Química

CHAPTER 4 THE PROPERTIES OF GASES

4.1

(a) 8 × 109 Pa;

(b) 80 kilobars;

(c) 6 × 107 Torr;

(d) 1 × 106 lb ⋅ in −2

4.3

(a) The difference in column height will be equal to the difference in pressure between atmospheric pressure and pressure in the gas bulb. If the pressures were equal, the height of the mercury column on the air side and on the apparatus side would be the same. The pressure in the gas bulb is
0.890 atm or 0.890 × 760 Torr ⋅ atm −1 = 676 Torr. The difference would

be 762 Torr – 676 Torr = 86 Torr = 86 mm Hg. of the atmosphere.

(b) The side attached to

the bulb will be higher because the neon pressure is less than the pressure (c) If the student had recorded the level in the atmosphere arm to be higher than the level in the bulb arm by 86 mm Hg then the pressure in the bulb would have been reported as 762 Torr + 86 Torr = 848 Torr. 4.5 d1 h1 = d 2 h 2 73.5 cm × 13.6 g ⋅ cm −3 = 909 cm or 9.09 m 1.10 g ⋅ cm −3

4.7

(20. in)(10. in)(14.7 lb ⋅ in 2 ) = 2.9 × 103 lb

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4.9

(a) Volume, L 0.01 0.02 0.03 0.04 0.05

nR , atm ⋅ K −1 V

8.21 4.10 2.74 2.05 1.64 nR V

(b) The slope is equal to

(c) The intercept is equal to 0.00 for all the plots. 4.11 (a) from PV1 = P2V2 , we have 1
(2.0 × 105 kPa) (7.50 mL) = (P2 ) (1000 mL); solving for P2 we get

1.5 × 103 kPa;

(b) similar to (a),

PV1 = P2V2 or (643 Torr) (54.2 cm 3 ) = ( P2 )(7.8 cm 3 ), P2 = 4.5 × 103 Torr 1 4.13 Using

P P2 P2 1.10 atm 1 and expressing T in Kelvins, ; P2 = 3.31 atm = = 298 K 898 K T1 T2
4.15 Using

P P2 P2 1.5 atm 1 and expressing T in Kelvins, ; P2 = 1.6 atm = = 283 K 303 K T1 T2 V1 V2 V1 V2 = . = , or 0.100 mol 0.110 mol n1 n2 n2V1 0.110V1 = = 1.10V1 . 0.10 n1

4.17

If P and T are constant, then

Solving for V2 in terms of V1 , we obtain V2 =

So the volume must be increased by 10.% to keep P and T constant.

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4.19

(a) Because P, V, and T all change, we use the relation Substituting for the