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Solutions Manual – Physical Chemistry 10th edition by Paula & Atkins
Solutions Manual – Physical Chemistry 10th edition by Paula & Atkins
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Solutions Manual –Physical Chemistry 10th edition by Paula &
Atkins
1
, 1 The Properties Of
Gases1A The Perfect
Gas Answers To Discussion
Questions
1A.2 The Partial Pressure Of A Gas In A Mixture Of Gases Is The Pressure The Gas Would Exert If
It
Occupied Alone The Same Container As The Mixture At The Same Temperature. Dalton’s
Law Is A Limiting Law Because It Holds Exactly Only Under Conditions Where The Gases
Have No Effect Upon Each Other. This Can Only Be True In The Limit Of Zero Pressure
Where The Molecules Of The Gas Are Very Far Apart. Hence, Dalton’s Law Holds Exactly
Only For A Mixture Of Perfect Gases; For Real Gases, The Law Is Only An
Approximation.
Solutions To Exercises
1A.1(B) The Perfect Gas Law [1A.5] Is Pv = Nrt, Implying That The Pressure Would Be
Nrt
P=
V
All Quantities On The Right Are Given To Us Except N, Which Can Be Computed From
The Given Mass Of Ar.
25 G
N= = 0626 Mol
3995 G
−1
Mol
(0626 Mol) (831 10−2 Dm3 Bar K−1 Mol−1) (30 + 273) K
So P =10.5bar
= 15 Dm3
So No, The Sample Would Not Exert A Pressure Of 2.0 Bar.
1A.2(B) Boyle’s Law [1A.4a] Applies.
Pv = Constant So Pfvf = Pivi
Solve For The Initial Pressure:
Pv (197 Bar) (214 Dm3 )
(i) Pi = f f = = 1.07 bar
Vi (214 + 180) Dm3
(ii) The Original Pressure In Torr Is
= (1.07 Bar) 1
P 760 Torr
Atm =
I 1.013 Bar 1 Atm 803 Torr
1A.3(B) The Relation Between Pressure And Temperature At Constant Volume Can Be Derived
From ThePerfect Gas Law, Pv = Nrt [1A.5]
Pi Pf
So P T And =
Ti Tf
The Final Pressure, Then, Ought To Be
Pt (125 Kpa) (11 + 273)K
Pf = i f = = 120 kPa
Ti (23 + 273)K
1A.4(B) According To The Perfect Gas Law [1.8], One Can Compute The Amount Of Gas From
Pressure,Temperature, And Volume.
Pv = Nrt
So N = (1.00 Atm) (1013 105 Pa Atm−1) (400 103 M3
Pv = = 166 105 Mol
)(8.3145 J K−1mol−1) (20 + 273)K
RT
Once This Is Done, The Mass Of The Gas Can Be Computed From The Amount And The
MolarMass:
2
, −1
M = (166 105 Mol) (16.04 G Mol ) = 267 106 G =2.67 103 kg
1A.5(B) The Total Pressure Is The External Pressure Plus The Hydrostatic Pressure [1A.1],
Making TheTotal Pressure
3
, P = Pex + Gh .
Let Pex Be The Pressure At The Top Of The Straw And P The Pressure On The Surface Of The
Liquid(Atmospheric Pressure). Thus The Pressure Difference Is
−3 3
P− P = Gh = (10 G Cm ) 1 Kg 1 (981 M S−2 ) (015 M)
Cm
Ex
10 3 10−2 M
G
= 1.5 103 Pa = 1.5 10−2 Atm
1A.6(B) The Pressure In The Apparatus Is Given By
P = Pex + Gh [1A.1]
Where Pex = 760 Torr = 1 Atm = 1.013105 Pa, 3
1 Kg 1 Cm
And Gh = 13.55 G Cm−3 3 −2 0.100 M 9.806 M S−2 = 1.33 104 Pa
10 G 10 M
P = 1.013 105 Pa + 1.33 104 Pa = 1.146 105 Pa = 115 kPa
Pv Pvm
1A.7(B) Rearrange The Perfect Gas Equation [1A.5] To Give R = =
Nt T
All Gases Are Perfect In The Limit Of Zero Pressure. Therefore The Value Of Pvm/T
ExtrapolatedTo Zero Pressure Will Give The Best Value Of R.
The Molar Mass Can Be Introduced Through
M
Pv = Nrt = RT
M
M RT RT
Which Upon Rearrangement Gives M = =
V P P
The Best Value Of M Is Obtained From An Extrapolation Of /P Versus P To Zero Pressure;
TheIntercept Is M/RT.
Draw Up The Following Table:
P/Atm (Pvm/T)/(Dm3 Atm K–1 Mol– (/P)/(G Dm–3 Atm–
1 1
) )
0.750 000 0.082 0014 1.428 59
0.500 000 0.082 0227 1.428 22
0.250 000 0.082 0414 1.427 90
From Figure 1A.1(A), R = Lim PvM =
0.082 062 dm3 atm K−1 mol−1
P→0 T
Figure 1A.1
(A)
4
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