Thermodynamics Quiz-11

Thermodynamics Quiz-11

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. A cylinder of gas is assumed to contain 11.2 kg of butane (C₄H₁₀). If a normal family needs 20000 kJ of energy per day, the cylinder will last ( Given that ΔH for combustion of butane is −2658 kJ )

•  20 days
•  25 days
•  26 days
•  24 days

Solution:-
Cylinder contains 11.2kg or 193.10 mole butane. (∵molecular mass of butane = 58)
∵ Energy released by 1 mole of butane = −2658
∵ Energy released by 193.10 mole of butane = −2658 × 193.10
= 5.13 × 10⁵ kJ
∴ 5.13×10⁵ ÷ 20000 = 25.66 or 26 days

Q2. Heat of combustion of CH₄, C₂H₄, C₂H₆ are -890, -1411 and -1560 kJ/mol respectively. Which has the lowest calorific fuel value in kJ/g ?

•  CH₄
•  C₂H₄
•  C₂H₆
•  All same

Solution:-
calorific value = Heat of combustion per g of fuel,
i.e. for C₂H₄, it is −1411÷28, the lowest value.

Q3. 16 g oxygen gas expands at STP to occupy double of its original volume. The work done during the process is :

•  260 kcal
•  180 kcal
•  130 kcal
•  272.8 kcal

Solution:-
At STP, 16g O₂ or 1/2 mole O₂will occupy 11.2 litre. Thus if volume is doubled, it means
(V₂ − V₁) = 22.4−11.2 = 11.2 litre
Now, W = P × (V₂ − V₁) = 1×11.2 litre atm
= (1×11.2×2)÷0.0821 = 272.84 kcal

Q4. When two moles of hydrogen expands isothermally against a constant pressure of 1 atm, at 25 °C from 15L to 50L, the work done (in litre atm) will be :

•  17.5
•  35
•  51.5
•  70

Solution:-
p = 1 atm
ΔV=(50−15)=35L
∴  W = −p.ΔV = −1×35 = −35 atm L
Hence work done by the system on the surroundings is equal to 35 L-atm

Q5. The enthalpy changes of formation of the gaseous oxides of nitrogen(N₂O and NO) are positive because of :

•  The high bond energy of the nitrogen molecule
•  The high electron affinity of the oxygen atoms
•  The high electron affinity of the nitrogen atoms
•  The tendency of oxygen to form O²⁻

Solution:-
Due to high bond energy of N≡N, more heat is absorbed to breakup N₂ molecule.

Q6. The enthalpy change is negative for :

•  Cl⁻(g) + aq → Cl⁻(aq)
•  Cl(g) → Cl⁺(g)+ e
•  1/2 Cl₂(g) → Cl(g)
•  Cl₂(l) → Cl₂(g)

Solution:-
Rest all are endothrmic process. Hydration is always exothermic.

Q7. A carnot engine operates between temperature T and 400K (T>400K). If efficiency of engine is 25%, the temperature T is :

•  400 K
•  500 K
•  533.3 K
•  600 K

Solution:-
η_efficiency=(T₂ − T₁)/T₂, here T₂ = T and T₁ = 400K
∴ T=533.3 K

Q8. A reaction occur spontaneously if :

•  TΔS = ΔH and both ΔH and ΔS are positive.
•  TΔS > ΔH and both ΔH and ΔS are positive.
•  TΔS < ΔH and both ΔH and ΔS are positive.
•  TΔS > ΔH and ΔH is positive and ΔS is negative.

Solution:-
For spontaneous reaction ΔG = −ve.
ΔG = ΔH − TΔS
ΔH = +ve, ΔS = +ve and TΔS > ΔH

Q9. The resultant heat change in a reaction is same whether it take place in one or several stages. This statement is called

•  Lavoisier and Laplace law
•  Hess's law
•  Joule's law
•  Le-chatelier's principle

Solution:-
Hess's law

Q10. The heat of combustion of ethanol determined by a bomb calorimeter is −670.48 kcal/mol at 25°C. What is ΔU at 25°C for the following reaction ?
C₂H₅OH(l) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(g)

•  −335.24 kcal
•  −669.28 kcal
•  −670.48 kcal
•  +670.48 kcal

Solution:-
Experimental determination of heat of reaction by bomb calorimeter represents its value at constant volume i.e. ΔU.