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

1

, 1

The properties of gases
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1A
@




The perfect gas Answe z@ z@ z @




rs to discussion questions
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1A.2 z @ z @ z @ z @ The partial pressure of a gas in a mixture of gases is the pressure the gas would ex
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ert if it
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occupied alone the same container as the mixture at the same temperature. Dalton‘s la
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w is a limiting law because it holds exactly only under conditions where the gases hav
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e no effect upon each other. This can only be true in the limit of zero pressure where
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the molecules of the gas are very far apart. Hence, Dalton‘s law holds exactly only for
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@a mixture of perfect gases; for real gases, the law is only an approximation.
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Solutions to exercises z@ z@




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




V
All quantities on the right are given to us except n, which can be computed from the
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given mass of Ar. z@ z@ z@




n 
 0626 mol
z@
z@ z@




25 g 39 z@ z@




1
95 g mol z@ z@










so (0626 mol)  (831  102 dm3 bar K1 mol1)  (30  273) K

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10.5bar
p z@
15 dm3 z@






So no, the sample would not exert a pressure of 2.0 bar.
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1A.2(b) Boyle‘s law [1A.4a] applies. z@ z@ z@




2

, pV = constantz@ z@ so pfVf = piVi z@ z@




Solve for the initial pressure:
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pfVf (197 bar)  (214 dm3 )
pi  
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z@ z@
(i) z@



1.07 bar z@

z @
 (214  180) dm3
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Vi
(ii) The original pressure in Torr is
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p  (1.07 bar) 1 atm  803 Torr
 760 Torr 
z@


 
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z@





 z@





 




i  1.013 bar  
z@ z@ z@ z @ 1 atmz@ z @



1A.3(b)

The relation between pressure and temperature at constant volume can be derived from
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the perfect gas law, pV = nRT [1A.5]
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pi pf
so p  T and
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z @
 z @




Ti Tf
The final pressure, then, ought to be
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piTf (125 kPa)  (11  273)K
 
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p z @ z @ z@




Ti 
120 kPaz@


(23  273)K z@ z@




1A.4(b)

According to the perfect gas law [1.8], one can compute the amount of gas from press
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ure, temperature, and volume.
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pV = nRT z@ z@




so (1.00 atm)  (1013  105

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z@




Pa atm1)  (400
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n
 103 m3 ) (8.314
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@ z@ z@ z@ z@





5 J K1mol1) 
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pV
(20  273)K
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z@

R

T
3

,  166  105 mol
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Once this is done, the mass of the gas can be computed from the amount and
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@ the molar mass:
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1
m  (166  105 mol)  (16.04 g mol )  267  106 g 2.67
z@ z@ z@ z@ z@ z@ z@   103 k z@ z@ z@ z@ z@ z@ z@ z@ z@ z@ z@




1A.5(b)

The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making
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the total pressure
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4