250+ TOP MCQs on Modeling and Predicting Vapor Pressure and Answers

Basic Chemical Engineering Multiple Choice Questions on “Modeling and Predicting Vapor Pressure”.

1. What is the pressure of a gas at T = 227^oC, it is given that A = 5, B = 4 X 104 and C = 0 if T is in kelvin and p in mm Hg?
a) 0.024 mm Hg
b) 0.049 mm Hg
c) 0.075 mm Hg
d) 0.096 mm Hg

Answer: b
Clarification: ln p* = A – B/(C + T), => ln p = 5 – 4 X 10³/(273 +227) = -3, => p = 0.049 mm Hg.

2. A material’s mass is given by the equation, m = 32P^0.02/T^0.5, where T is in Kelvin, m is in Kg and P is in mm Hg, what is the mass of material at 127^oC, use A = 10, B = 6 X 10³ and C = 0?
a) 1.44 Kg
b) 1.69 Kg
c) 1.96 Kg
d) 2.25 Kg

Answer: a
Clarification: ln p* = A – B/(C + T) = 10 – 6 X 10³/(400) = -5, => p = 0.0067 mm Hg, => m = 32*(0.0067)^0.02/400^0.5 = 1.44 Kg.

3. If the pressure of a gas at 27^oC is 15 mm Hg, what will be its pressure at 127^oC, use B = 5 X 10³, C = 0, T is in kelvin and P is in mm Hg?
a) 579 mm Hg
b) 775 mm Hg
c) 961 mm Hg
d) 999 mm Hg

Answer: c
Clarification: ln P₁/P₂ = B (1/T₂ – 1/T₁), => ln 15/P₂ = 5 X 10³ (1/400 – 1/300), P₂ = 961 mm Hg.

4. Pressure of a gas at 127^oC is 10 mm Hg and the pressure at 527^oC is 20 Pa, what is the value of B?
a) 225.1
b) 365.5
c) 499.2
d) 554.5

Answer: d
Clarification: ln 10/20 = B (1/800 – 1/400), => B = 554.5.

5. Pressure of a gas at temperature T is P, What is the pressure of the gas at 2T, use B = 1?
a) Pe^1/T
b) Pe^1/2T
c) Pe^1/3T
d) Pe^1/4T

Answer: b
Clarification: ln P/P₂ = 1*(1/2T – 1/T), => P₂ = Pe^1/2T.

6. What is the pressure of CO₂ at 325 K, if it is given that pressure at 330 K is 6 Pa, and pressure at 320 K is 4 Pa?
a) 4.5 Pa
b) 5 Pa
c) 5.5 Pa
d) 6 Pa

Answer: b
Clarification: Using linear interpolation, P_325K = 4 + (325 – 320)*(6 – 4)/(330 – 320) = 4 + 5*2/10 = 5 Pa.

7. Temperature of SO₂ at 5 Pa is 395 K and at 10 Pa temperature is 420 K, what is the temperature at 6 Pa?
a) 400 K
b) 405 K
c) 410 K
d) 415 K

Answer: a
Clarification: Using linear interpolation, T_6Pa = 395 + (6 – 5)*(420 – 395)/(10 – 5) = 395 + 5 = 400 K.

8. At which of the following values of pressure a solid melts first?
a) 5 atm
b) 720 mm Hg
c) 500 KPa
d) 40 Torr

Answer: a
Clarification: A solid melts first at the highest pressure which is 5 atm.

9. A liquid evaporates first at which of the following conditions?
a) High attraction forces
b) Low attraction forces
c) Low vapor pressure
d) High boiling point

Answer: b
Clarification: A liquid evaporates first at low attraction forces.

10. At triple point, vapor pressure of ice is given by ln p* = 18 – 2560/T, and vapor pressure of water is given by ln p* = 28 – 5120/T, what is the temperature at triple point?
a) 225 K
b) 256 K
c) 320 K
d) 400 K

Answer: b
Clarification: At triple point pressure and temperature of all the phases becomes same, => 18 – 2560/T = 28 – 5120/T, => 10 = 2560/T, => T = 256 K.

11. The vapor pressure of liquid benzene is given by ln p* = 15 – 1280/T, and that of vapor benzene is given by ln p* = 23 – 2560/T, what is the boiling point of benzene?
a) 160 K
b) 200 K
c) 240 K
d) 300 K

Answer: a
Clarification: At boiling point pressure and temperature of both the phases will be same, => 23 – 2560/T = 15 – 1280/T, => 8 = 1280/T, => T = 160K.

12. The vapor pressure of a compound in a liquid phase is given by ln p* = 21 – 2800/T, and in solid phase it is given by ln p* = 28 – 3500/T, what is the sublimation temperature of the compound?
a) 100 K
b) 120 K
c) 150 K
d) 200 K

Answer: a
Clarification: At sublimation point pressure and temperature of both the phases will be same, => 21 – 2800/T = 28 – 3500/T, => T = 100 K.

13. For CaOCl₂, the rate of evaporation is given by W = 325p^0.5M/T^1.5 g/liter, where T is in K, p in mm Hg, and M is molecular weight, if its pressure is given by ln p* = 25 – 2700/T, what is the evaporation rate at T = 150 K?
a) 256 g/liter
b) 499 g/liter
c) 571 g/liter
d) 744 g/liter

Answer: d
Clarification: ln p* = 25 – 2700/150 = 7, => p = 1096.6 mm Hg, => W = 325(1096.6)^0.5*127/150^1.5 = 744 g/liter.

14. What is the vapor pressure of ammonia at 305 K, if it is given that the vapor pressure of ammonia at 315 K is 5 mm Hg and at 325 K it is 6 mm Hg?
a) 1 mm Hg
b) 2 mm Hg
c) 3 mm Hg
d) 4 mm Hg

Answer: d
Clarification: p305K = 5 – (315 – 305)*(6 – 5)/(325 – 315) = 5 – 1 = 4 mm Hg.

15. The pressure of a compound at 200 K is 10 mm Hg and at 400 K it is 50 mm Hg, what is the pressure at T = 250 K?
a) 18.1 mm Hg
b) 24.2 mm Hg
c) 36.3 mm Hg
d) 48.4 mm Hg

Answer: a
Clarification: Let the pressure of the compound is given by ln p* = a + b/T, => ln 10 = a + b/200, ln 50 = a + b/400, => ln 5 = -b/400, => b = -643.7, => a = 5.5, => At T = 250 K, ln p* = 5.5 – 643.7/250 = 2.9, p = 18.1 mm Hg.