynamics? law of
Ans. The most important conclusion drawn from the second law of thermodynamics is the simple fact that it is impossible for heat to flow by itself from a colder to a hotter body or region.

Q. 13. Why is a spark produced sometimes when two substances are struck hard against each other?
Ans. When two substances are struck hard against each other, mechanical work is done against frictional force acting between the substances. This work done is transformed into heat. If amount of heat produced is quite large, a spark is produced between the two substances.

Q. 14. Water at the base of a waterfall is slightly warmer than that at the top. Explain why.
Ans. When water falls from a height, its gravitational potential energy is converted into kinetic energy at the base of waterfall. A part of this kinetic energy is further transformed into heat, due to which water at the base of waterfall becomes slightly warmer.

Q. 15. On removing valve of a bicycle tube, the escaping air becomes cool. Why?
Ans. Inside a bicycle tube, air is filled at a pressure greater than the atmospheric pressure. When valve of tube is removed, the air present inside the tube suddenly expands adiabatically. As a result of adiabatic expansion of air, the temperature of air falls and as a result, the air escaping from the bicycle tube becomes cool.

Q. 16. Can we decide whether change in internal energy of a system is due to its heating or on account of work done on the system by the surroundings? Give reason too.
Ans. We cannot decide the cause of change in internal energy of a system because internal energy of a system changes on supplying heat as well as on doing external work on the system.

Read also: Class 11 Physics Chapter 12 MCQs with Answer Thermodynamics

Q. 17. An electric fan is switched on in a closed room. Will the air of the room be cooled or heated? Justify.
Ans. When an electric fan is switched on in a closed room, electric energy is being continuously consumed and kinetic energy is produced. The kinetic energy of the fan is being consumed to do work against dissipative forces like friction and viscous drag of air, etc. This work done against dissipative forces is converted into heat. As a result, the air of the room is heated up.

Q. 18. Is it possible for a ship to utilise the internal energy of sea water to operate its engine? Give reason too.
Ans. No, it is not possible for a ship to utilise the internal energy of sea water to operate its engine. If the ship is to utilise internal energy of sea water, then the internal energy of sea water is to be converted into mechanical work. As no engine can convert 100% energy into work, hence some part of energy has to be rejected to a colder environment. Since no such colder environment is available there, hence ship cannot use internal energy of sea water to run its own engine.

Q. 19. Although mechanical work can be completely converted into heat but whole of heat cannot be converted into work. Why?
Ans. Yes, mechanical work can be completely converted into heat. However, heat given to a system cannot be completely converted into work. This restriction is on account of second law of thermodynamics. When heat is continuously converted into work by employing a heat engine, then a part of heat absorbed must be rejected to a colder surroundings and hence, 100% of the heat cannot be converted into work.

Q. 20. An electric refrigerator transfers heat from a colder region to warmer surroundings. Does it not violate the second law of thermodynamics? Comment.
Ans. No, the working of an electric refrigerator does not violate the second law of thermodynamics. As per second law of thermodynamics, heat cannot be transferred itself from a colder body to a hotter body but, of course, we can transfer heat from a colder region to a hotter region provided that some work is being done on the system for this. In an electric refrigerator, external electrical energy works on the system (refrigerant system) and enables heat to flow from a colder portion of refrigerator to outer warm surroundings.

Q. 21. If efficiency of a Carnot engine is independent of the nature of working substance, then perhaps real engines should also be similarly independent upto a certain extent. Why then, for real engines, we are so concerned to find a suitable fuel like coal, petrol, etc. Why do we not use stones as fuel?
Ans. Of course, efficiency of a heat engine may be independent of the nature of working substance, but to convert heat into work by an engine, we need initial absorption of heat by the engine from a heat source. Fuel used in engine on combustion provides the requisite amount of heat. Hence, we want to find a suitable fuel which provides large amount of heat on combustion. A stone cannot produce heat by combustion process.

Q. 22. Is coefficient of performance of a domestic refrigerator a constant quantity? Why?
Ans. No, the coefficient of performance of a domestic refrigerator is not a constant quantity. Coefficient of performance, `\alpha=\frac{T_2}{T_1-T_2}` From the relation, it is clear that the coefficient of performance of a domestic refrigerator gradually decreases as the temperature of food items (i.e., `T_2`) stored in a refrigerator gradually falls.

Q. 23. Two reversible heat engines operate between same pair of hot and cold reservoirs. Will the two be equally efficient or have different, efficiencies? Why?
Ans. Both the reversible heat engines will have exactly the same efficiency when operating between same pair of hot and cold reservoirs. It is because the efficiency of a reversible engine simply depends on the temperatures of hot and cold reservoirs and is independent of the nature or the quantity of working substance.

Q. 24. Name a thermodynamic process in which:
(a) internal energy remains constant,
(b) total heat content remains constant, and
(c) no work is being done on or by the gas.
Ans. (a) In an isothermal process, temperature and hence internal energy of a gas remains constant, i.e.,

U = a constant

or ΔU = 0

(b) In an adiabatic process, total heat content of the system remains constant and there is no exchange of heat with its surroundings, i.e.,

Q = a constant

or ΔQ = 0

(c) In an isochoric process, work done by a gas is zero, i.e.,

W = 0

Q. 25. Surface temperature of the Sun is about 6000 K. Can we produce a temperature of 8000 K (say) by focussing the Sun's rays at an object with the help of a large convex lens?
Ans. No, we cannot expect to achieve a temperature greater than the surface temperature of the Sun, i.e., 6000 K because second law of thermodynamics does not allow transfer of heat from a colder body to a hotter body.

Q. 26. Is it possible to increase the temperature of a gas without giving it heat? If yes, under what condition?
Ans. In an adiabatic compression of a gas, although no heat is given to a gas, yet its temperature rises because work done on the gas during its compression leads to increase in internal energy and consequently, the temperature of the gas.

Q. 27. A thermos flask contains coffee. The flask is vigorously shaken. Consider the coffee as a system.

  • (a) Does its temperature rise?
  • (b) Has heat been added to it?
  • (c) Has work been done on it?
  • (d) Has its internal energy changed?

Ans. (a) Yes, temperature of coffee rises because work has been done on it in shaking and work is converted into heat. As a result, temperature of coffee rises.

(b) No, heat has not been added to the system.

(c) Yes, work has been done on the system.

(d) Yes, due to rise in temperature, the internal energy of the system rises.

Q. 28. A matchstick can be lighted by rubbing it against a rough surface. How?
Ans. When a matchstick is rubbed against a rough surface, some mechanical work is done against friction and this work is transformed into heat. Consequently, temperature of matchstick rises and matchstick is lighted up.

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