This set of Basic Chemical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Modeling and Predicting Vapor Pressure”.
1. What is the pressure of a gas at T = 227oC, 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
View answer
Explanation: 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 = 32P0.02/T0.5, where T is in Kelvin, m is in Kg and P is in mm Hg, what is the mass of material at 127oC, 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
View answer
Explanation: ln p* = A – B/(C + T) = 10 – 6 X 103/(400) = -5, => p = 0.0067 mm Hg, => m = 32*(0.0067)0.02/4000.5 = 1.44 Kg.
3. If the pressure of a gas at 27oC is 15 mm Hg, what will be its pressure at 127oC, 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
View answer
Explanation: 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 127oC is 10 mm Hg and the pressure at 527oC is 20 Pa, what is the value of B?
a) 225.1
b) 365.5
c) 499.2
d) 554.5
View answer
Explanation: 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) Pe1/T
b) Pe1/2T
c) Pe1/3T
d) Pe1/4T
View answer
Explanation: ln P/P2 = 1*(1/2T – 1/T), => P2 = Pe1/2T.
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
View answer
Explanation: Using linear interpolation, P325K = 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
View answer
Explanation: Using linear interpolation, T6Pa = 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
View answer
Explanation: 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
View answer
Explanation: 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
View answer
Explanation: 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
View answer
Explanation: 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
View answer
Explanation: 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 = 325p0.5M/T1.5 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
View answer
Explanation: ln p* = 25 – 2700/150 = 7, => p = 1096.6 mm Hg, => W = 325(1096.6)0.5*127/1501.5 = 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
View answer
Explanation: 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
View answer
Explanation: 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.
Sanfoundry Global Education & Learning Series – Basic Chemical Engineering.
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