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.
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
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
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
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?
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.
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?
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)
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?
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
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?
Answer: Statements (b) and (c) are incorrect regarding the first law of thermodynamics.
Below mentioned are some important thermodynamics formulae related to the laws of the thermodynamics for the quick revision.
Zeroth Law of Thermodynamics | (TA = TB) ˄ (TB = TC) → (TA = TC) (systems in thermal equilibrium) |
First Law of Thermodynamics | ∆U = ∆Q + ∆W |
second Law of Thermodynamics | ∆S≥U |
Third Law of Thermodynamics | S = SSTRUCTURAL + CT |
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.
*The article might have information for the previous academic years, please refer the official website of the exam.