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50 2 INTERNAL ENERGY
�e van der Waals parameter a of sulfur dioxide is found in Table 1C.3, and
needs to be converted to SI units
a = (6.775 dm6 atmmol
−2
) × (10
−6 m6
1 dm6
) × (1.01325 × 10
5 Pa
1 atm
)
= 0.686... m6 Pamol−2
�erefore the internal pressure is
πT =
a
V 2m
= (0.686... m6 Pamol−2)
(2.47... × 10−2 m3mol−1)2
= 1.12 × 103 Pa
E2D.2(b) �e internal energy of a closed system of constant composition is a function of
temperature and volume. For a change in V and T , dU is given by [2D.5–61],
dU = πTdV + CVdT . At constant temperature, this reduces to dU = πTdV .
Substituting in the given expression for πT for a van der Waals gas and using
molar quantities
dUm = a
V 2m
dVm
�is expression is integrated between Vm,i and Vm,f to give
∫
Vm,f
Vm,i
dUm = ∫
Vm,f
Vm,i
a
V 2m
dVm
hence
∆Um = − a
V 2m
∣
Vm,f
Vm,i
= −a ( 1
Vm,f
− 1
Vm,i
)
�e van der Waals parameter a for argon is found in Table 1C.3, and needs to
be converted to SI units
a = (1.337 dm6 atmmol
−2
) × (10
−6 m6
1 dm6
) × (1.01325 × 10
5 Pa
1 atm
)
= 0.135... m6 Pamol−2
∆Um = −(0.135... m6 Pamol−2) ( 1
30.00 × 10−3 m3mol−1
− 1
1.00 × 10−3 m3mol−1
)
= +1.30... × 102 Jmol−1 = +131 Jmol−1
�e work done by an expanding gas is given by [2A.5a–39], dw = −pexdV . For
a reversible expansion pex is the pressure of the gas, hence
w = −∫ pdVm
Substituting in the expression for the pressure of a van der Waals gas, [1C.5b–
24]
w = −∫
RT
Vm − b
− a
V 2m
dVm = −∫
RT
Vm − b
dVm + ∫
a
V 2m
dVm
= −∫
RT
Vm − b
dVm + ∆Um