Important Questions Class 11 Physics Chapter 12

Important Questions Class 11 Physics Chapter 12

Important Questions for CBSE Class 11 Physics Chapter 12 – Thermodynamics

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CBSE Class 11 Physics Chapter 12 Important Questions – Free Download

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Study Important Questions for Class 11 Physics Chapter 12 – Thermodynamics

Very Short Questions and Answers                                                              

[1 or 2 marks]

Q1.) What is the ratio of slopes of P-V graphs of the adiabatic and isothermal processes?

Ans. Let the slope of the P-V graph be  dPdV.

We know that for an isothermal process, (PV = constant)

So, dPdV=PV(1).

For an adiabatic process (PVY=constant)

dPdV=YPV(2)

Divide equation (2) by (1)

So, the ratio of adiabatic slope to the isothermal slope is Y.

Q2.) Why do animals curl into a ball after they are cold?

Ans: The surface area reduces as the animals curl because the energy radiated changes directly with the surface area, and the heat loss due to radiation is also reduced.

Q3.) Why do you feel warmer when clouds cover the sky than when the sky is cloudless on a winter night?

Ans: During the day, the earth absorbs heat and emits it at night. The heat discharged by Earth is reversed when the sky is covered by clouds, and Earth turns warmer; when the sky is clear, the heat radiated by Earth is released into space.

Q4.) If a body has infinite heat capacity. What does it imply?

Ans: Infinite heat capacity implies the ability of a substance to keep up its temperature despite how much heat it receives or loses.

Q5.) A Carnot engine absorbs 6105Cal at 227℃. Evaluate the work done per cycle by the engine when it sinks at 127℃ .

Ans. As per the data given,

Heat absorbed =Q1=6105Cal

Initial temperature =T1=227℃=227+273=500K

Final temperature =T2=127℃=127+273=400K

As, for Carnot engine,

Q2Q1=T2T1

Q2=Q1T2T1

Q2=4005006105

Q2=4.8105Cal

Q2=Final heat emitted

As W=Q1-Q2

=6105Cal-4.8105Cal

=1.2105Cal

Work =w=1.251054.2 J

Dore =5.04105J

Q6.) Calculate the temperature in Kelvin at which a perfectly black body radiates at the rate of  5.67w/cm2.

Ans. Given rate,

E=5.67w/cm2; E=energy radiated

=5.67107erg/s/cm2

=Stefan’s constant=5.6710-5erg/s/cm2/K4, from Stefan’s law

E=T4

T=E14

T=5.671075.6710-514

T=101214=103=1000K

Q7.) Why does the tile floor feel colder than the wooden floor when both floor materials are at a similar temperature?

Ans: Since a tile is a good conductor of heat when compared with wood, the heat from our feet transferredto the wood does not disperse rapidly. Therefore, the wood heats up to the temperature of the foot as it comes in contact with its surface, whereas tile transfers heat away quickly and can thus absorb more heat from our feet, lowering its surface temperature. Thus, a tile floor feels colder than a wooden floor when both floor materials are at similar temperatures.

Q8.) A steam engine delivers 5.4108J of work per minute and services 3.6109J of heat per minute from its boiler. What is the efficiency of the engine, and how much heat is wasted per minute?

Ans. As per the data given,

Work done by the steam engine per minute, W=5.4108J

Heat supplied from the boiler, H=3.6109J

Efficiency of the engine =Output energyInput energy

∴=WH=5.41083.6109=0.15

Hence, the percentage efficiency of the engine is 15%.

Amount of heat wasted =3.6109-5.4108

=30.6108=3.6109J

Therefore, the amount of heat wasted per minute is 3.6109J .

Q9.) A thermometer has the wrong calibration. It reads the melting point of ice as -10℃ . It reads 60℃ in the place of 50℃ . What is the temperature of the boiling point of water on the scale? 

Ans. Given,

Lower fixed point on the wrong scale, =-10℃

Let ‘n’ = number of the division between upper and lower fixed points on this scale. And Q= reading on his scale,

Then C-O100=Q-(-10)n

Now, C= Incorrect Reading =60℃

2= Correct Reading =50℃

So, 50-0100=60-(-10)n

50100=70n

n=7010050

n=140

Now, C-O100=Q-(-10)140

On, the Celsius scale, the boiling point of water is 100℃.

So, 100-0100=Q+10140

Q=140-10

Q=130℃

Short Questions and Answers                                                                           

[3 or 4 Marks]

Q1.) 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. If the gas is compressed to half its original volume, what factors increase the pressure of the gas?

Ans: The cylinder is entirely closed off from the rest of the world which leads to no heat movement between the system (cylinder) and the environment. Hence, 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, Y=1.4

For an adiabatic process, we get,

P1V1y=P1V2y

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

∴ V2=V12

P1(V1)y=P2V12y

P2P1=V1yV12y=2y=21.4=2.639

Hence, the pressure increases by a factor of 2.639.

Q2.) Calculate the difference in the efficiency of a Carnot energy working between 

a.) 400K and 350K and b.) 350K and 300K.

Ans. The efficiency of heat engine =n=1-T2T1

T2= Final temperature

T1= Initial temperature

a.) 400K and 350K

T2=350K, and T1=400K

n=1-350400

=50400

n1=18 or  100%8=12.5%

b.) 350K and 300K

T2=300K, and T1=350K

n=1-300350

=50350

n1=17=100%7=14.3%

Change in efficiency =n2-n1=14.3%-12.5%=1.8%.

Q3.) Calculate the amount of heat necessary to raise the temperature of 2 moles of HE gases from 20℃ to 50℃ using: 

a.) Constant-Volume Process 

b.) Constant Pressure Process

Here for, He; Cv=1.5R and CP=2.94R

Ans. a) The amount of heat required for the constant-volume process is:

Qv=nCVT

Here, n= 2 moles, CV=1.5R=1.58.314 J/mol/℃

T2= Final temperature

T1= Initial temperature

T=T2-T1

=50-20

=30℃

QV=21.58.31430

QV=748 J

b.) The amount of heat required for the constant-pressure process is:

Qp=nCPT

Here, n=2 moles, CP=2.49R=2.498.314

T=30

QP=22.498.31430QP=1242 J

As the temperature increase is similar, i.e., 748 J, the change in internal energy is similar in both scenarios. However, extra heat is obtained in the constant-pressure process, which is further used in gas expansion.

Q4.) What amount of heat must be supplied to 2.010-2kg of nitrogen (at room temperature) to raise its temperature by 45℃ at constant pressure? (Molecular mass of N2=28; R=8.3 J mol-1K-1.)

Ans: Given data,

Mass of nitrogen, m=2.010-2kg=20g

Rise in temperature, T=45℃

Molecular mass of N2, M=28.

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

Number of moles, n=mM

=2.010-210328=0.714

Molar specific heat at constant pressure for nitrogen,

CP=72R

=728.3

=29.05 J mol-1K-1

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

Q=nCPT

=0.71429.0545

=933.38 J

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

Q5.) One kilogram molecule of a gas at 400k expands isothermally until its volume is doubled. Find the amount of work done and heat produced.

Ans. Let us consider,

Initial volume, V1=V

Final volume, V2=2V

Initial temperature T=400k

Find temperature =400k (∴process is isothermal)

Gas constant, R=8.3kj/mole/K=8.310-3 J/mole/k done during is thermal process

W=2.3026RT log 10V1V2

W=2.30268.310-3400log102VV

W=2.30268.310-3400log102

W=2.3016

If H is the amount of heat produced than,

H=WJ=2.30164.2=0.548Cal

Long Questions and Answers                                                                               

[5 or 6 Marks]

Q1.) State an equation for the work done in an isothermal expansion.

Ans: Let us assume that one gramme mole of any ideal gas has certain properties like volume, pressure, and temperature.

Let the gas expand to its full volume when the pressure falls and the temperature remains unchanged.

If A= Area of the cross-section of piston

Force = Pressure Area

Force=PressureArea

F=PA

If we assume that piston moves a displacement dx,

the work done: dw=Fdx

dx=PAdxdw=pdv

Total work done in increasing the volume from V1 to V2.

So W=2.3026RT log 10P1P2.