O ar está expandindo em um sistema de pistão–cilindro de modo que P V^{2}=2, \operatorname{com} P \mathrm{em} \mathrm{kPa} \mathrm{e} V em \mathrm{m}^{3}. O trabalho, quando o volume muda de 0,04 \mathrm{~m}^{3} para 0,2 \mathrm{~m}^{3} é mais próximo de:
(A) 20 kJ
(B) 25 kJ
(C) 40 kJ
(D) 55 kJ

Passo 1

O enunciado dá a relação P V^{2}=2 com P em \text{kPa} e V em \text{m}^{3}. O volume muda de V_{1}=0,04\\ \text{m}^{3} para V_{2}=0,2\\ \text{m}^{3}. A pergunta é: qual o trabalho realizado durante essa transformação? (escolhas: (A) 20\\ \text{kJ}, (B) 25\\ \text{kJ}, (C) 40\\ \text{kJ}, (D) 55\\ \text{kJ}).

Passo 2

Recordando a definição física do trabalho em um processo quasistático de pistão:
W=\int_{V_{1}}^{V_{2}} P\,dV.
Aqui W tem unidades de pressão vezes volume. Como P está em \text{kPa} e V em \text{m}^{3}, a unidade resultante \text{kPa}\cdot\text{m}^{3} é igual a \text{kJ}. Isso significa que, ao integrar com P em \text{kPa} e V em \text{m}^{3}, o resultado já sai em \text{kJ} — bom para a prova, sem conversões extras.

Passo 3

Usando a relação fornecida, isolamos P:
P=\frac{2}{V^{2}}=2\,V^{-2}.
Agora substituímos na integral do trabalho.

Passo 4

Montamos a integral e explico a técnica de integração que vou usar. A integral é
W=\int_{0,04}^{0,2} 2\,V^{-2}\,dV.
Regra de potência para integrais: para n\neq -1,
\int V^{n}\,dV=\frac{V^{n+1}}{n+1}.
Aqui n=-2, então \int V^{-2}\,dV=\int V^{ -2}\,dV=\frac{V^{-1}}{-1}=-V^{-1}. Multiplicando pelo fator 2, o antiderivado é -2\,V^{-1}=-\dfrac{2}{V}.

Passo 5

Avalio a primitiva nos limites:
W=\Big[-\frac{2}{V}\Big]_{0,04}^{0,2}=-\frac{2}{0,2}-\Big(-\frac{2}{0,04}\Big).
Calculo passo a passo:
\frac{1}{0,2}=5\quad\Rightarrow\quad \frac{2}{0,2}=10,
\frac{1}{0,04}=25\quad\Rightarrow\quad \frac{2}{0,04}=50.
Portanto
W=-10-(-50)=-10+50=40.
Como explicado no Passo 2, o valor 40 está em \text{kPa}\cdot\text{m}^{3}, ou seja em \text{kJ}. Logo o trabalho é 40\\ \text{kJ}, que corresponde à alternativa (C)."]))])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])])+) ]
\\ \text{O valor \(40 está em \text{kPa}\cdot\text{m}^{3}, ou seja em \text{kJ}. Logo o trabalho é 40\\ \text{kJ}, que corresponde à alternativa (C)."}))]))
"]
"))
print(default_api.SolutionSteps(solution_steps=[)

Resposta

(C) 40 kJ

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