Ch E102 - Tutorial 5 Solutions PDF

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Tutorial 5 Solutions...

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ChE102 Chemistry for Engineers - Tutorial 5

Solve the following problems either individually or in small groups (3 students maximum). Submit one copy of the solutions per group. Ask the TA for help if you need assistance solving the problems.

Problem 1 A) Explain how the drinking bird works. Inside of the drinking bird is a volatile fluid, methylene chloride (CH2Cl2). After the bird has a drink, water evaporates from his head. When the water evaporates, the head cools to a lower temperature. Keep in mind during evaporation the fastest moving molecules (i.e. those with the most energy) leave the liquid phase. Because of this the average kinetic energy, which is directly related to the average temperature, of the remaining water molecules decreases. Heat is energy that is being transferred from a warmer to a colder object. If the water is colder than the head of the drinking bird energy is transferred from the head of the bird to the water until they are at the same temperature, a condition known as thermal equilibrium. Because the head cools, the vapour pressure at the head becomes lower than the vapour pressure at the base. This pressure differential causes the liquid to rise from the base. As the liquid rises, the bird becomes top heavy and tips over. You’ll notice that once the bird tips over a bubble of vapour rises up the tube, and the liquid returns to the bottom of the tube. B) How can you speed up the drinking bird? 

Heat up the base of the bird so that the vapour pressure at the base increases, thus increasing the pressure differential between the base and the head.



Blow a fan over the head of the bird to increase the rate of evaporation.



Use rubbing alcohol or any liquid more volatile than water, to increase the rate of evaporation.



Lower the temperature of the liquid the bird is drinking; this will further decrease the vapour pressure in the head.

C) Why won’t the drinking bird work when it’s raining? On a rainy day the air is at 100% humidity which means the water can no longer evaporate from the head of the drinking bird.

Problem 2 The vapour pressure of benzene is 1.34 atm at 90. °C and its boiling point at 1.00 atm is 80. °C. Assuming that Hvap is constant over this temperature range, calculate the pressure at which benzene will boil at 0. °C. Apply the Clausius-Clapeyron equation twice: 𝑃

ln 𝑃2 = − 1

Δ 𝐻vap 𝑅

1

1

(𝑇 − 𝑇 ) 2

1

1.34

Δ 𝐻vap

( 90+273.15 −

𝑃

Δ 𝐻vap

( 0+273.15 −

ln 1.00 = − ln 1.00 = −

𝑅

𝑅

1

1

1

80+273.15

)

1 ) 80+273.15

Use equation (1) to solve for

Δ 𝐻vap 𝑅

(1) (2)

and then use equation (2) to solve for the unknown pressure.

Alternatively, divide equation (2) by equation (1) to eliminate pressure. (2) divided by (1): 𝑃 1.00 1.34 ln 1.00

ln

𝑃

=

1 ) 80+273.15 1 1 ( − ) 90+273.15 80+273.15

ln 1.00 =

1

( 0+273.15−

1 ) 1.34 80+273.15 1 1 1.00 (90+273.15− 80+273.15) 1 1 − ( 0+273.15 ) 80+273.15 1 1 ( 90+273.15− 80+273.15) 1

( 0+273.15−

𝑃 = 1.00exp [

ln

ln

1.34 ] 1.00

= 𝟎. 𝟎𝟒𝟒𝟓 𝐚𝐭𝐦

Δ 𝐻vap 𝑅

and solve for the unknown

Problem 3 The vapour pressure of diethyl ether (C2H5OC2H5) at 25 °C is 545 mmHg. A closed 5 US gal drum contains 50. g of diethyl ether. Approximately how many grams of diethyl ether in the container would be in the liquid phase at 25 °C? Data 264.17 US gal = 1000L, 1gal = 3.786L Pvap(Et2O) = 545 mmHg = 0.7171 atm T = 25 °C = 298.15 K 𝑀(C2 H5)2 O = 74.12 g/mol Calculate the mass of Et2O in the vapour phase using the ideal gas law and assuming that the volume occupied by the liquid is negligible: 𝑚=

𝑃𝑉𝑀 𝑅𝑇

=

(0.7171 atm)(5 gal)(3.786 L/gal)(74.12 g/mol) L atm

(0.08206 mol K)(298.15 K)

= 41 g

mass Et2O in liquid = 50 g – 41 g = 9g

Problem 4 Identify whether each of the following statements regarding the phase diagram for sulfur is TRUE or FALSE. Explain your reasoning. Note that sulfur has two solid phases: monoclinic and rhombic. A) AT STP the sulfur will be rhombic solid. True, at 1 bar and 25 °C sulfur will be rhombic. B) There is only one triple point on the diagram. False, any point where three lines come together is a triple point. This phase diagram has three triple points. C) There are two triple points where the gas phase is in equilibrium with two other phases. True, there is a triple point where the gas phase is in equilibrium with the rhombic and monoclinic phase. There is a second triple point where the gas phase is in equilibrium with the monoclinic and the liquid phase.

D) Sulfur boils only at a temperature of 444.6 °C. False, the normal boiling point is at 444.6 °C. If the atmospheric pressure is greater than or less than 1 atm sulfur will boil at a different temperature. E) Rhombic sulfur cannot be melted at pressures below 1420 atm. True, rhombic sulfur is in equilibrium with the liquid phase at pressures above 1420 atm. Problem 5 Identify whether each of the following statements regarding the phase diagram below is TRUE or FALSE. Explain your reasoning.

A) Region A represents the solid phase. True, A is solid, B liquid, C vapour/gas. B) Point e is the critical point. False, e is the triple point. C) Only a liquid phase exists to the right of point f. False, f is the critical point. To the right and above point f supercritical fluid (having properties similar to both liquids and gases) exists. D) Energy is absorbed by the system in the transition from region B to region C. True, in the transition from liquid (B) to vapour (C) the system must absorb energy. E) Solid and gas phases may exist simultaneously anywhere along the line de. True, the lines represent P-T conditions where multiple phases may exist simultaneously. de is along the boundary between the solid and gas phases....