Mostre que as capacidades térmicas podem ser escritas das seguintes maneiras:
C_{V}=-T\left(\frac{\partial^{2} F}{\partial T^{2}}\right)_{V} \quad \text { e } \quad C_{P}=-T\left(\frac{\partial^{2} G}{\partial T^{2}}\right)_{P}

Passo 1

O enunciado pede para mostrarmos duas expressões para as capacidades térmicas em função das derivadas segundas dos potenciais termodinâmicos F (energia livre de Helmholtz) e G (energia livre de Gibbs):
C_{V}=-T\left(\frac{\partial^{2}F}{\partial T^{2}}\right)_{V} \quad\text{e}\quad C_{P}=-T\left(\frac{\partial^{2}G}{\partial T^{2}}\right)_{P}.
Para isso vamos usar as definições
F = U - T S, \quad G = H - T S,
e as formas diferenciais
dF = -S\,dT - P\,dV, \quad dG = -S\,dT + V\,dP.

Passo 2

Derivação de C_{V}:
Por definição
C_{V} = \Bigl(\frac{\partial U}{\partial T}\Bigr)_{V}.
Como F = U - T S, temos U = F + T S. Logo, em volume fixo,
\Bigl(\frac{\partial U}{\partial T}\Bigr)_{V} = \Bigl(\frac{\partial F}{\partial T}\Bigr)_{V} +T\,\Bigl(\frac{\partial S}{\partial T}\Bigr)_{V} +S.
Mas da diferencial de F,
dF = -S\,dT - P\,dV \quad\Longrightarrow\quad S = -\Bigl(\frac{\partial F}{\partial T}\Bigr)_{V}
e
\Bigl(\frac{\partial S}{\partial T}\Bigr)_{V} =-\Bigl(\frac{\partial^2 F}{\partial T^2}\Bigr)_{V}.
Substituindo em \partial U/\partial T:
\Bigl(\frac{\partial U}{\partial T}\Bigr)_{V} = \Bigl(\frac{\partial F}{\partial T}\Bigr)_{V} +T\bigl[-\Bigl(\frac{\partial^2 F}{\partial T^2}\Bigr)_{V}\bigr] + \Bigl[-\Bigl(\frac{\partial F}{\partial T}\Bigr)_{V}\Bigr] =-\,T\Bigl(\frac{\partial^2 F}{\partial T^2}\Bigr)_{V}.
Portanto
C_{V}=-\,T\Bigl(\frac{\partial^2 F}{\partial T^2}\Bigr)_{V}.

Passo 3

Derivação de C_{P}:
Por definição
C_{P} = \Bigl(\frac{\partial H}{\partial T}\Bigr)_{P},
onde H=U+P\,V. Como G = H - T S, segue H = G + T S. Em pressão fixa,
\Bigl(\frac{\partial H}{\partial T}\Bigr)_{P} = \Bigl(\frac{\partial G}{\partial T}\Bigr)_{P} +T\,\Bigl(\frac{\partial S}{\partial T}\Bigr)_{P} +S.
Da diferencial de G,
dG = -S\,dT + V\,dP \quad\Longrightarrow\quad S = -\Bigl(\frac{\partial G}{\partial T}\Bigr)_{P}, \quad \Bigl(\frac{\partial S}{\partial T}\Bigr)_{P} =-\Bigl(\frac{\partial^2 G}{\partial T^2}\Bigr)_{P}.
Logo
\Bigl(\frac{\partial H}{\partial T}\Bigr)_{P} = \Bigl(\frac{\partial G}{\partial T}\Bigr)_{P} +T\bigl[-\Bigl(\frac{\partial^2 G}{\partial T^2}\Bigr)_{P}\bigr] + \Bigl[-\Bigl(\frac{\partial G}{\partial T}\Bigr)_{P}\Bigr] =-\,T\Bigl(\frac{\partial^2 G}{\partial T^2}\Bigr)_{P}.
Assim
C_{P}=-\,T\Bigl(\frac{\partial^2 G}{\partial T^2}\Bigr)_{P}.

Resposta

C_{V}=-T\left(\frac{\partial^{2} F}{\partial T^{2}}\right)_{V}, \quad C_{P}=-T\left(\frac{\partial^{2} G}{\partial T^{2}}\right)_{P}.

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