UNIVERSITY OF TORONTO
FACULTY OF APPLIED SCIENCE AND ENGINEERING
CHE 112F – PHYSICAL CHEMISTRY
MID-TERM EXAM,
OCTOBER 16 2014
6:10 pm to 8:00 pm
EXAMINER – P.V. Yaneff
1. Answer all questions. The marks add up to 90. The marks for each question are
indicated in [square brackets].
2. The mid-term counts for 20% of your final grade.
3. University of Toronto Approved calculators only.
4. LOOK OVER THE USEFUL INFORMATION BELOW TO ASSIST IN
SOLVING THE QUESTIONS BEFORE YOU BEGIN.
5. ALL WORK IS TO BE DONE IN EXAM BOOKLETS
6. PUT YOUR NAME AND STUDENT ID # ON YOUR EXAM BOOKLET AND IF
USING MORE THAN ONE INDICATE THE TOTAL (EG., 1 OF 2, 2 OF 2 ON
THE COVER.
PUT A SQUARE BOX AROUND ALL ANSWERS
Some Useful Data:
SATP = 0oC, 1 bar; STP = 0oC, 1 atm; 1 bar = 105 Pa = 0.9869 atm; 1 atm = 101.3
kPa = 760 mmHg; 1 J = 1 Pa/m3
R = 8.3145 J mol-1 K-1 = 62.37 L torr mol-1 K-1 = 0.08314 L bar mol-1 K-1 = 0.08206 L
atm mol-1 K-1
Molar mass (g/mol): C (12.01); H (1.01); He (4.00); N (14.01); O (16.0); Cl (35.46): K
(39.1); S (32.06)
Acceleration due to gravity: g = 9.807 m.s-2
Power is the rate of doing work; 1 Watt = 1J s-1. Density of water is 1000 kg m3
Some Useful Formulae:
Pi = xiP; P= PA + PB + PC
Wmech = mgΔz; WPV = -PextΔVsyst ; Wrev = -nRT ln(V2/V1)
vdw: [P+ a(n/v)2 ](V-nb) = nRT; Virial: Pv/RT = 1 + B(T)/v + C(T)/v2
ΔU= Q + W; ΔUv= Q; Q= CΔT = ncΔT ; Q= mcΔT, where c = specific heat
capacity, cp = cv + R
ΔHp = ΔU + PΔV; ΔHp = Qp = Cp ΔT = ncpΔT ; ΔHp= mcpΔT, where cp = specific
heat capacity @ const P; ΔHfus = nΔhfus; Specific heat of fusion: ΔHfus = mΔhfus
Page 1 of 11
Solution to question 3 is at the end. 1. Phosphorus trichloride (PCl3) reacts with water to form phosphorous acid (H3PO3) and hydrochloric acid: PCl3(liq) + 3H2O(liq) → H3PO3(aq) + 3HCl(aq) (a) Which is the limiting reactant when 12.4 g of phosphorus trichloride is mixed with 10.0 g of water? (b) What masses of phosphorous acid and hydrochloric acid are formed? Solution (a) PCl3 = 30.97 + 3(35.45) = 137.35 g mol–1 H2O = 2(1.008) + 16.00 = 18.02 g mol–1 H3PO3 = 3(1.008) + 30.97 + 3(16.00) = 81.99 g mol–1 HCl = 1.008 + 35.45 = 36.46 g mol–1 From the stoichiometry of the reaction, 3 moles of H2O are required for every mole of PCl3. 12.4 g
No. moles of PCl3 added to mixture = = 0.0903 mol 137.35 g / mol
No. moles of H2O added to mixture = 10.0 g
= 0.555 mol 18.02 g/ mol
! moles of H2 O $
0.555
&
The actual ratio of H2O to PCl3 is #
= = 6.146 > 3 " moles of PCl3 % actual 0.0903
There is too much water; i.e., there is not enough PCl5 for the amount of H2O present, so all the PCl3 will be reacted, but not all the H2O; PCl3 is the limiting reactant. Ans: PCl3 is the limiting reactant (b) From the stoichiometry of the reaction, Moles H3PO3 produced = moles of PCl3 consumed = 0.0903 mol Page 2 of 11
Therefore, mass of H3PO3 produced = (0.0903
mol H3 PO 3
(
) 81.99
g H3 PO 3
mol H3 PO 3
) = 7.403 g H3PO3 Similarly, Moles of HCl produced = 3 times the number of moles of PCl3 reacted; therefore: Mass of HCl produced = (0.0903
mol PCl3
(
)3
mol HCl
mol PCl3
)(36.46
g HCl
mol HCl
) = 9.877 g HCl Ans: 7.40 g H3PO3 9.88 g HCl 2. A company sells oxygen-­‐enriched air, which they claim contains at least 50% by volume pure oxygen. To test their claim, a 1.00 liter glass bulb was filled with the gas to a total pressure of 750 Torr at 20.0°C. The weight of the gas sample was found to be 1.236 grams. What is the volume percent of oxygen gas in the mixture? Assume ordinary air consists of 20.0% by volume O2 and 80.0% by volume N2. Molar masses: O = 16.00, N = 14.01. Solution Molar masses: O2 = 2 16.00 = 32.00 g mol–1 N2 = 2 14.01 = 28.02 g mol–1 Use Word 6.0c or later to
view Macintosh picture.
Use Word 6.0c or later to
view Macintosh picture.
Method 1: 750
PV ( 760 ! 101325)( 0.001)
=
Total moles n of gas in bulb: n = = 0.041024 mol (8.3145)(293.15)
RT
Let there be x moles of O2, and (0.041024 – x) moles of N2 in the sample. Total mass = mass of O2 + mass of N2 1.236 g = ( x mol) 32.00 mol + (0.041024 ! x mol) 28.02 mol Solving: x = 0.021736 mol (
g
)
(
g
)
x
0.021736
Therefore, % O2 = ! 100% =
! 100% = 52.98% n
0.041024
Page 3 of 11
Ans: 53.0 % The company is telling the truth! Method 2: Let the mass of O2 in the bulb be m grams; therefore the mass of the N2 is (1.236 – m) grams 750
PV ( 760 ! 101325)( 0.001)
=
Total moles n of gas in bulb: n = = 0.041024 mol (8.3145)(293.15)
RT
Therefore n O 2 + n N 2 = 0.041024 i.e., m g
(1.236 ! m) g
+
= 0.041024 32.00 g / mol 28.02 g / mol
28.02m + 32.00(1.236 ! m)
= 0.041024 (32.00)(28.02)
28.02m + 32.00(1.236 ! m) = (0.041024)(32.00)(28.02) 39.552 – 3.98m = 36.78376 m = Therefore, and the % by volume O2 is 39.552 ! 36.78376
= 0.69554 3.98
nO2 =
nO2
n
m
0.69554
= 0.02174 mol =
32.00
32.00
! 100% =
0.02174
! 100% = 52.99% 0.041024
Ans: 53.0% 4. Calculate the total work for one cycle consisting of the isothermal expansion (A → B) and compression (B → A) of one mole of ideal gas at 273.15 K between one atm and 0.500 atm. The expansion involves lifting 5.0 kg through 22 m. The compression is accomplished by a mass of 40 kg falling through 22 m. Friction may be neglected. The acceleration due to gravity is 9.807 m s–2 Page 4 of 11
Solution Take the gas as the system: When the gas expands, it delivers work equal to wAB = mg∆z = (5.0)(9.807)(22) = 1078.8 J When the gas is compressed, work must be done on the gas; the amount is wBA = mg∆z = (40)(9.807)(22) = 8630.2 J More work is done to compress the gas than is delivered by the gas when it expands. The net work done on the gas is thus wnet = 8630.2 – 1078.8 = 7551.4 J = 7.551 kJ Ans: Net work input of 7.55 kJ 5. One kilogram of O2 is heated in a frictionless cylinder-­‐piston apparatus at a constant pressure of one atm from 20°C to 30°C. Assuming ideal gas behaviour, calculate: (a) ∆H for the process, (b) ∆U for the process, and (c) cV for oxygen gas. The molar mass and molar heat capacity of O2 are, respectively, 32.00 g mol–1 and cP = 29.355 J mol–1 K–1. Solution 1000 g
nO2 =
(a) No. of moles of O2: = 31.25 mol 32.00 g / mol
Therefore ∆H = ncP ∆T = (31.25)(29.355)(30 – 20) = 9173.4 J (b) ∆U Ans: 9.173 kJ = ∆H – ∆(PV) = ∆H – ∆(nRT) = ∆H – nR∆T = 9173.4 – (31.25)(8.3145)(30 – 20) = 9173.5 – 2598.3 = 6575.2 J Ans: 6.575 J (c) For an ideal gas, even if the volume changes,∆U = ncV ∆T Page 5 of 11
Therefore, cV = !U
6575.2
=
= 21.04 J mol–1 K–1 n!T (31.25)(10)
Ans: Alternatively, 21.04 J mol–1 K–1 cV = cP – R = 29.355 – 8.3145 = 21.04 J mol–1 K–1 [as above] 6. 4.00 L of nitrogen gas at 100 kPa pressure and 300 K is Patm = 100 kPa
located in a cylinder/piston assembly as shown. The piston weighs 100 kg and the surrounding atmosphere is at a m = 100 kg
pin
pressure of 100 kPa. The cross-­‐sectional area of the piston face is 100 cm2. For the following calculations, nitrogen 100 kPa
may be assumed to behave as an ideal gas and any friction 300 K
may be neglected. The acceleration due to gravity is 9.8 m 4.00 L
s–2. (a) The retaining pin is removed and the gas is Area = 0.01 m2
spontaneously compressed until the piston comes to rest and the system re-­‐attains thermal equilibrium at 300 K. How much work will have been done on the gas? (b) If the gas were compressed isothermally and reversibly between the same initial and final states as in (a), how much work would have to be done on the gas? Solution (a) The piston will compress the gas until the final pressure of the gas (P2) is the same as the external pressure (Pext ), which is the sum of the atmospheric pressure plus the force exerted by the mass of the piston: Thus P2 = Pext = Patm + Ppiston = Patm + mg
(100)(9.80)
= 100 ! 103 + A
0.01
= 105 + 0.98 ! 105 = 1.98 ! 105 Pa Therefore: ∆V = V2 – V1 = P1V1
(10 5 )(0.004)
– V1 = – 0.00400 P2
1.98! 105
= 0.00202 – 0.00400 = –0.00198 m3 Page 6 of 11
W = –Pext∆V = – (1.98 ! 105)(–0.00198) = + 392 J The positive sign indicates that work must be done on the gas. Ans: + 392 J (b) At constant temperature: P1V1 = P2 V2 = nRT " V2 %
" P1 %
" 10 5 %
5
= +273.2 J W = ! nRT ln $ ' = ! P1V1 ln $ ' = ! (10 )(0.004)ln $
# 1.98(10 5 &
# V1 &
# P2 &
Ans: + 273 J 7. The molar heat capacity at constant pressure for ammonia (NH3) vapour is c P = 29.75 + 0.0251T , where cP is in J K–1 mol–1 and T is in kelvin. Assuming ammonia behaves as an ideal gas, calculate the change in internal energy ∆U when 10.0 moles of ammonia vapour is heated at a constant pressure of 0.100 bar from 25.0°C to 75.0°C. Solution T1 = 25.0 + 273.15 = 298.15 K T2 = 75.0 + 273.15 = 348.15 K For an ideal gas, T2
∆U = n ! cV dT T
1
cV = cP – R = (29.75 + 0.0251T ) – 8.314 = 21.436 + 0.0251T 0.0251 2
Therefore, ∆U = n "#21.436(T2 ! T1 ) +
(T2 ! T12 )%&' $
2
0.0251
= (10.0) "#21.436( 348.15 ! 298.15) +
348.152 ! 298.152 )%& (
$
'
2
= (10.0) {1071.8 + 405.55} = (10.0) {1477.35} = + 14 773.5 J Ans: + 14.77 kJ Alternatively: ∆H = 18 930.5 J ∆U = ∆H – ∆(PV) = ∆H – ∆(nRT) = ∆H – nR∆T = 18 930.5 – (10)(8.314)(50.0) = 18 930.5 – 4157 = 14 773.5 J = 14.77 kJ [As before.] Page 7 of 11
8. A 350 g metal bar initially at 1000 K is removed from a furnace and quenched by quickly immersing it in a well-­‐insulated closed tank containing 10.00 kg of water initially at 300 K. The heat capacities of the metal and the water are constant at 0.45 J g–
1 K–1 and 4.18 J g–1 K–1, respectively. Neglecting any heat loss from the tank or water lost by vaporization, what is the final equilibrium temperature in the tank? Solution Thermal energy leaves the metal and enters the water until both are at the same final temperature Tf . At constant pressure the heat transferred is just the change in enthalpy. Furthermore, since the tank is insulated, no thermal energy leaves the tank. Thus ∆Hmetal + ∆Hwater = 0 i.e., (mcP !T )metal + (mcP !T )water = 0 (350)(0.45)(Tf – 1000) + (10 000)(4.18)(Tf – 300) = 0 157.5Tf – 157 500 + 41 800Tf – 12 540 000 = 0 41 957.5Tf = 12 697 500 Tf = 12 697 500
= 302.6277 K = 29.4777°C 41 957.5
Ans: 29.48°C 9. One mole of nitrogen gas at 250 K occupies a volume of 10.0 L in the frictionless piston/cylinder arrangement shown. The mass of the piston may be assumed to be negligible. The cross-­‐sectional area of the piston face is 300 cm2, and the pressure of the external atmosphere is 100 kPa. When a 500 kg weight is placed on the piston, the piston lowers, compressing the gas. The process may be assumed to be isothermal, and nitrogen may be assumed to behave as an ideal gas. The acceleration due to gravity is 9.80 m s–2. Patm = 100 kPa
N2 gas
1.0 mol
250 K
10.0 L
restraining
pin
(a) How much work is done on the gas? (b) What is the minimum work required to carry out this compression? Page 8 of 11
Solution (a) W = –Pext ∆Vsyst The external pressure is the sum of the atmospheric pressure plus the pressure resulting from the 500 kg weight being placed on the piston. The initial pressure of the gas is nRT (1.0)(8.3145)(250)
=
= 207 863 Pa V1
0.0100
The compression will stop when the final gas pressure P2 is in mechanical equilibrium with the external pressure causing the compression. Therefore, P1 =
P2 mg
(500)(9.8)
= 100 000 +
A
0.0300
= 100 000 + 163 333 = 263 333 Pa = Patm + Pweight = Patm +
Isothermal process: Therefore, P1V1 = P2V2 V2 =
P1 V1 (207 863)(0.0100)
=
= 0.007894 m3 P2
263 333
For the gas, ∆V = V2 – V1 = 0.007894 – 0.0100 = – 0.002106 m3 Therefore, W = – Pext ∆V = – P2 ∆V = – (263 333)(– 0.002106) = + 554.58 J The positive sign indicates a work input. Ans: + 555 J (b) The minimum work for the isothermal compression of an ideal gas is W "V %
" 0.007894 %
= ! nRT ln $ 2 ' = ! (1.00)(8.3145)(250) ln
# 0.0100 &
# V1 &
= + 491.56 J A reversible compression always requires less work than a real compression. Ans: + 492 J Page 9 of 11
10. Problem 8–17 (SOLUTION GIVEN IN MIDTERM PREPARATION SOLUTIONS) A closed system undergoes a cycle consisting of two processes. During the first process, which starts at an initial system pressure of 1.00 atm and system temperature of 300 K, 60 kJ of heat is transferred from the system while 25 kJ of work is done on the system. At the end of the first process, the system pressure is 0.50 atm. During the second process, 15 kJ of work is done by the system. The temperature and pressure of the surroundings are constant throughout at 300 K and one atm, respectively. (a) Determine Qsyst for the second process. (b) Determine Qsyst for the cycle. (c) Determine Wsyst for the cycle. (d) Determine ∆Usyst for the cycle. (e) Determine ∆Hsyst for the cycle. Solution For a closed system going through a cycle, ∆Ucycle = ∑ (Q + W) = 0 (a) Let the first process be process “A” and the second process be process “B.” ∆Ucycle = 0 = ∑ (Q + W) QA + QB + WA + WB = ( – 60) + QB + (+ 25) + (– 15) = 0 Therefore, QB = + 60 – 25 + 15 = + 50 kJ (b) Ans: Qcycle = Qnet = QA + QB = (– 60) + (50) = – 10 kJ (c) + 50 kJ Ans: – 10 kJ Wcycle = Wnet = WA + WB = (+ 25) + (– 15) = + 10 kJ Ans: + 10 kJ (d) For a cyclic process, the system is restored to its original state, so that all state properties of the system have no net change. Therefore, ∆Usyst = 0. Ans: 0 Ans: 0 (e) Similarly, ∆Hsyst = 0. Page 10 of 11
3. Part A: Calculate the pressure in MPa using the following gas laws: (a) the ideal gas law (b) the van der Waals equation (c) the virial equation Part B: Assuming the pressure calculated using the virial equation is the “true” value, determine the % error in the pressure by using the ideal gas law and the van der Waals equation. For neon, a = 0.02135 Pa m6 mol-­‐2, b = 17.09 x 10-­‐6 m3 mol-­‐1. For neon at 25oC, values for B(T) and C(T) are 11.42 cm3 mol-­‐1 and 221 cm6 mol-­‐2, respectively. Page 11 of 11