**NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics**– Here are all the NCERT solutions for Class 11 Physics Chapter 12. This solution contains questions, answers, images, explanations of the complete chapter 12 titled Of Thermodynamics taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 12 Thermodynamics After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics in one place.

## NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics

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For a better understanding of this chapter, you should also see summary of Chapter 12 Thermodynamics , Physics, Class 11.

Class | 11 |

Subject | Physics |

Book | Physics Part I |

Chapter Number | 12 |

Chapter Name |
Thermodynamics |

### NCERT Solutions Class 11 Physics chapter 12 Thermodynamics

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### Question & Answer

Q.1:\({\text { A geyser heats water flowing at the rate of } 3.0 \text { litres per minute from } 27^{\circ} \mathrm{C} \text { to } 77^{\circ} \mathrm{C} \text { . If }} {\text { the geyser operates on a gas burner, what is the rate of consumption of the fuel if its }} {\text { heat of combustion is } 4.0 \times 10^{4} \mathrm{J} / \mathrm{g} \text { ? }}\)

Ans :\(\begin{array}{l}{\text { Water is flowing at a rate of } 3.0 \text { litre/min. }} \\ {\text { The geyser heats the water, raising the temperature from } 27^{\circ} \mathrm{C} \text { to } 77^{\circ} \mathrm{C} \text { . }} \\ {\text { Initial temperature, } T_{1}=27^{\circ} \mathrm{C}} \\ {\text { Final temperature, } T_{2}=77^{\circ} \mathrm{C}} \\ {\therefore \text { Rise in temperature, } \Delta T=T_{2}-T_{1}} \\ {=77-27=50^{\circ} \mathrm{C}}\end{array}\) \(\begin{array}{l}{\text { Heat of combustion }=4 \times 10^{4} \mathrm{Jg}} \\ {\text { Specific heat of water, } c=4.2 \mathrm{g}^{-1} \mathrm{c}^{-1}} \\ {\text { Mass of flowing water, } m=3.0 \text { litre/min }=3000 \mathrm{g} / \mathrm{min}} \\ {\text { Total heat used, } \Delta Q=m c \Delta T} \\ {=3000 \times 4.2 \times 50} \\ {=6.3 \times 10^{5} \mathrm{J} / \mathrm{min}}\end{array}\) \(\therefore \text { Rate of consumption }=\frac{6.3 \times 10^{5}}{4 \times 10^{4}}=15.75 \mathrm{g} / \mathrm{min}\)

Q.2:\({\text { What amount of heat must be supplied to } 2.0 \times 10^{-2} \text { kg of nitrogen (at room }} {\text { temperature) to raise its temperature by } 45^{\circ} \mathrm{C} \text { at constant pressure? (Molecular mass of }} {\mathrm{N}_{2}=28 ; R=8.3 \mathrm{Jmol}^{-1} \mathrm{K}^{-1} \text { . }}\)

Ans :\(\begin{array}{l}{\text { Mass of nitrogen, } m=2.0 \times 10^{-2} \mathrm{kg}=20 \mathrm{g}} \\ {\text { Rise in temperature, } \Delta T=45^{\circ} \mathrm{C}} \\ {\text { Molecular mass of } \mathrm{N}_{2}, \mathrm{M}=28} \\ {\text { Universal gas constant, } R=8.3 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}}\end{array}\) \(\begin{array}{l}{n=\frac{m}{M}} \\ {\text { Number of moles, }^{n=\frac{m}{M}}} \\ {=\frac{2.0 \times 10^{-2} \times 10^{3}}{28}=0.714}\end{array}\) \(\begin{array}{l}{\text { Molar specific heat at constant pressure for nitrogen, }^{C_{p}=\frac{7}{2} R}} \\ {=\frac{7}{2} \times 8.3} \\ {\text { The total amount of heat to be supplied is given by the relation: }} \\ {\Delta Q=n C_{P} \Delta T}\end{array}\) \(\begin{array}{l}{=0.714 \times 29.05 \times 45} \\ {=933.38 \mathrm{J}} \\ {\text { Therefore, the amount of heat to be supplied is } 933.38 \mathrm{J} \text { . }}\end{array}\)

Q.3:Explain why (a) Two bodies at different ter-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.

Ans :(a) When two bodies at different temperatures Tl and T2 are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures Of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T1+T2)/2 only when the thermal capacities of both the bodies are equal. (b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a hi$l specific heat 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. (c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles' law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase. (d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Q.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?

Ans :The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic. \(\begin{array}{l}{\text { Initial pressure inside the cylinder }=P_{1}} \\ {\text { Final pressure inside the cylinder }=P_{2}} \\ {\text { Initial volume inside the cylinder }=V_{1}} \\ {\text { Final volume inside the cylinder }=V_{2}} \\ {\text { Ratio of specific heats, } Y=1.4} \\ {\text { For an adiabatic process, we have: }} \\ {\quad P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}}\end{array}\) The final volume is compressed to half of its initial volume. \(V_{2}=\frac{V_{1}}{2}\) \(\begin{array}{l}{P_{1}\left(V_{1}\right)^{\gamma}=P_{2}\left(\frac{V_{1}}{2}\right)^{\gamma}} \\ {\frac{P_{2}}{P_{1}}=\frac{\left(V_{1}\right)^{\gamma}}{\left(\frac{V_{1}}{2}\right)^{T}}=(2)^{\gamma}=(2)^{1,4}=2.639} \\ {\text { Hence, the pressure increases by a factor of } 2.639 .}\end{array}\)

Q.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 cistern. 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)

Ans :\(\begin{array}{l}{\text { The work done (W) on the system while the gas changes from state } A \text { to state } B \text { is } 22.3} \\ {\text { J. }} \\ {\text { This is an adiabatic process. Hence, change in heat is zero. }} \\ {\therefore \Delta Q=0} \\ {\Delta W=-22.33 \text { (Since the work is done on the system) }} \\ {\text { From the first law of thermodynamics, we have: }}\end{array}\) \(\begin{array}{l}{\Delta Q=\Delta U+\Delta W} \\ {\text { Where, }} \\ {\Delta U=\text { Change in the internal energy of the gas }} \\ {\therefore \Delta U=\Delta Q-\Delta W=-(-22.3]} \\ {\Delta U=+22.3 \mathrm{J}}\end{array}\) \(\begin{array}{l}{\text { When the gas goes from state } A \text { to state } B \text { via a process, the net heat absorbed by the }} \\ {\text { system is: }} \\ {\Delta Q=9.35 \text { cal }=9.35 \times 4.19=39.1765 \mathrm{J}} \\ {\text { Heat absorbed, } \Delta Q=\Delta U+\Delta Q} \\ {\therefore \Delta W=\Delta Q-\Delta U} \\ {=39.1765-22.3} \\ {=16.8765 J}\end{array}\)

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