SaraNextGen.Com
Updated By SaraNextGen
On March 11, 2024, 11:35 AM

INTRODUCTION - Chapter 12 Thermodynamics class 11 ncert solutions Physics - SaraNextGen [2024]


Question 12.1:

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?

Answer:

Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water, raising the temperature from 27°C to 77°C.

Initial temperature, T1 = 27°C

Final temperature, T2 = 77°C

∴Rise in temperature, ΔT = T2 – T1

= 77 – 27= 50°C

Heat of combustion = 4 × 104 J/g

Specific heat of water, c = 4.2 J g–1 °C–1

Mass of flowing water, m = 3.0 litre/min = 3000 g/min

Total heat used, ΔQ = mc ΔT

= 3000 × 4.2 × 50

= 6.3 × 10J/min

∴Rate of consumption =https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4667/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_m3f863c4b.gif  = 15.75 g/min

Question 12.2:

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; = 8.3 J mol–1 K–1.)

Answer:

Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g

Rise in temperature, ΔT = 45°C

Molecular mass of N2M = 28

Universal gas constant, R = 8.3 J mol–1 K–1

Number of moles,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4668/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_m50cdfa06.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4668/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_5d1cf111.gif

Molar specific heat at constant pressure for nitrogen,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4668/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_3129db53.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4668/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_m193a0c5e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4668/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_6137b3f9.gif

The total amount of heat to be supplied is given by the relation:

ΔQ = nCΔT

= 0.714 × 29.05 × 45

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

Question 12.3:

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.

Answer:

(a) When two bodies at different temperatures T1 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 high 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.

Question 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?

Answer:

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.

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 heats, γ = 1.4

For an adiabatic process, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4670/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_m473af606.gif

The final volume is compressed to half of its initial volume.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4670/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_m4b5b7692.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/190/4670/NS_14-10-08_Sravana_11_Physics_12_10_NRJ_LVN_html_30709b38.gif

Hence, the pressure increases by a factor of 2.639.

Question 12.5:

In changing the state of a gas adiabatically from an equilibrium state 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 to 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)

Answer:

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. Hence, change in heat is zero.

∴ ΔQ = 0

ΔW = –22.3 J (Since the work is done on the system)

From the first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where,

ΔU = Change in the internal energy of the gas

∴ ΔU = ΔQ – ΔW = – (– 22.3 J)

Δ= + 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 × 4.19 = 39.1765 J

Heat absorbed, ΔQ = Δ+ ΔQ

∴ΔW = ΔQ – ΔU

= 39.1765 – 22.3

= 16.8765 J

Therefore, 16.88 J of work is done by the system.

Question 12.6:

Two cylinders and of equal capacity are connected to each other via a stopcock. contains a gas at standard temperature and pressure. 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 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?

Answer:

Answer:

(a) 0.5 atm

(b) Zero

(c) Zero

(d) No

Explanation:

(a) The volume available to the gas is doubled as soon as the stopcock between cylinders A and B is opened. Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5 atm.

(b) The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

(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.

(d) The given process is a case of free expansion. It is rapid and 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 PVT surface of the system.

Also Read : Page-No-317:-Chapter-12-Thermodynamics-class-11-ncert-solutions-Physics

SaraNextGen