Chemical Engineering Questions and Answers – Application of the Energy Balance to Closed Systems

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This set of Chemical Engineering Aptitude Test focuses on “Application of the Energy Balance to Closed Systems”.

1. What is the energy balance of a closed system?
a) Matter in the system is constant
b) Energy in the system is constant
c) Neither energy nor matter is constant
d) Both of energy and matter are constant
View answer

Answer: a
Explanation: For closed system energy balance, matter in the system is constant.
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2. What is the heat transferred to the system, if a change in internal energy is 10 J?
a) 0
b) -10 J
c) 10 J
d) Cannot be determined
View answer

Answer: c
Explanation: Q = ∆U = 10 J.

3. What is the heat transferred to the system, if change in enthalpy is 15 J and change in volume is 2 m3 at 5 Pa?
a) 5 J
b) 8 J
c) 10 J
d) 12 J
View answer

Answer: a
Explanation: Q = ∆H – P∆V = 15 – 5*2 = 5 J.

4. What is the heat transferred to the system, if a change in enthalpy is 25 J and change in pressure is 5 Pa at volume 3 m3?
a) 5 J
b) 10 J
c) 15 J
d) 20 J
View answer

Answer: b
Explanation: Q = ∆H – V∆P = 25 – 3*5 = 10 J.

5. What is the heat transferred to the system, if change in enthalpy is 30 J and pressure, volume changed from 8 Pa, 5 m3 to 4 Pa, 12 m3?
a) 12 J
b) 15 J
c) 18 J
d) 22 J
View answer

Answer: d
Explanation: Q = ∆H – V∆P = 30 – (4*12 – 8*5) = 22 J.
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6. The specific volume of SO2 is 10 m3/Kg, at 5 Pa, H = 100 J/Kg and at 10 Pa, H = 250 J/Kg, what is the heat transferred to the system?
a) 100 J/Kg
b) 150 J/Kg
c) 250 J/Kg
d) 350 J/Kg
View answer

Answer: a
Explanation: Q = ∆H – V∆P = (250 – 100) – 10(10 – 5) = 100 J/Kg.

7. The specific volume of a gas is 5 m3/Kg, at 10 Pa, H = 50 J/Kg and at 20 Pa, H = 100 J/Kg, what is the heat transferred to the system?
a) 0
b) 50 J/Kg
c) 100 J/Kg
d) 150 J/Kg
View answer

Answer: a
Explanation: Q = ∆H – V∆P = (100 – 50) – 5(20 – 10) = 0.

8. The specific volume of a gas is 5 m3/Kg, at 25 Pa, H = 40 J/Kg and at 40 Pa, H = 90 J/Kg, what is the heat transferred to the system?
a) -5 J/Kg
b) 5 J/Kg
c) -25 J/Kg
d) 25 J/Kg
View answer

Answer: c
Explanation: Q = ∆H – V∆P = (90 – 40) – 5(40 – 25) = -25 J/Kg.

9. The specific volume of a gas is 2 m3/Kg, at 12 Pa, H = 30 J/Kg and at 15 Pa, H = 45 J/Kg, what is the heat transferred to the system?
a) 6 J/Kg
b) 9 J/Kg
c) 12 J/Kg
d) 15 J/Kg
View answer

Answer: b
Explanation: Q = ∆H – V∆P = (45 – 30) – 2(15 – 12) = 15 – 6 = 9 J/Kg.
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10. The specific volume of a gas is 3 m3/Kg, at 6 Pa, H = 4 J/Kg and at 9 Pa, H = 14 J/Kg, what is the heat transferred to the system?
a) 0
b) 1 J/Kg
c) 5 J/Kg
d) 25 J/Kg
View answer

Answer: b
Explanation: Q = ∆H – V∆P = (14 – 4) – 3(9 – 6) = 1 J/Kg.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn