NCERT 11th Physics Chapter 12, class 11 Physics chapter 12 solutions<\/p>\n\n\n\n
Page No: 316<\/p>\n\n\n\n
Excercises<\/h4>\n\n\n\n
12.1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 \u00b0C to 77 \u00b0C. 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 \u00d7 104<\/sup> J\/g?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Water is flowing at a rate of 3.0 litre\/min. 12.2. What amount of heat must be supplied to 2.0 \u00d7 10\u20132<\/sup> kg of nitrogen (at room temperature) to raise its temperature by 45 \u00b0C at constant pressure? (Molecular mass of N2<\/sub> = 28; R <\/em>= 8.3 J mol\u20131<\/sup> K\u20131<\/sup>.)<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Mass of nitrogen, m<\/em> = 2.0 \u00d7 10\u20132<\/sup> kg = 20 g 12.3. Explain why Answer<\/strong><\/p>\n\n\n\n (a) In thermal contact, heat flows from the body at higher temprature to the body at lower temprature till tempratures becomes equal. The final temprature can be the mean temprature (T<\/em>1<\/sub> + T<\/em>2<\/sub>)\/2only when thermal capicities of the two bodies are equal.<\/p>\n\n\n\n (b) This is bcause heat absorbed by a substance is directly proportional to the specific heat of the substance.<\/p>\n\n\n\n (c) During driving, the temprature of air inside the tyre increases due to moion. Accordingto Charle’s law, P<\/em> \u221d T. <\/em>Therefore, air pressure inside the tyre increases.<\/p>\n\n\n\n (d) This is because in a harbour town, the relative humidity is more than in a desert town. hence, the climate of a harbour town is without extremes of hot and cold.<\/p>\n\n\n\n NCERT 11th Physics Chapter 12, class 11 Physics chapter 12 solutions<\/p>\n\n\n\n 12.4. 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?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n 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. 12.5. In changing the state of a gas adiabatically from an equilibrium state A <\/em>to another equilibrium state B<\/em>, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A <\/em>to B <\/em>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)<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n The work done (W<\/em>) on the system while the gas changes from state A<\/em> to state B<\/em> is 22.3 J. 12.6. Two cylinders A <\/em>and B <\/em>of equal capacity are connected to each other via a stopcock. A <\/em>contains a gas at standard temperature and pressure. B <\/em>is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: Answer<\/strong><\/p>\n\n\n\n (a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atmosphere.<\/p>\n\n\n\n (b) There will be no change in the internal energy of the gas as no work is done on\/by the gas.<\/p>\n\n\n\n (c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all. 12.7. A steam engine delivers 5.4\u00d7108 <\/sup>J of work per minute and services 3.6 \u00d7 109 <\/sup>J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Work done by the steam engine per minute, W<\/em> = 5.4 \u00d7 108<\/sup> J NCERT 11th Physics Chapter 12, class 11 Physics chapter 12 solutions<\/p>\n\n\n\n 12.8.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?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Heat is supplied to the system at a rate of 100 W. NCERT 11th Physics Chapter 12, class 11 Physics chapter 12 solutions<\/p>\n\n\n\n 12.9. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13).<\/strong><\/p>\n\n\n\n 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.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Total work done by the gas from D to E to F = Area of \u0394DEF 12.10. A refrigerator is to maintain eatables kept inside at 9\u00b0C. If room temperature is 36\u00b0 C, calculate the coefficient of performance.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Temperature inside the refrigerator, T<\/em>1<\/sub> = 9\u00b0C = 282 K NCERT 11th Physics Chapter 12, class 11 Physics chapter 12 solutions<\/p>\n\n\n\n NCERT Solutions for 11th Class Physics: Chapter 12-Thermodynamics<\/p>\n\n\n\n Download PDF<\/strong>: NCERT Solutions for 11th Class Physics: Chapter 12-Thermodynamics PDF<\/a><\/p>\n\n\n\n
The geyser heats the water, raising the temperature from 27\u00b0C to 77\u00b0C.
Initial temperature, T<\/em>1<\/sub> = 27\u00b0C
Final temperature, T<\/em>2<\/sub> = 77\u00b0C
\u2234Rise in temperature, \u0394T<\/em> = T<\/em>2<\/sub> \u2013 T<\/em>1<\/sub>
= 77 \u2013 27 = 50\u00b0C
Heat of combustion = 4 \u00d7 104<\/sup> J\/g
Specific heat of water, c <\/em>= 4.2 J g\u20131 <\/sup>\u00b0C\u20131<\/sup>
Mass of flowing water, m<\/em> = 3.0 litre\/min = 3000 g\/min
Total heat used, \u0394Q<\/em> = mc<\/em> \u0394T<\/em>
= 3000 \u00d7 4.2 \u00d7 50
= 6.3 \u00d7 105 <\/sup>J\/min
\u2234 Rate of consumption = 6.3 \u00d7 105<\/sup> \/ (4 \u00d7 104<\/sup>) = 15.75 g\/min.<\/p>\n\n\n\n
Rise in temperature, \u0394T<\/em> = 45\u00b0C
Molecular mass of N2<\/sub>, M<\/em> = 28
Universal gas constant, R<\/em> = 8.3 J mol\u20131<\/sup> K\u20131<\/sup>
Number of moles, n = m\/M
= (2 \u00d7 10-2<\/sup> \u00d7 103<\/sup>) \/ 28
= 0.714
Molar specific heat at constant pressure for nitrogen, Cp<\/sub> = (7\/2)R
= (7\/2) \u00d7 8.3
= 29.05 J mol-1<\/sup> K-1<\/sup>
The total amount of heat to be supplied is given by the relation:
\u0394Q = nC<\/em>P <\/em><\/sub>\u0394T<\/em>
= 0.714 \u00d7 29.05 \u00d7 45
= 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J.<\/p>\n\n\n\n
(a) Two bodies at different temperatures T<\/em>1<\/sub> and T<\/em>2<\/sub> if brought in thermal contact do not necessarily settle to the mean temperature (T<\/em>1<\/sub> + T<\/em>2<\/sub>)\/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.<\/strong><\/p>\n\n\n\n
Initial pressure inside the cylinder = P<\/em>1<\/sub>
Final pressure inside the cylinder = P<\/em>2<\/sub>
Initial volume inside the cylinder = V<\/em>1<\/sub>
Final volume inside the cylinder = V<\/em>2<\/sub>
Ratio of specific heats, \u03b3 = 1.4
For an adiabatic process, we have:
P<\/em>1<\/sub>V<\/em>1<\/sub>\u03b3<\/sup> =P<\/em>2<\/sub>V<\/em>2<\/sub>\u03b3<\/sup>
The final volume is compressed to half of its initial volume.
\u2234 V<\/em>2<\/sub> = V<\/em>1<\/sub>\/2
P<\/em>1<\/sub>V<\/em>1<\/sub>\u03b3<\/sup> = P<\/em>2<\/sub>(V<\/em>1<\/sub>\/2)\u03b3<\/sup>
P<\/em>2<\/sub>\/P<\/em>1<\/sub> =V<\/em>1<\/sub>\u03b3<\/sup> \/ (V<\/em>1<\/sub>\/2)\u03b3<\/sup>
= 2\u03b3<\/sup> = 21.4<\/sup> = 2.639
Hence, the pressure increases by a factor of 2.639.<\/p>\n\n\n\n
This is an adiabatic process. Hence, change in heat is zero.
\u2234 \u0394Q<\/em> = 0
\u0394W<\/em> = \u201322.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
\u0394Q<\/em> = \u0394U<\/em> + \u0394W<\/em>
Where,
\u0394U<\/em> = Change in the internal energy of the gas
\u2234 \u0394U<\/em> = \u0394Q<\/em> \u2013 \u0394W<\/em> = \u2013 (\u2013 22.3 J)
\u0394U <\/em>= + 22.3 J
When the gas goes from state A<\/em> to state B<\/em> via a process, the net heat absorbed by the system is:
\u0394Q<\/em> = 9.35 cal = 9.35 \u00d7 4.19 = 39.1765 J
Heat absorbed, \u0394Q<\/em> = \u0394U <\/em>+ \u0394Q<\/em>
\u2234\u0394W<\/em> = \u0394Q<\/em> \u2013 \u0394U<\/em>
= 39.1765 \u2013 22.3
= 16.8765 J
Therefore, 16.88 J of work is done by the system.<\/p>\n\n\n\n
(a) What is the final pressure of the gas in A <\/em>and B?<\/em>
(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 <\/em>surface?<\/strong><\/p>\n\n\n\n
(d) No, because the process called free expansion is rapid and cannot be controlled. the intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.<\/p>\n\n\n\n
Heat supplied from the boiler, H<\/em> = 3.6 \u00d7 109<\/sup> J
Efficiency of the engine = Output energy \/ Input energy
\u2234 \u03b7 = W<\/em> \/ H<\/em> = 5.4 \u00d7 108<\/sup> \/ (3.6 \u00d7 109<\/sup>) = 0.15
Hence, the percentage efficiency of the engine is 15 %.
Amount of heat wasted = 3.6 \u00d7 109<\/sup> \u2013 5.4 \u00d7 108<\/sup>
= 30.6 \u00d7 108<\/sup> = 3.06 \u00d7 109<\/sup> J
Therefore, the amount of heat wasted per minute is 3.06 \u00d7 109<\/sup> J.<\/p>\n\n\n\n
\u2234Heat supplied, Q<\/em> = 100 J\/s
The system performs at a rate of 75 J\/s.
\u2234Work done, W<\/em> = 75 J\/s
From the first law of thermodynamics, we have:
Q<\/em> = U<\/em> + W<\/em>
Where,
U<\/em> = Internal energy
\u2234U<\/em> = Q \u2013 W<\/em>
= 100 \u2013 75
= 25 J\/s
= 25 W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/03\/fig-1-chapter-12-class-11th.png\")
<\/figure>\n\n\n\n
Area of \u0394DEF = (1\/2) DE \u00d7 EF
Where,
DF = Change in pressure
= 600 N\/m2<\/sup> \u2013 300 N\/m2<\/sup>
= 300 N\/m2<\/sup>
FE = Change in volume
= 5.0 m3<\/sup> \u2013 2.0 m3<\/sup>
= 3.0 m3<\/sup>
Area of \u0394DEF = (1\/2) \u00d7 300 \u00d7 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.<\/p>\n\n\n\n
Room temperature, T<\/em>2<\/sub> = 36\u00b0C = 309 K
Coefficient of performance = T1<\/sub> \/ (T2<\/sub> – T1<\/sub>)
= 282 \/ (309 – 282)
= 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.<\/p>\n\n\n\nNCERT Solutions for 11th Class Physics: Chapter 12: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise NCERT Solutions for Class 11 Physics<\/strong>:<\/h2>\n\n\n\n
\n