Solutions Manual – Atkins' Physical Chemistry, 12th Edition by Atkins | All 19 Chapters Covered

1

, 1

The properties of gase
x@ x@ x@




s 1A
x@




The perfect gas Answ
x@ x@ x @




ers to discussion questions
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1A.2 x @ x @ x @ x @ The partial pressure of a gas in a mixture of gases is the pressure the gas w
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@




ould exert if it
x@ x@ x@




occupied alone the same container as the mixture at the same temperature. Dalton‘
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s law is a limiting law because it holds exactly only under conditions where the
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gases have no effect upon each other. This can only be true in the limit of zero
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@




pressure where the molecules of the gas are very far apart. Hence, Dalton‘s law h
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olds exactly only for a mixture of perfect gases; for real gases, the law is only a
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@




n approximation.
x@




Solutions to exercises x@ x@




1A.1(b) x @ x @ The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@




nRT
p  x@ x@




V
All quantities on the right are given to us except n, which can be computed from
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@




x@ the given mass of Ar.
x@ x@ x@ x@




n 
 0626 mol
x@
x@ x@



25 g 39
x@ x@



1
95 g mol x@ x@









so (0626 mol)  (831  102 dm3 bar K1 mol1)  (30  273) K

x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@



10.5bar
p x@
15 dm3 x@





So no, the sample would not exert a pressure of 2.0 bar.
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2

, 1A.2(b) Boyle‘s law [1A.4a] applies.
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pV = constant
x@ x@ so pfVf = piVi x@ x@




Solve for the initial pressure:
x@ x@ x@ x@




pfVf (197 bar)  (214 dm3 )
pi  
x@ x@ x@ x@ x@ x @
x@
(i) x@


1.07 ba x@

x@x @
 (214  180) dm3 x@ x@ x@




Vi
(ii) The original pressure in Torr is
x@ x@ x@ x@ x@




p  (1.07 bar) 1 atm  803 Tor
 760 Torr 
x@

 
x @ x@ x@ x x@ x@ x@ x@ x @
x@




@ x@




 




i  1.013 bar 
x@ x@ x@ x @ 1 atm
x@ x @



1A.3(b)

The relation between pressure and temperature at constant volume can be derived
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from the perfect gas law, pV = nRT [1A.5]
x@ x@ x@ x@ x@ x@ x@ x@




pi pf
so p  T and
x@ x@
x @
 x @




Ti Tf
The final pressure, then, ought to be
x@ x@ x@ x@ x@ x@




piTf (125 kPa)  (11  273)K
 
x@ x@ x@ x@ x@
p x @ x @ x@





x@
Ti
120 kPx@

(23  273)K
x@ x@




1A.4(b)

According to the perfect gas law [1.8], one can compute the amount of gas from
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x




@pressure, temperature, and volume.
x@ x@ x@




pV = nRT x@ x@




V
RT  (1.00 atm)  (1013  1
x@
so x@
x@ x@ x@ x@ x@




05 Pa atm1)  (
x@ x@ x@ x@



n
400  103 m3 ) (
x@
x@ x@ x@ x@ x@




8.3145 J K1mol
x
x@ x@



p 1
@ )  (20  273
x@ x@ x@ x@




3

, )K
 166  105 mol
x@ x@ x@ x@




Once this is done, the mass of the gas can be computed from the amount
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x@ and the molar mass:
x@ x@ x@




1
m  (166  105 mol)  (16.04 g mol )  267  10
x@ x@ x@ x@ x@
6
2.67 g 103
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@
x@ x@ x@




1A.5(b)

The total pressure is the external pressure plus the hydrostatic pressure [1A.1], ma
x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@ x@




king the total pressure
x@ x@ x@




4