Thermodynamics is a crucial portion in JEE Mains Physics syllabus and usually covers 8-16 marks of the entire paper with 2-4 questions on the topic. In order to secure a meritorious position in JEE Main 2021, the candidates are required to prepare the section on Thermodynamics with the help of the previous year question papers. The thorough understanding of the topic might help the candidates to qualify a better position. 

Physics is usually the most difficult section in JEE Main Question Paper and Thermodynamics is one of the topics that is always covered. Therefore, JEE Main Test Series on Thermodynamics can be helpful in better understanding and preparation of JEE Main Physics section. 

Thermodynamics is a branch of science that deals with the study of the relationship between heat, work, temperature and energy. Literally, meaning, thermodynamics deals with the transfer of energy from one source to another, transforming into different forms. This branch of physics deals with the impacts of thermal energy and its transformations into varying matters. 

Some of the important topics covered in Thermodynamics include Thermal equilibrium, Zeroth Law of Thermodynamics, Concept of temperature. heat, work and internal energy, First Law of Thermodynamics, Second Law of Thermodynamics, reversible and irreversible processes, Carnot engine and its efficiency. Check JEE Main Physics Syllabus

For reference, a few of previous year questions and solutions are given below on Thermodynamics which are more frequently asked in JEE Main exam every year.

---

JEE Main Thermodynamics Questions and Solutions

Majorly all the questions from Thermodynamics in JEE Main paper carry 3 mark each. Mentioned below some of the questions from JEE Main previous year papers on Thermodynamics: 

Q1. Even a Carnot engine cannot give 100% efficiency because we cannot

1. prevent radiation
2. find ideal sources
3. reach absolute zero temperature
4. eliminate friction.

Answer: (c): We cannot reach absolute zero temperature

Q2. “Heat cannot by itself flow from a body at a lower temperature to a body at a higher temperature” is a statement or consequence of

1. The second law of thermodynamics
2. conservation of momentum
3. conservation of mass
4. The first law of thermodynamics

Answer: (a) Second law of thermodynamics

Q3. A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6 When T2 is lowered by 62 K, its efficiency increases to ⅓. Then T1 and T2 are, respectively

1. 372 K and 310 K
2. 372 K and 330 K
3. 330 K and 268 K
4. 310 K and 248 K

Solution

The efficiency of Carnot engine,

η =1- (T2/T1)

η = ⅙

T2/T1 = ⅚

T1 = 6T2/5 ————–(1)

As per the question, when T2 is lowered by 62 K, then its efficiency becomes 1/3

⅓ = [1 – (T2 -62/T1)] [T2 -62/T1 ] = 1-(⅓) (Using equation (1))

5(T2 -62)/6T2= ⅔

5T2 – 310 = 4T2 ⇒ T2= 310 K

From equation (1) T1 = (6 x 310)/5 = 372 K

Answer: (a) 372 K and 310 K

Q4. Which of the following statements is correct for any thermodynamic system?

1. The internal energy changes in all processes
2. Internal energy and entropy are state functions
3. The change in entropy can never be zero
4. The work done in an adiabatic process is always zero

Answer: (b) Internal energy and entropy are state functions

Q5. From the following statements, concerning ideal gas at any given temperature T, select the correct.

1. (a)The coefficient of volume expansion at constant pressure is the same for all ideal gases
2. (b)The average translational kinetic energy per molecule of oxygen gas is 3kT, k being Boltzmann constant
3. (c)The mean-free path of molecules increases with an increase in the pressure
4. (d)In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different

Solution

γ = dV/(V0 x dT) at a constant temperature

γ = 1/V0(dV/dT)p since PV = RT

PdV = RdT or (dV/dT) = R/P0

Therefore, γ = (1/V0)(R/P0) = R/RT0

γ = 1/T0

γ = 1/273

Answer: (a)The coefficient of volume expansion at constant pressure is the same for all ideal gases

Q6. Which of the following parameters does not characterize the thermodynamic state of matter?

1. temperature
2. pressure
3. work
4. volume

Answer: (c): The work does not characterize the thermodynamic state of matter

Q7.100 g of water is heated from 30 °C to 50 °C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J kg–1 K–1)

1. 4.2 kJ
2. 8.4 kJ
3. 84 kJ
4. 2.1 kJ

Solution

ΔQ = msΔT

Here m = 100 g = 100 x10-3 Kg

S = 4184 J kg-1K-1 and ΔT = (50 – 30) = 20 0C

ΔQ = 100 x 10-3 x 4184 x 20 = 8.4 x 103 J

ΔQ = ΔU + ΔW

Change in internal energy

ΔU = ΔQ = 8.4 x 103 J = 8.4 kJ

Answer: (b) 8.4 kJ

Q8. Calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 °C and it is defined under which of the following conditions?

1. From 14.5 °C to 15.5 °C at 760 mm of Hg
2. From 98.5 °C to 99.5 °C at 760 mm of Hg
3. From 13.5 °C to 14.5 °C at 76 mm of Hg
4. From 3.5 °C to 4.5 °C at 76 mm of Hg

Solution

1 calorie is the amount of heat required to raise the temperature of 1gm of water from 14.5 0C to 15.5 0C at 760 mm of Hg

Answer: (a) From 14.5 °C to 15.5 °C at 760 mm of Hg

Q9. An ideal gas heat engine is operating between 227 °C and 127 °C. It absorbs 104J of heat at the higher temperature. The amount of heat converted into work is

1. 2000 J
2. 4000 J
3. 8000 J
4. 5600 J

Solution

η =1- (T2/T1)

η =1- (127 + 273)/(227 + 273) = 1 – (400/500) = ⅕

W = ηQ1 = ⅕ x 104 = 2000 J

Answer: (a) 2000 J

Q10. Which of the following is incorrect regarding the first law of thermodynamics?

1. It introduces the concept of internal energy
2. It introduces the concept of entropy
3. It is not applicable to any cyclic process
4. It is a restatement of the principle of conservation of energy

Answer: Statements (b) and (c) are incorrect regarding the first law of thermodynamics.

---

JEE Main Thermodynamics Quick Formulas

Below mentioned are some important thermodynamics formulae related to the laws of the thermodynamics for the quick revision.

Above mentioned are some of the important formulas that will help you crack questions from thermodynamics in JEE Main Physics section. Since Physics is usually the most difficult section to crack, JEE Main test series on Thermodynamics can help students in a quick revision. Apart from these, students must often practice solving JEE Main Sample Papers that will help in the preparations as well.