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**Unformatted text preview: **14 STUDENT'S SOLUTIONS MANUAL: PHYSICAL CHEMISTRY
Solutions to exercises
E1.1(a)
(a) The perfect gas equation [1.8] is pV = nRT
Solving for the pressure gives p = -
ART
V
The amount of xenon is n =
131g
131 g mol-1
7 = 1.00 mol
p =
(1.00 mol) x (0.0821 dm3 atm K-1 mol-1) x (298.15 K)
1.0 dm3
24 atm
That is, the sample would exert a pressure of 24 atm if it were a perfect gas, not 20 atm.
(b) The van der Waals equation [1.21a] for the pressure of a gas is p =
nRT
an2
V - nb
V2
For xenon, Table 1.6 gives a = 4.137 dm atm mol-2 and b = 5.16 x 10-2 dm3 mol-1.
Inserting these constants, the terms in the equation for p become
nRT
(1.00 mol) x (0.08206 dm' atm K-1 mol-) x (298.15 K)
V - nb
1.0 dm' - {(1.00 mol) x (5.16 x 10-2 dm' mol-1)}
2 =25.8 atm
an? _(4.137 dm atm mol-2) x (1.00 mol)2
V 2
(1.0 dm3)2
- = 4.137 atm
Therefore, p = 25.8 atm - 4.137 atm = 22 atm
E1.2(a)
Boyle's law [1.5] applies.
pV = constant so pfVr = p;Vi
This equation can be solved for either initial or final pressure, hence
Pi = V.
V. = 4.65 dm3, V; = (4.65 + 2.20) dm' = 6.85 dm', pf = 5.04 bar
Therefore,
(a) pi =
4.65 dm3
6.85 dm3
x (5.04 bar) = 3.42 bar
(b) Since 1 atm = 1.013 bar, p; = 3.42 bar x
1 atm
1.013 bar
3.38 atm
E1.3(a)
The perfect gas law, PV = nRT [1.8], can be rearranged to P = nk
constant. Hence, ~
Pr = Pi or, solving for Po Pr= T.
It xpi
V
= constant, if n and V are...

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- Fall '19