Chemical Engineering Questions and Answers – Compressibility

This set of Basic Chemical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Compressibility”.

1. What is the compressibility of ideal gases?
a) 1
b) 2
c) 3
d) 4
View answer

Answer: a
Explanation: Compressibility = PV/nRT, but for ideal gases PV = nRT, => Compressibility = 1.

2. Which of the following is true about compressibility of real gases?
a) Z = 1
b) Z < 1
c) Z > 1
d) Both Z < 1 and Z > 1
View answer

Answer: d
Explanation: For any value of compressibility except 1, the gas is real.

3. What is the pressure of CO2 gas with compressibility z = 0.8, if the ideal gas pressure of CO2 is 10 Pa, all other variables are same?
a) 5
b) 8
c) 10
d) 14
View answer

Answer: b
Explanation: Pressure of CO2 = 0.8*10 = 8 Pa.
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4. What is the pressure of H2 with compressibility 0.6, molar volume 5 liter/mole at 27oC?
a) 2.9 atm
b) 5.4 atm
c) 9.6 atm
d) 14.5 atm
View answer

Answer: a
Explanation: P = znRT/V, => P = 0.6*0.0821*300/5 = 2.9 atm.

5. At what temperature, does SO2 with compressibility 2 has pressure 5 atm and molar volume 10 mole/s?
a) 300.5 K
b) 304.5 K
c) 310 K
d) 312.5 K
View answer

Answer: b
Explanation: T = PV/zR = 5*10/(2*0.0821) = 304.5 K.

6. What is the compressibility factor of water with density 1 g/liter at 27oC and 1 atm?
a) 0.25
b) 0.59
c) 0.73
d) 0.91
View answer

Answer: c
Explanation: z = PM/dRT = 1*18/(1*0.0821*300) = 0.73.

7. What is the compressibility factor of a liquid of molar mass 90 g and density 10 g/liter at 27oC and 2 atm?
a) 0.14
b) 0.39
c) 0.51
d) 0.73
View answer

Answer: d
Explanation: z = PM/dRT = 2*90/(10*0.0821*300) = 0.73.
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8. What is the pressure of 80 grams of CH4 with compressibility 5, with 10 liter volume at 27oC?
a) 10.5 atm
b) 28.9 atm
c) 44.2 atm
d) 61.5 atm
View answer

Answer: d
Explanation: P = 5*5*0.0821*300/10 = 61.5 atm.

9. What is the compressibility of methane with density 10 g/liter at 27oC and 500 atm?
a) 32.5
b) 34.8
c) 39.6
d) 44.1
View answer

Answer: a
Explanation: z = PM/dRT = 500*16/(10*0.0821*300) = 32.5.
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10. What is the compressibility of neon (molar mass = 20) with density 20 g/liter at 27oC and 100 atm?
a) 2.1
b) 3.5
c) 4
d) 6.8
View answer

Answer: c
Explanation: z = PM/dRT = 100*20/(20*0.0821*300) = 4.

11. 90 g glucose is filled in a container at 10 atm, if the volume of container is 2 liter, what is the temperature of glucose (z = 1.5)?
a) 25.4oC
b) 51.7oC
c) 76.9oC
d) 103.5oC
View answer

Answer: b
Explanation: T = PV/znR = 10*2/(1.5*0.5*0.0821) = 324.8 K = 51.7oC.

12. How many grams of C2H6 (z = 1.4) is present in the container at 27oC and 5 atm and with volume 10 liter?
a) 20 g
b) 30.4 g
c) 40.6 g
d) 50.8 g
View answer

Answer: c
Explanation: n = 5*10/(1.4*0.0821*300) = 1.45, mass of C2H6 = 1.45*28 = 40.6 g.

13. What is the volume of 21 g of propane (z = 1.9) at 1 atm and 27oC?
a) 23.4 liter
b) 37.5 liter
c) 42.1 liter
d) 59.6 liter
View answer

Answer: a
Explanation: V = 1.9*0.5*0.0821*300/1 = 23.4 liter.

14. 1 mole of cyclohexane (z = 1.2) is filled in a container at 27oC and 4 atm, what is the volume of container?
a) 3.8 liter
b) 4.4 liter
c) 5.6 liter
d) 7.4 liter
View answer

Answer: d
Explanation: V = 1.2*1*0.0821*300/4 = 7.4 liter.

15. 39 grams of benzene (z = 1.4) is filled in a container of volume 10 liter at 27oC, what is the pressure of container?
a) 1.7 atm
b) 1.9 atm
c) 2.2 atm
d) 2.5 atm
View answer

Answer: a
Explanation: P = 1.4*0.5*0.0821*300/10 = 1.7 atm.

Sanfoundry Global Education & Learning Series – Basic Chemical Engineering.

To practice all areas of Basic Chemical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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