pchem11e-student-answers-a4

Student Solutions
Manual to
Accompany Atkins’
Physical Chemistry
ELEVENTH EDITION
Peter Bolgar
Haydn Lloyd
Aimee North
Vladimiras Oleinikovas
Stephanie Smith
and
James Keeler
Department of Chemistry
University of Cambridge
UK
Numerical solutions to the problems
compiled by Jack Entwistle
Selwyn College and the Department of Chemistry
University of Cambridge

Preface
is document is a compilation of the numerical solutions to the (a) Exercises and the
odd-numbered Discussion questions and Problems from the th edition of Atkins’ Physical
Chemistry. Where a problem requests the derivation of a result or expression, and provided
that expression is not too complex, we have also included such results.
Errors and omissions
In such a complex undertaking some errors will no doubt have crept in, despite the authors’
best eorts. Readers who identify any errors or omissions are invited to pass them on to us
by email to .
Jack Entwistle
James Keeler
Cambridge, August 

1
1 The properties of gases
1A The perfect gas
E1A.1(a)  Torr . atm
E1A.2(a) no . atm
E1A.3(a) . bar . atm
E1A.4(a) 30 lb in−2.
E1A.5(a) . bar 4.27 ×105Pa
E1A.6(a) S8.
E1A.7(a) . kg
E1A.8(a) xO2=0.240 xN2=0.760 pO2=0.237 bar pN2=0.750 bar xN2=0.780
xO2=0.210 pN2=0.770 bar pO2=0.207 bar
E1A.9(a) . kg mol−1
E1A.10(a) θ=−273 ○C
E1A.11(a) xH2=2
3xN2=1 3pH2=2.0×105Pa pN2=1.0×105Pa ptot =3.0×105Pa
P1A.1 1.15 ×105Pa . J K−1mol−1
P1A.3 0.082062 atm dm3mol−1K−1
P1A.5 p=ρRTM. g mol−1
P1A.7 24.5 Pa 9.14 kPa 24.5 Pa
P1A.9 between . km3and . km3
P1A.11 −2 Pa 0.25 atm
P1A.13 cCCl3F=1.1 ×10−11 mol dm−3cCCl2F2=2.2 ×10−11 mol dm−3cCCl3F=8.0 ×
10−13 mol dm−3cCCl2F2=1.6 ×10−12 mol dm−3
1B The kinetic model
E1B.1(a) .
E1B.2(a) υrms,H2=1.90 km s−1υrms,O2=478 m s−1
E1B.3(a) 6.87 ×10−3
E1B.4(a) 1832 m s−1
E1B.5(a) υmp =333 m s−1υmean =376 m s−1υrel =531 m s−1
E1B.6(a) 1.7 ×1010 s−1
E1B.7(a) 475 m s−1. nm 8.10 ×109s−1
E1B.8(a) . Pa
E1B.9(a) 1.4 ×10−6m = . µm
P1B.3 υmean, new ≈0.493 υmean
P1B.5 3.02 ×10−3for n=3 4.89 ×10−6for n=4
P1B.7 1.12 ×104m s−15.04 ×103m s−1

2
P1B.9 . at  K . at  K
P1B.11 9.7 ×1010 s−1
1C Real gases
E1C.1(a) 0.99 atm 1.8 ×103atm
E1C.2(a) a=0.0761 kg m5s−2mol−2b=2.26 ×10−5m3mol−1
E1C.3(a) . . dm3mol−1
E1C.4(a) 140 atm
E1C.5(a) . atm 35.2 atm .
E1C.6(a) . atm dm6mol−2. dm3mol−1 pm
E1C.7(a) 1.41 ×103K  pm
E1C.8(a) . atm 3.6 ×103K . atm 2.6 ×103K . atm  K
E1C.9(a) 4.6 ×10−5m3mol−1.
P1C.1 . atm
P1C.3 . . dm3mol−1
P1C.5 . K
P1C.7 . dm3mol−1. .
P1C.9 . dm3mol−1. atm dm6mol−2. . atm
P1C.11 . dm3mol−1. atm dm6mol−2
P1C.13 Vm=3CB T =B23CR p =B327C2
P1C.15 B′=0.0868 atm−1B=2.12 dm3mol−1
P1C.19 1+bp
RT 1.11
P1C.21 −0.01324 dm3mol−11.063 ×10−3dm6mol−2
P1C.23 Vm=13.6 dm3mol−1%
I1.1 υ=2RT
M1/2
I1.3 . dm3mol−1

3
2 Internal energy
2A Internal energy
E2A.1(a) 8.7 kJ mol−17.4 kJ mol−17.4 kJ mol−1
E2A.3(a) −76 J
E2A.4(a) q=+2.68 kJ w=−2.68 kJ ∆U=0q=+1.62 kJ w=−1.62 kJ ∆U=0
q=0w=0 ∆U=0
E2A.5(a) pf=1.33 atm ∆U=+1.25 kJ q=+1.25 kJ w=0
E2A.6(a) −88 J −1.7 ×102J
P2A.1 6.2 kJ mol−1
P2A.3 1
2al2−2 5bl 5
2
P2A.7 −1.7 kJ −1.8 kJ −1.5 kJ
P2A.9 −1.5 kJ −1.6 kJ
2B Enthalpy
E2B.1(a) Cp,m =30 J K−1mol−1CV,m =22 J K−1mol−1
E2B.2(a) −5.0 kJ mol−1
E2B.3(a) qp=+10.7 kJ w=−624 J ∆U=+10.1 kJ ∆H=+10.7 kJ qV=+10.1 kJ
w=0 ∆U=+10.1 kJ ∆H=+10.7 kJ
E2B.4(a) qp=∆H=+2.2 kJ ∆U=+1.6 kJ
P2B.1 11 min
P2B.3 62.2 kJ
P2B.5 w=0 ∆U=qV=+2.35 kJ ∆H=3.0 kJ
2C Thermochemistry
E2C.1(a) q=∆H=+22.5 kJ w=−1.6 kJ ∆U=+21 kJ
E2C.2(a) −4.57 ×103kJ mol−1
E2C.3(a) −167 kJ mol−1
E2C.4(a) 1.58 kJ K−1+3.07 K
E2C.5(a) ∆rH−○ (3)=−114.40 kJ mol−1∆rU−○ =−112 kJ mol−1∆fH−○ (HCl, g)=−92.31 kJ mol−1
∆fH−○ (H2O, g)=−241.82 kJ mol−1
E2C.6(a) −1368 kJ mol−1
E2C.7(a) ∆rH−○ (298 K)=+131.29 kJ mol−1∆rU−○ (298 K)=+128.81 kJ mol−1∆rH−○ (478 K)=
+134.1 kJ mol−1∆rU−○ (478 K)=+130 kJ mol−1
E2C.8(a) −394 kJ mol−1
P2C.1 37 K 4.1 kg