Thermodynamics Quiz-20

Thermodynamics Quiz-20

Dear Readers,

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

•  40.3 kJ/mol
•  43.2 kJ/mol
•  16.7 kJ/mol
•  180.4 kJ/mol

Solution:-

Q2. The energy absorbed by each molecule (A₂) 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 :

•  4.0 × 10⁻²⁰ J
•  2.0 × 10⁻²⁰ J
•  2.2 × 10⁻¹⁹ J
•  2.0 × 10⁻¹⁹ J

Solution:-

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

•  ΔS_v = ^ΔH_v ⁄ _T
•  ΔS_v = ^ΔU_v ⁄ _T
•  ΔS_v = ^ΔH_v ⁄ _ΔT
•  None of these

Solution:-
This is derived formula.

Q4. Heat of fusion of a molecular solid is :

•  Very high
•  High
•  Low
•  None of these

Solution:-
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

•  1.15 kJ
•  15.1 kJ
•  937.2 kJ
•  1511 kJ

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

Q6. Hess’s law is used to calculate

•  Enthalpy of reaction
•  Entropy of reaction
•  Work done in reaction
•  All of these

Solution:-
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:

•  5.260 cal mol⁻¹K⁻¹
•  0.526 cal mol⁻¹K⁻¹
•  10.52 cal mol⁻¹K⁻¹
•  21.04 cal mol⁻¹K⁻¹

Solution:-
ΔS = ΔH/T = 5.260 cal mol⁻¹K⁻¹

Q8. Work done in reversible adiabatic process is given by :

•  2.303 RT log( ^V₂⁄_{V1 )}
•  ^{nR(T₂−T₁)} ⁄ _(γ−1)
•  2.303 RT log( ^V₁⁄_{V2 )}
•  None of these

Solution:-
This is the derived formula for W_rev in adiabatic process.

Q9. Equal volume of C₂H₂ and H₂ are combusted under identical condition. The ratio of their heat of combustion is :
H₂(g) + ½O₂(g) → H₂O(g);   ΔH = −241.8 kJ
C₂H₂(g) + 2½O₂(g) → 2CO₂(g) + H₂O(g);   ΔH = −1300 kJ

•  5.37/1
•  1/5.37
•  1/1
•  None of these

Solution:-
1300/241.8 = 5.37/1

Q10. ΔC_p for : N₂(g) + 3H₂(g) → 2NH₃(g) is :

•  C_pNH₃ + C_pN₂ + C_pH₂
•  C_pNH₃ − C_pN₂ − C_pH₂
•  2C_pNH₃ − C_pN₂ − 3C_pH₂
•  2C_pNH₃ + C_pN₂ − 3C_pH₂

Solution:-
ΔC_p = ΣC_pproduct − ΣC_preactant, note C_p is for 1 mole.