## 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:-
Î”Svap = Î”Hvap / T = 9710/373 = 26 cal mol-1 K-1

Q3. Cdiamond + O2(g) → CO2(g);   Î”H = -395 kJ.....(i)
Cgraphite + O2(g) → CO2(g);   Î”H = -393.5 kJ.....(ii)
The Î”H, when diamond is formed from graphite is :
•  -1.5 kJ
•  +1.5 kJ
•  +3.0 kJ
•  -3.0 kJ
Solution:-
By eq. (ii) − (i)
CG → CD;   Î”H = +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:-
Î”G = Î”H − TÎ”S; at equilibrium, Î”G = 0
∴ Î”H = TÎ”S
or Î”H = 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 Î”G 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Negative specific heat refers that in order to rise the temperature, certain quantity of heat is to be withdrawn from the body.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:-
Î”H/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 loge(V2 / V1)
•  nR loge(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) ;    Î”H = −26.8 kJ
FeO(s) + CO(g) → Fe(s) + CO2(g) ;    Î”H = −16.5 kJ
The value of Î”H 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:-

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