Assuma que um gás satisfaz
p=\frac{m R T}{V-m b}-\frac{m^{2} a}{V^{2}}, \quad U=\frac{3}{2} m R T-\frac{m^{2} a}{V} .
Usando a regra da cadeia, encontre \left(\frac{\partial U}{\partial p}\right)_{m, T}.

Passo 1

O enunciado nos dá as dependências
p = \frac{mRT}{V - m b} \;-\;\frac{m^2 a}{V^2}, \quad U = \frac{3}{2}\,mRT \;-\;\frac{m^2 a}{V},
e pede o valor de \bigl(\tfrac{\partial U}{\partial p}\bigr)_{m,T}. Note que, para m e T fixos, tanto p quanto U são funções de V apenas. Vamos usar a regra da cadeia via V.

Passo 2

Pela regra da cadeia, mantendo m,T constantes, temos
\Bigl(\frac{\partial U}{\partial p}\Bigr)_{m,T} = \frac{\bigl(\partial U/\partial V\bigr)_{m,T}}{\bigl(\partial p/\partial V\bigr)_{m,T}}.

Passo 3

Calculamos \displaystyle \bigl(\frac{\partial U}{\partial V}\bigr)_{m,T}. Como
U(V) = \frac{3}{2}\,mRT - \frac{m^2 a}{V},
vem
\Bigl(\frac{\partial U}{\partial V}\Bigr)_{m,T} =0 \;-\; \bigl(-m^2a\bigr)\,V^{-2} = \frac{m^2a}{V^2}.

Passo 4

Agora \displaystyle \bigl(\frac{\partial p}{\partial V}\bigr)_{m,T}. De
p(V) = \frac{mRT}{V-mb} - \frac{m^2a}{V^2}
temos
\Bigl(\frac{\partial p}{\partial V}\Bigr)_{m,T} = -\,\frac{mRT}{(V - mb)^2} \;+\;2\,\frac{m^2a}{V^3}.

Passo 5

Substituindo na razão da cadeia,
\Bigl(\frac{\partial U}{\partial p}\Bigr)_{m,T} = \frac{\displaystyle \frac{m^2a}{V^2}} {\displaystyle -\frac{mRT}{(V-mb)^2} + \frac{2\,m^2a}{V^3}}.

Resposta

\Bigl(\frac{\partial U}{\partial p}\Bigr)_{m,T} = \frac{\tfrac{m^2a}{V^2}} {-\tfrac{mRT}{(V-mb)^2} \;+\;\tfrac{2\,m^2a}{V^3}}.

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