Thermodynamics is a very important branch of both chemistry and physics. It deals with the study of energy, the conversion of energy between different forms and the ability of energy to do work. Thermodynamics is an important chapter for JEE Mains, JEE Advance NEET and many others exams. As this chapter is present in both chemistry and physics and there is only a minor difference between them hence it becomes more important topic. It is not a tough topic, it only requires practice. This topic has been given good weightage in all engineering exams. So don't skip this and prepare well. All the best !.

**Q1.**50 mL of water takes 5 min to evaporate from a vessel on a heater connected to an electric source which delivers 400 W. The enthalpy of vaporisation of water is

Solution:-

**Q2.**The energy absorbed by each molecule (A

_{2}) of a substance is 4.4 × 10

^{−19}J and bond energy per molecule is 4.0 × 10

^{−19}J. The kinetic energy of the molecule per atom will be :

Solution:-

**Q3.**Entropy change of vaporisation at constant pressure is given by:

Solution:-

This is derived formula.

This is derived formula.

**Q4.**Heat of fusion of a molecular solid is :

Solution:-

Molecular solids are covalent compounds having low m.p.

Molecular solids are covalent compounds having low m.p.

**Q5.**Heat of dissociation of benzene of elements is 5335 kJ/mol. The bond enthalpies of bonds are 347.3, 615 and 416.2 kJ respectively. Resonance energy of benzene is

Solution:-

First we calculate the expected bond dissociation energy of benzene molecules as

3 × C − C + 3 × C = C + 6 × C − H

∴ calculated value = 3(347.3) + 3(615) + 6(412.2) = 4397.8

Resonance energy = Experimental value − calculated value

5335 − 4397.8 = 937.2 kJ/mol

First we calculate the expected bond dissociation energy of benzene molecules as

3 × C − C + 3 × C = C + 6 × C − H

∴ calculated value = 3(347.3) + 3(615) + 6(412.2) = 4397.8

Resonance energy = Experimental value − calculated value

5335 − 4397.8 = 937.2 kJ/mol

**Q6.**Hess’s law is used to calculate

Solution:-

Hess’s law states that enthalpy changes during and process are independent of path. So, this law is used in calculating enthalpy.

Hess’s law states that enthalpy changes during and process are independent of path. So, this law is used in calculating enthalpy.

**Q7.**The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is:

Solution:-

Î”S = Î”H/T = 5.260 cal mol

Î”S = Î”H/T = 5.260 cal mol

^{−1}K^{−1}**Q8.**Work done in reversible adiabatic process is given by :

Solution:-

This is the derived formula for W

This is the derived formula for W

_{rev}in adiabatic process.**Q9.**Equal volume of C

_{2}H

_{2}and H

_{2}are combusted under identical condition. The ratio of their heat of combustion is :

H

_{2}(g) + ½O

_{2}(g) → H

_{2}O(g); Î”H = −241.8 kJ

C

_{2}H

_{2}(g) + 2½O

_{2}(g) → 2CO

_{2}(g) + H

_{2}O(g); Î”H = −1300 kJ

Solution:-

1300/241.8 = 5.37/1

1300/241.8 = 5.37/1

**Q10.**Î”C

_{p}for : N

_{2}(g) + 3H

_{2}(g) → 2NH

_{3}(g) is :

Solution:-

Î”C

Î”C

_{p}= Î£C_{p}product − Î£C_{p}reactant, note C_{p}is for 1 mole.