Thermodynamics Quiz-12

Thermodynamics Quiz-12

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. And with regular practice one can dominate over this topic very easily.This topic has been given good weightage in all engineering exams. So don't skip this and prepare well. All the best !.

Q1. The heat required to raise the temperature of a body by 1 K is called

•  Specific heat
•  Thermal capacity
•  Water equivalent
•  None of these

Solution:-
Thermal capacity

Q2. Entropy of vaporisation of water at 100°C, if molar heat of vaporisation is 9710 cal/mol will be:

•  20 cal mol^-1 K^-1
•  26 cal mol^-1 K^-1
•  24 cal mol^-1 K^-1
•  28 cal mol^-1 K^-1

Solution:-
ΔS_vap = ΔH_vap / T = 9710/373 = 26 cal mol^-1 K^-1

Q3. C_diamond + O₂(g) → CO₂(g);   ΔH = -395 kJ.....(i)
  C_graphite + O₂(g) → CO₂(g);   ΔH = -393.5 kJ.....(ii)
  The ΔH, when diamond is formed from graphite is :

•  -1.5 kJ
•  +1.5 kJ
•  +3.0 kJ
•  -3.0 kJ

Solution:-
By eq. (ii) − (i)
C_G → C_D;   ΔH = +1.5 kJ

Q4. One mole of ice is converted into water at 273 K. The entropies of H₂O(s) and H₂O(l) are 38.20 and 60.01 J mol^-1K^-1 respectively. The enthalpy change for the conversion is:

•  59.54 J mol^-1
•  5954 J mol^-1
•  595.4 J mol^-1
•  320.6 J mol^-1

Solution:-
ΔG = ΔH − TΔS; at equilibrium, ΔG = 0
∴ ΔH = TΔS
or ΔH = 273 ∴(60.01 − 38.20) = 5954.13 J/mol

Q5. Which is not characteristic of thermochemical equation?

•  It indicates physical state of reactants and products.
•  It indicates whether the reaction is endothermic or exothermic.
•  It indicates allotrope of reactants if present.
•  It indicates whether the reaction would occur or not.

Solution:-
The spontaneity of reaction cannot be decided by simply looking the chemical change. We need ΔG value for it.

Q6. Which species have negative value of specific heat?

•  Ice
•  Water
•  Vapour
•  Saturated vapour

Solution:-
Negative specific heat refers that in order to rise the temperature, certain quantity of heat is to be withdrawn from the body.

Q7. A thermally isolated gaseous system can exchange energy with the surroundings. The mode of transference of energy can be :

•  Heat
•  Work
•  Heat and radiation
•  None of these

Solution:-
Only work can be done by a thermally isolated system between it and surroundings.

Q8. 2.1 g of Fe combine with S evolving 3.77 kJ. The heat of formation of FeS in kJ/mol is :

•  −1.79
•  −100.5
•  −3.77
•  None of these

Solution:-
ΔH/mol of FeS = (3.77×56)/2.1 = 100.5

Q9. Change in entropy for a reaction is given by:

•  2.303 nR log₁₀(V₂ / V₁)
•  nR log_e(V₂ / V₁)
•  nR log_e(P₁ / P₂)
•  All of these

Solution:-
These are derived formulae.

Q10. The following two reaction are known :
 Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) ;    ΔH = −26.8 kJ
 FeO(s) + CO(g) → Fe(s) + CO₂(g) ;    ΔH = −16.5 kJ
 The value of ΔH for the following reaction...
Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g)   is

•  +10.3 kJ
•  −43.3 kJ
•  −10.3 kJ
•  +6.2 kJ

Solution:-