Thermodynamics is the study of how heat or any other form of energy flows into and out of a system as it goes through a physical or chemical transformation. While examining and evaluating the flow of energy into and out of the system, it will be important to evaluate changes in specific system parts. These parameters include the system’s temperature, pressure, volume, and concentration. Measuring changes in these qualities from the start to the end state can reveal information about energy and related quantities like heat and work. In this article we will look at some of the tricky questions for Class 11 and Class 12 from Thermodynamics.

Laws of Thermodynamics

There are three laws of thermodynamics.The first law is the law of conservation of energy which states that energy can neither be created nor destroyed. It can only be converted from one form to another.

The second law states that for a spontaneous process, the entropy of the universe increases. Or in other words, heat at a given temperature cannot be converted entirely into work. Consequently, the entropy of a closed system, or heat energy per unit temperature, increases over time toward some maximum value.

The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches zero. This allows an absolute scale for entropy to be established,, determining the degree of randomness or disorder in a system.

Concepts of thermodynamics is a significant area of study, especially for students preparing for their board exams and competitive exams such as JEE. Aspirants appearing for these exams can check out the questions and answers given below for better understanding and application of various thermodynamics concepts.

Laws of Thermodynamics Tricky Questions

Let us look at some of the important questions about laws of thermodynamics in detail:

1. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

(a) possible at high temperature

(b) possible only at low temperature

(c) not possible at any temperature

(d) possible at any temperature

Ans: d: For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.

2. The enthalpies of all elements in their standard states are

(a) unity

(b) zero

(c) <0

(d) different for each element.

Ans: (b): By convention, the standard enthalpy of formation of every element in its standard state is zero.

3. For an isolated system, ΔU = 0, what will be ΔS?

Ans: When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.

4. Why is the entropy of a diamond less than that of graphite?

Ans: Diamond is more compact than graphite.

5. Which of the following is a state function?

(i) height of a hill

(ii) distance traveled in climbing the hill

(iii) energy consumed in climbing the hill

Ans: Energy is consumed while climbing the hill

6. Why is it essential to mention the physical state of reactants and products in thermochemical reactions?

 Ans: It is because the physical state of reactants and products also contributes          significantly to the value ΔU or ΔH.

7. How are internal energy change, energy change and entropy change related to one another?

Ans: ∆G = ∆H – T∆S (At constant pressure)

8. Why is the standard entropy of an elementary substance not zero whereas the standard enthalpy of formation is taken as zero?

Ans: A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

9. Arrange the following fuels in order of increasing fuel efficiency: kerosene, diesel oil, wax, natural gas.

Ans: Wax < diesel oil < kerosene < natural gas (Lower hydrocarbons have higher calorific value and therefore are more efficient).

10. Many thermodynamically feasible reactions do not occur under ordinary conditions. Why?

Ans: Under ordinary conditions, the average energy of reactants may be less than the threshold energy. They require some activation energy to initiate the reaction.

11.  Ice is lighter than water. However, the entropy of ice is less than that of water. Explain.

Ans: Water is the liquid form while ice is its solid form. Molecular motion in ice is restricted than in water, i.e. a disorder in ice is more restricted than in water. Therefore, disorder in ice is less than in water.

12.  What is the most important condition for a process to be reversible in thermodynamics?

Ans: The process should be carried out infinitesimally slowly or the driving force should be infinitesimally greater than the opposing force.

13. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Ans: Heat absorbed by the system, q = 701 J Work done by the system = – 394 J Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

14.  For the reaction; 2Cl(g) ———-> Cl2(g); what will be the signs of ∆H and ∆S?

Ans: ∆H : negative (- ve) because energy is released in bond formation

∆S : negative (- ve) because entropy decreases when atoms combine to form molecules.

15.  The entropy change for the vaporization of water is 109 JK-1 mol-1. Calculate the enthalpy change for the vaporization of water at 373 K.

Ans: Δvap H = TΔvap S

= 373 × 109 = 40657 Jmol-1

= 40.657 kJ mol-1

Thermodynamics is an important concept for JEE Mains and JEE Advanced. The chapter is present in both Physics and Chemistry. Practicing a lot of questions and interpreting the laws and concepts logically can help in grasping the subject better. A little motivation and a lot of effort can help in acing the concepts!