Energia Interna e Expansão Adiabática

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56 2 INTERNAL ENERGY
(i) Without any vibrational contribution: CV ,m = 1
2 × (3 + 2 + 0) × R = 5
2R.
For a perfect gas, [2B.9–49], Cp ,m = CV ,m + R, therefore
γ = CV ,m + R
CV ,m
=
5
2R + R
5
2R
= 7
5
≈ 1.40
(ii) With the vibrational contribution: �e number of vibrational modes is
νv = 3N − 5 for a linear molecule, where N is the number of atoms in the
molecule.�erefore νv = 3N − 5 = 4.�is gives CV ,m = 1
2 × (3 + 2 + 2 ×
4) × R = 13
2 R and thus Cp ,m = 15
2 R and so γ = 15/13 ≈ 1.15.
�e experimental value of γ for carbon dioxide is
γ =
Cp ,m
CV ,m
=
Cp ,m
Cp ,m − R
= (37.11 JK−1mol−1)
(37.11 JK−1mol−1) − (8.3145 JK−1mol−1)
= 1.29
which is somewhat closer to the value expected if vibration is neglected, but
does not match that value closely. �is indicates that vibrations may be con-
tributing somewhat - for example the lower frequency bending modes.
E2E.2(b) For 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. For a perfect gas Cp ,m −
CV ,m = R, [2B.9–49], so c = (Cp ,m − R)/R. Using the value of Cp ,m from the
Resource section gives
c = (37.11 JK−1mol−1) − (8.3145 JK−1mol−1)
(8.3145 JK−1mol−1)
= 3.463
�erefore
Tf = Ti (
Vi
Vf
)
1
c
= (298.15 K) × ( 500 cm3
2.00 × 103 cm3
)
1
3.463
= 200 K
E2E.3(b) For a reversible adiabatic expansion the initial and �nal states are related by
[2E.3–68], piV γ
i = pfV γ
f , where γ is the ratio of heat capacities, γ = Cp ,m/CV ,m.
�e initial volume of the sample is
Vi =
nRT
p
= (2.5 mol) × (8.3145 JK−1mol−1) × (325 K)
(2.40 × 105 Pa)
= 2.81... × 10−2 m3
= 28.1... dm3
For a perfect gas Cp ,m − CV ,m = R, hence
γ =
Cp ,m
CV ,m
=
Cp ,m
Cp ,m − R
= (20.8 JK−1mol−1)
(20.8 JK−1mol−1) − (8.3145 JK−1mol−1)
= 1.66...
�e �nal volume is given by
Vf = (28.1... dm3)×(2.40×105 Pa/1.50×105 Pa)1/1.66.. . = 37.3... dm3 = 37 dm3