4.15 Admitindo que U é uma função de P e V, deduza as seguintes equações:
(a)
\mathrm{d} Q=\left(\frac{\partial U}{\partial P}\right)_{V} d P+\left[\left(\frac{\partial U}{\partial V}\right)_{P}+P\right] d V
(b)
\left(\frac{\partial U}{\partial P}\right)_{V}=\frac{C_{V K}}{\beta}
(c)
\left(\frac{\partial U}{\partial V}\right)_{P}=\frac{C_{P}}{V \beta}-P

Passo 1

O enunciado pede para assumir que a energia interna U é função das variáveis termodinâmicas pressão P e volume V, e daí deduzir três resultados:
(a) a expressão de dQ em termos de derivadas parciais de U;
(b) \displaystyle\bigl(\frac{\partial U}{\partial P}\bigr)_{V}=\frac{C_{V}K}{\beta};
(c) \displaystyle\bigl(\frac{\partial U}{\partial V}\bigr)_{P}=\frac{C_{P}}{V\beta}-P.
Aqui vamos usar a primeira lei dU=dQ-dW, cadeias de derivadas parciais, e as definições
\beta=\frac{1}{V}\Bigl(\frac{\partial V}{\partial T}\Bigr)_{P}, \quad K=-\frac{1}{V}\Bigl(\frac{\partial V}{\partial P}\Bigr)_{T},
além de
C_{V}=\Bigl(\frac{\partial U}{\partial T}\Bigr)_{V}, \quad C_{P}=\Bigl(\frac{\partial Q}{\partial T}\Bigr)_{P}.

Passo 2

Para (a), começamos pelo diferential de U como função de (P,V):
dU = \Bigl(\frac{\partial U}{\partial P}\Bigr)_{V}dP + \Bigl(\frac{\partial U}{\partial V}\Bigr)_{P}dV.
Pela primeira lei temos dU=dQ - dW e, para trabalho de expansão, dW = P\,dV. Logo
dQ = dU + P\,dV = \Bigl(\frac{\partial U}{\partial P}\Bigr)_{V}dP + \Bigl(\frac{\partial U}{\partial V}\Bigr)_{P}dV + P\,dV = \Bigl(\frac{\partial U}{\partial P}\Bigr)_{V}dP + \Bigl[\Bigl(\frac{\partial U}{\partial V}\Bigr)_{P} + P\Bigr]dV.
Isto prova a parte (a).

Passo 3

Para (b), fixamos V. Então em um processo isocórico dQ = dU e também dQ = C_{V}\,dT. De (a) com dV=0 vem
dQ = \Bigl(\frac{\partial U}{\partial P}\Bigr)_{V}dP.
Igualando as expressões de dQ:
\Bigl(\frac{\partial U}{\partial P}\Bigr)_{V}dP = C_{V}\,dT \quad\Longrightarrow\quad \Bigl(\frac{\partial U}{\partial P}\Bigr)_{V} = C_{V}\,\Bigl(\frac{\partial T}{\partial P}\Bigr)_{V}.

Passo 4

Ainda para (b), calculamos \bigl(\frac{\partial T}{\partial P}\bigr)_{V} via relação cíclica
\Bigl(\frac{\partial T}{\partial P}\Bigr)_{V} \Bigl(\frac{\partial P}{\partial V}\Bigr)_{T} \Bigl(\frac{\partial V}{\partial T}\Bigr)_{P} = -1.
Mas
\Bigl(\frac{\partial V}{\partial T}\Bigr)_{P}=V\,\beta, \quad \Bigl(\frac{\partial P}{\partial V}\Bigr)_{T} = \bigl[(\partial V/\partial P)_{T}\bigr]^{-1} =(-V\,K)^{-1} =-\frac{1}{V\,K}.
Substituindo na cíclica:
\Bigl(\frac{\partial T}{\partial P}\Bigr)_{V} \bigl(-\tfrac{1}{V\,K}\bigr) (V\,\beta) = -1 \quad\Longrightarrow\quad \Bigl(\frac{\partial T}{\partial P}\Bigr)_{V} = \frac{K}{\beta}.
Logo de \bigl(\frac{\partial U}{\partial P}\bigr)_{V}=C_{V}(\partial T/\partial P)_{V} resulta
\Bigl(\frac{\partial U}{\partial P}\Bigr)_{V} = C_{V}\,\frac{K}{\beta},
que é exatamente a parte (b).

Passo 5

Para (c), escrevemos U=U(T,V) e usamos a regra de cadeias em (\partial U/\partial V)_{P}:
\Bigl(\frac{\partial U}{\partial V}\Bigr)_{P} = \Bigl(\frac{\partial U}{\partial T}\Bigr)_{V} \Bigl(\frac{\partial T}{\partial V}\Bigr)_{P} + \Bigl(\frac{\partial U}{\partial V}\Bigr)_{T}.
Sabemos que
\displaystyle (\partial U/\partial T)_{V}=C_{V} e, da termodinâmica,
\displaystyle (\partial U/\partial V)_{T}=T\bigl(\tfrac{\partial P}{\partial T}\bigr)_{V}-P.
Além disso, análoga à parte anterior,
\Bigl(\frac{\partial T}{\partial V}\Bigr)_{P} = -\frac{(\partial P/\partial V)_{T}}{(\partial P/\partial T)_{V}} = \frac{1/(V\,K)}{\beta/K} = \frac{1}{V\,\beta}, \quad \Bigl(\frac{\partial P}{\partial T}\Bigr)_{V} = \frac{\beta}{K}.
Substituindo tudo:
\Bigl(\frac{\partial U}{\partial V}\Bigr)_{P} = C_{V}\,\frac{1}{V\,\beta} + \bigl[T\,\frac{\beta}{K}-P\bigr] = \frac{C_{V}}{V\,\beta} + \frac{T\,\beta}{K} - P.
Por fim usamos a identidade bem conhecida
C_{P}-C_{V} = \frac{T\,V\,\beta^{2}}{K} \quad\Longrightarrow\quad \frac{C_{V}}{V\,\beta} + \frac{T\,\beta}{K} = \frac{C_{V} + \tfrac{T\,V\,\beta^{2}}{K}}{V\,\beta} = \frac{C_{P}}{V\,\beta}.
Isso leva à forma final
\Bigl(\frac{\partial U}{\partial V}\Bigr)_{P} = \frac{C_{P}}{V\,\beta} \;-\; P,
que é exatamente a parte (c).

Resposta

(a) \displaystyle dQ=\Bigl(\frac{\partial U}{\partial P}\Bigr)_{V}dP+\bigl[\bigl(\frac{\partial U}{\partial V}\bigr)_{P}+P\bigr]dV
(b) \displaystyle \bigl(\frac{\partial U}{\partial P}\bigr)_{V}=\frac{C_{V}K}{\beta}
(c) \displaystyle \bigl(\frac{\partial U}{\partial V}\bigr)_{P}=\frac{C_{P}}{V\,\beta}-P

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