School of Science and Mathematics

Lecture 2

Problem 5

a) 1 mmHg = 101325 Pa/760 = 133.3 Pa

b) 28.3 mmHg = 3773 Pa

Problem 6

a) 2200 Pa

b) 220 000 Pa

c) 330.3 Pa

Problem 7

a) 800.5 kJ

b)
D V = 40
L = 0.040 m³
p = 0.25 x 101325 Pa = 25331 Pa
\ w = 1013.25 J

Problem 8

a) final speed = 96 kmh = 96000 m /3600 s = 26.67 m/s
acc’n = 26.67 m/s / 4 s = 6.67 m/s² = 0.68 "gravities"

b) 250 kg x 9.81 m.s^-2 x 2 m = 4905 N.m = 4905 J

c) w = 2 x 101325 Pa x 5 x 10^-4 m² x 0.3 m = 30.4 J

d) 706.5 kPa : 6.97 i.e. 7 atm

Problem 9

done in class

Problem 10

a)
10 dm = 1 m
100 dm² = 1 m²
1000 L = 1 m³

b) 20

c) 10⁴cm² = 1 m²
10⁶ cm³ = 1 m³

Problem 11

R = 8.313 J.mol^-1K^-1

Problem 12

n = 0.0157 mol

Problem 13

a) 62.4 J
b) 334.4 kJ
c) 16.72 GJ

Problem 14

Note use of units

heat cap. = 9.201 J / 1.652 ° C = 5.5696 J/° C
spec. heat cap. = (5.5696 J/° C) / 2.311 g = 2.410 J.° C^-1.g^-1 = 2.410 J.K^-1.g^-1

molar heat cap = 2.410 J.K^-1.g^-1 x 46.07 g.mol^-1 = 111.0 J.K^-1.mol^-1

Problem 15

2C_graphite + 3H_2,g + ½ O_2,g = C₂H₅OH_l
4C_graphite + 4H_2,g + O_2,g = CH₃.COOC₂H_5,l
H_2,g + ½ O_2,g = H₂O_l
Ba_s + S_s + 2 O_2,g = BaSO_4,s

Problem 16

2C_graphite + 3H_2,g = C₂H_6,g
D H_combust = 2 x D H_f(CO_2,g) + 3 x D H_f(H₂O_l) - D H_f(C₂H_6,g)
gives D H_f(C₂H_6,g) = -84.6 kJ/mol

Problem 16

CH₃CHO_l

formation: 2C_graphite + 2H_2,g + ½ O_2,g = CH₃CHO_l

combustion: CH₃CHO_l + 2½ O_2,g = 2 CO_2,g + 2 H₂O_l etc.

Problem 17

D H = 43.05 kJ/mol

Problem 18

a) D H_f(glycine) = -536.6 kJ/mol
b) D H_Rx = + 41.4 kJ/mol (endothermic)

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