Thermodynamics Class 11 Physics Exam Questions

Exam Questions Class 11

Please refer to Thermodynamics Class 11 Physics Exam Questions provided below. These questions and answers for Class 11 Physics have been designed based on the past trend of questions and important topics in your class 11 Physics books. You should go through all Class 11 Physics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 11 Physics Exam Questions Thermodynamics

Class 11 Physics students should read and understand the important questions and answers provided below for Thermodynamics which will help them to understand all important and difficult topics.

Very Short Answer type Questions

Question. What is the amount of work done in the Cyclic process?
Answer : It is numerically equal to the area of the cyclic process.

Question. Which Thermodynamical variable is defined by the first law of thermodynamics?
Answer : Internal energy.

Question. What is the nature of P-V diagram for isobaric and isochoric process? 
Answer : The P-V diagram for an isobaric process is a straight line parrel to the volume axis while that
For an isochoric process is a straight line parallel to pressure axis.

Question. Out of the parameters- temperature, pressure,work and volume, which parameter does not Characterize the thermodynamics state of matter?
Answer : Work

Question. Can we increase the temperature of gas without supplying heat to it?
Answer : Yes, the temperature of gas can be by compressing the gas under Adiabatic condition.

Question. Can the whole of work be converted into heat?
Answer : Yes ,Through friction.

Question. On what factors does the efficiency of Carnot engine depends?
Answer : Temperature of the source of heat and sink.

Question. Which thermodynamic variable is defined by Zeroth law of thermodynamics?
Answer : Temperature

Question. Why does the gas get heated on compression?
Answer : Because the work done in compressing the gas increases the internal energy of the gas.

Question. In a Carnot engine, temperature of the sink is increased. What will happen to its efficiency?
Answer : We know ƞ = 1 – T2/T1 
On increasing the temperature of the sink (T2) , the efficiency of the Carnot engine will decrease

Short Answer type Questions

Question. Air pressure in a car increases during driving. Explain Why? 
Answer : During driving as a result of the friction between the tyre and road ,the temperature of
The tyre and the air inside it increases. Since volume of the tyre does not change, due to increase in temperature ,pressure of the increases (due to pressure law ) .

Question. If hot air rises , why is it cooler at the top of mountain than near the sea level ?
Answer : Since atmospheric pressure decreases with height, pressure at the top of the mountain is lesser. When the hot air rises up,it suffer adiabatic expansion at the top of the mountain.For an adiabatic change,first law of thermodynamics may be express as
                           dU + dW = 0            (dQ = 0)
                          dW = -dU
Therefore work done by the air in rising up (dW = +ve ) result in decrease in the internal Energy of the air (dU = -ve) and hence a fall in the temperature.

Question. Why cannot the Carnot’s engine be realised in practice?
Answer : Because of the following reasons
(i) The main difficulty is that the cylinder should come in contact with the source,sink and stand again and again over a complete cycle which is very difficult to achieve in practice.
(ii) The working substance should be an ideal gas however no gas fulfils the ideal gas behaviour.
(iii) A cylinder with a perfectly frictionless piston cannot be realised

Question. The efficiency of a heat engine cannot be 100%. Explain why ?
Answer : The efficiency of heat engine ƞ = 1 – (T2/T1)  
The efficiency will be 100% or 1, if T2 = 0 K. 
Since the temperature of 0 K cannot be reached, a heat engine cannot have 100% efficiency.

Question. Can water be boiled without heating ?
Answer : Yes, water can be boil without heating. This is done by increasing the pressure on the surface of water inside a closed insulated vessel. By doing so, the boiling point of the water decreases to the room temperature and hence starts boiling.

Question. In an effort to cool a kitchen during summer, the refrigerator door is left open and the kitchen door and windows are closed. Will it make the room cooler ?
Answer : The refrigerator draws some heat from the air in front of it. The compressor has to do some Mechanical work to draw heat from the air at lower temperature. The heat drawn from the air together with the work done by the compressor in drawing it, is rejected by the refrigerator with the help of the radiator provided at the back to the air. IT follows that in each cycle, the amount of heat rejected to the air at the back of the refrigerator will be greater than that is drawn from the air in front of it. Therefore temperature of the room will increase and make hotter.

Question. A slab of ice at 273K and at atmospheric pressure melt.(a) What is the nature of work done on The ice water system by the atmosphere?(b) What happen to the internal energy of the ice- Water system?
Answer : (a) The volume of the ice decreases on melting. Hence the work done by the atmosphere on The ice – water system is positive in nature.
(b) Since heat is absorbed by the ice during melting, the internal energy of the icewater system increases.

Question. What are the limitations of the first law of thermodynamics ?
Answer : The limitations are — (i) It does not tells us the directions of heat transfer
(ii) it does not tell us how much of the heat is converted into work.
(iii) it does not tell us under what conditions heat is converted into work.

Question. What happen to the internal energy of a gas during (i) isothermal expansion
(ii) adiabatic Expansion?
Answer : In isothermal expansion ,temperature remains constant.Therefore internal energy which is a function of temperature will remain constant.
(ii) for adiabatic change dQ = 0 and hence first law of thermodynamics becomes
0 = dU + dW
dW = – dU
During expansion, work is done by the gas i.e. dW is positive. Hence ,dU must be negative.
Thus ,in an adiabatic expansion , the internal energy of the system will decrease.

Question. Why is the conversion of heat into work not possible without a sink at lower temperature? 
Answer :For converting heat energy into work continuosly, a part of the heat energy absorbed from the source has to be rejected.The heat energy can be rejected only if there is a body whose
Temperature is less than that of the source. This body at lower temperature is called sink.

Long Answer type Questions

Question. If at 50oC and 75 cm of mercury pressure, a definite mass of gas is compressed (i) slowly (iii) suddenly, then what will be the final pressure and temperature of the gas in each case, if the final volume is one fourth of the initial volume? Given γ = 1.5
Answer : (I) When the gas is compressed slowly, the change is isothermal.
Therefore P2 V2 = P1 V1
P2 = P1V1/V2
=(75 x V1/V1) x 4 = 300 cm of mercury
Temperature remains constant at 50oC
(ii) When the gas is compressed suddenly, the change is adiabatic
As per P2 V2
γ = P1 V1γ
P2 = P1(V1/V2)γ
= 75 x (4)1.5 =600 cm of Hg
Also T2 V2¥-1 = T1 V1¥-1
T2 = T1 (V1/V2)¥-1 = 323 x (4)(1.5- 1) = 646K
= 646-273 =373 oC

Question. Calculate the fall in temperature when a gas initially at 72oC is expanded suddenly to eight times its original volume. Given γ = 5/3. 3
Answer : Let V1 = x cc V2 = 8x cc
T1 = 273+72 = 345 K ¥ = 5/3 , T2 = ?
Using the relation T1 V1¥-1 = T2 V2¥-1
Therefore T2 = T1 (V1/V2¥-1
                   = 345 x (1/8) 2/3
Taking log of both sides, we get
Log T2 = log 345 – 2/3 log 8
= 2.5378 – 2/3(0.9031)
= 2.5378 -0.6020 = 1.9358
Or T2 = 86.26 K
Therefore the fall in temperature = 345 -86.26 258.74 K

Question. A Carnot engine whose source temperature is at 400K takes 100 Kcal of heat at this temperature in each cycle and gives 70 Kcal to the sink. Calculate (i) the temperature of the sink
(ii) the efficiency of the engine.
Answer : Here T1 = 400K , Q1 = 100 Kcal , Q2 = 70 Kcal
T2 = ? ,¶ = ?
(i) Q1/Q2 = T1/T2
Or T2 = (Q2/Q1) T1
Or T2 = 70/100 x 400
Or T2 = 280 K
(ii) ƞ = 1- T2/T1
= 1 – 280/400
= 1- 0.7 = 0.3
Or % of ƞ= 0.3 x 100 =30 %

Question. Assuming a domestic refrigerator as a reversible heat engine working between melting point Of ice and the room temperature at 27oC, calculate the energy in joule that must be supplied to freeze 1Kg of water at 0oC
Answer : Here T1 = 27+273 =300K , T2 = 0 +273 = 273
Mass of water to be freezed , M = 1 Kg = 1000g
Amount of heat that should be removed to freeze the water
Q2 = ML =1000X 80 cal
= 1000X80 X4.2 =3.36 x 105 J
Now Q1 = (T1/T2 )X Q2 = (300/273)X3.36 x 105 = 3.692 x 105 J
Therefore energy supplied to freeze the water
W =Q1 – Q2 = 3.693×105 – 3.36 x 105
= 3.32 x 105 J

Question. Two engines A and B have their sources at 400K and 350 K and sink at 350K and 300K Respectively. Which engine is more efficient and by how much?
Answer : For engine A T1 =400K , T2 = 350 K
Efficiency ƞA =1 – T2/T1
=1-350/400 =1/8
% of ƞA = 1/8 X 100 = 12.5%
For Engine B T1 = 350 K , T2 = 300 K
Efficiency ƞB = 1 – T2/T1
= 1- 300/350 = 1/7
% of ƞB = 1/7 x 100 =14.3%
Since ƞB > ƞA so engine A is much more efficient than engine B by (14.3% – 12.5% ) = 1.8%

Question. The temperature T1 and T2 of two heat reserviour in an ideal carnot engine are 1500oC and
500oC. Which of this increasing the temperature T1 by 100oC or decreasing T2 by 100oC would result in greater improvement of the efficiency of the engine?
Ans: Using ¶ =1 – T2/T1 =(T1– T2)/T1
(1) increasing T1 by 100oC          ¶1 = (1600 -500)/(1600 + 273 )
                                                     = 1100/1873 = 59 %
(ii) Decreasing T2 by 1000C        ¶2 = 1500 –(500-100)/(1500 +273 )
                                                     =1100/1773 = 67%
Therefore decreasing T2 by 100oC results in greater improvement of efficiency.

Question. A refrigerator freezes 5Kg of water at 0oC into ice at 0oC in a time interval of 20 minutes. Assume that the room temperature is 20oC, calculate the minimum power needed to accomplish it.
Answer : Amount of heat required to convert water into ice at 0oC,
Q2 = mL = (5Kg) x (80)Kcal/Kg
= 400 Kcal
Now T1 =20oC =273+20 = 293K
T2 = 0oC 0+273 =273 K
We know that Q2/W = T2/(T1 – T2 )
Or W = Q2 x (T1 – T2)/T2
= 400x(293- 273 )/273
= 29.3 Kcal = 29.3×4.2 x 103J
=123 x 103 J
Time t = 20 min =20×60 =1200s
Power needed P = W/t =123 x 103/1200
=102.5 W