Topic: UNIT 7: Thermodynamics



Topic: UNIT 7: Thermodynamics
Q.1
The amount of heat exchanged with the surrounding at constant temperature and pressure is given by the quantity
A. ∆E
B. ∆H
C. ∆S
D. ∆G
Answer : Option B
Explaination / Solution:
No Explaination.


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Q.2
All the naturally occurring processes proceed spontaneously in a direction which leads to
A. decrease in entropy
B. increase in enthalpy
C. increase in free energy
D. decrease in free energy
Answer : Option D
Explaination / Solution:
No Explaination.


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Q.3
In an adiabatic process, which of the following is true ?
A. q = w
B. q = 0
C. ∆E = q
D. P ∆ V= 0
Answer : Option B
Explaination / Solution:
No Explaination.


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Q.4
In a reversible process, the change in entropy of the universe is
A. > 0
B. ≥ 0
C. < 0
D. = 0
Answer : Option D
Explaination / Solution:
No Explaination.


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Q.5
In an adiabatic expansion of an ideal gas
A. w = – ∆u
B. w = ∆u + ∆H
C. ∆u = 0
D. w = 0
Answer : Option A
Explaination / Solution:
No Explaination.


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Q.6
The intensive property among the quantities below is
A. mass
B. volume
C. enthalpy
D. mass/volume
Answer : Option D
Explaination / Solution:
No Explaination.


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Q.7
An ideal gas expands from the volume of 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure at 1 × 105 Nm–2. The work done is
A. – 900 J
B. 900 kJ
C. 270 kJ
D. – 900 kJ
Answer : Option A
Explaination / Solution:

w = – P V

w = – (1 × 105 Nm–2)

(1× 10–2 m3 – 1 × 10–3 m3)

w= –105 (10–2 – 10–3) Nm

w = – 105 (10 – 1) 10–3) J

w = – 105 (9 × 10–3) J

w = –9 × 102 J w = –900 J


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Q.8
Heat of combustion is always
A. positive
B. negative
C. zero
D. either positive or negative
Answer : Option B
Explaination / Solution:
No Explaination.


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Q.9
The heat of formation of CO and CO2 are – 26.4 kCal and – 94 kCal, respectively. Heat of combus-tion of carbon monoxide will be
A. + 26.4 kcal
B. – 67.6 kcal
C. – 120.6 kcal
D. + 52.8 kcal
Answer : Option B
Explaination / Solution:

CO (g )+1/2O2 (g )→CO 2 (g)

ΔHC0 (CO) = ΔHf (CO2) – ΔHf(CO) + ΔHf (O2)]

ΔHC0 (CO) = ΔHf (CO2) – ΔHf(CO) + ΔHf (O2)]

ΔHC0 (CO) = – 94 KCal – [– 26.4 KCal + 0]

ΔHC0 (CO) = – 94 KCal + 26.4

KCal ΔHC0 (CO) = – 67.4 Kcal


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Q.10
C(diamond) → C(graphite), ∆H = –ve, this indicates that
A. graphite is more stable than diamond
B. graphite has more energy than diamond
C. both are equally stable
D. stability cannot be predicted
Answer : Option A
Explaination / Solution:
No Explaination.


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