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58 2 INTERNAL ENERGY
Solutions to problems
P2E.2 In a reversible adiabatic expansion the initial and �nal states are related by
[2E.2a–68], (Tf/Ti) = (Vi/Vf)1/c where c = CV ,m/R. Taking the logarithm
of both sides and using the relationship ln xa = a ln x gives
ln(Tf
Ti
) = 1
c
ln(Vi
Vf
)
It follows that
c = ln (Vi/Vf)
ln (Ti/Tf)
= ln (1/2)
ln (248.44 K/298.15 K)
= 3.80...
CV ,m is calculated from c as follows
CV ,m = cR = (3.80...) × (8.3145 JK−1mol−1) = 31.5... J K−1mol−1
Likewise, the initial and �nal states are related by [2E.3–68], piV γ
i = pfV γ
f ,
where γ = Cp ,m/CV ,m. It follows that ln(pi/pf) = γ ln(Vf/Vi), hence
γ = ln (pf/pi)
ln (Vi/Vf)
= ln (81.840 kPa/202.94 kPa)
ln (1/2)
= 1.31...
Hence
Cp ,m = CV ,mγ = (31.5... J K−1mol−1) × (1.31...) = 41.397 JK−1mol−1
Integrated activities
I2.2 (a) �e table belowdisplays computed enthalpies of formation (semi-empirical,
PM3 level, PC Spartan ProTM), enthalpies of combustion based on them
(andon experimental enthalpies of formation ofH2O(l),−285.83 kJmol−1,
and CO2(g), −393.51 kJmol−1), experimental enthalpies of combustion
(from the Resource section), and the relative error in the enthalpy of com-
bustion.
compound ∆fH−○/kJmol−1 ∆cH−○/kJmol−1 ∆cH−○/kJmol−1 % error
(calc.) (expt.)
CH4(g) −54.45 −910.7 −890 2.33
C2H6(g) −75.88 −1568.6 −1560 0.55
C3H8(g) −98.84 −2225.0 −2220 0.23
C4H10(g) −121.6 −2881.6 −2878 0.12
C5H12(g) −142.1 −3540.4 −3537 0.10
�e combustion reactions can be expressed as
CnH2n+2(g)+ 12 (3n + 1)O2(g)ÐÐ→ nCO2(g)+(n+1)H2O(l)