Demonstre as seguintes expressões:
\frac{\left(\frac{\partial V}{\partial T}\right)_{S}}{\left(\frac{\partial V}{\partial T}\right)_{P}}=\frac{1}{1-\gamma}

Passo 1

O enunciado pede para mostrar que
\frac{\bigl(\tfrac{\partial V}{\partial T}\bigr)_{S}}{\bigl(\tfrac{\partial V}{\partial T}\bigr)_{P}} =\frac{1}{1-\gamma},
onde \gamma = C_{P}/C_{V}. Vamos usar relações termodinâmicas fundamentais: o diferencial de entropia e de pressão, além da identidade
C_{P}-C_{V} = T\bigl(\tfrac{\partial P}{\partial T}\bigr)_{V} \bigl(\tfrac{\partial V}{\partial T}\bigr)_{P}.

Passo 2

Começamos pelo diferencial de entropia a partir da energia interna U. A forma fundamental é
T\,dS = dU + P\,dV \quad\Longrightarrow\quad T\,dS = C_{V}\,dT + T\Bigl(\tfrac{\partial P}{\partial T}\Bigr)_{V}dV.

Passo 3

Em um processo isotérmico de entropia constante ( dS=0), da última equação vem
0 = C_{V}\,dT + T\Bigl(\tfrac{\partial P}{\partial T}\Bigr)_{V}dV \quad\Longrightarrow\quad \Bigl(\tfrac{\partial V}{\partial T}\Bigr)_{S} = -\frac{C_{V}}{T\;(\tfrac{\partial P}{\partial T})_{V}}.

Passo 4

Agora o diferencial de pressão em função de T e V é
dP = \Bigl(\tfrac{\partial P}{\partial T}\Bigr)_{V}dT + \Bigl(\tfrac{\partial P}{\partial V}\Bigr)_{T}dV.
Para processo isobárico ( dP=0) temos
0 = (\tfrac{\partial P}{\partial T})_{V}dT + (\tfrac{\partial P}{\partial V})_{T}dV \quad\Longrightarrow\quad \Bigl(\tfrac{\partial V}{\partial T}\Bigr)_{P} = -\frac{(\tfrac{\partial P}{\partial T})_{V}}{(\tfrac{\partial P}{\partial V})_{T}}.

Passo 5

A identidade bem conhecida entre C_{P} e C_{V} é
C_{P}-C_{V} = T\bigl(\tfrac{\partial P}{\partial T}\bigr)_{V} \bigl(\tfrac{\partial V}{\partial T}\bigr)_{P}.
Substituindo \bigl(\tfrac{\partial V}{\partial T}\bigr)_{P} do passo anterior, obtemos
C_{P}-C_{V} = -\,T\;\frac{\bigl(\tfrac{\partial P}{\partial T}\bigr)_{V}^{2}} {\bigl(\tfrac{\partial P}{\partial V}\bigr)_{T}} \quad\Longrightarrow\quad \frac{T\,(\tfrac{\partial P}{\partial T})_{V}^{2}} {(\tfrac{\partial P}{\partial V})_{T}} = -\,(C_{P}-C_{V}).

Passo 6

Agora formamos o quociente pedido:
\frac{\bigl(\tfrac{\partial V}{\partial T}\bigr)_{S}} {\bigl(\tfrac{\partial V}{\partial T}\bigr)_{P}} = \frac{-\dfrac{C_{V}}{T\,(\tfrac{\partial P}{\partial T})_{V}}} {-\dfrac{(\tfrac{\partial P}{\partial T})_{V}}{(\tfrac{\partial P}{\partial V})_{T}}} = \frac{C_{V}}{T}\;\frac{(\tfrac{\partial P}{\partial V})_{T}} {(\tfrac{\partial P}{\partial T})_{V}^{2}} = \frac{C_{V}}{-\,(C_{P}-C_{V})} = \frac{C_{V}}{C_{V}-C_{P}} = \frac{1}{1-\gamma}.

Resposta

\frac{\bigl(\tfrac{\partial V}{\partial T}\bigr)_{S}} {\bigl(\tfrac{\partial V}{\partial T}\bigr)_{P}} =\frac{1}{1-\gamma}.

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