Antwort
$W = nRT \ln\left(\frac{V_2}{V_1}\right) = nRT \ln\left(\frac{P_1}{P_2}\right)$.
Erklärung
Since temperature is constant, $\Delta U = 0$ for an ideal gas, meaning all absorbed heat equals work done ($Q = W$).
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What is the expression for work done in an isothermal expansion of $n$ moles of ideal gas from $V_1$ to $V_2$ at temperature $T$?
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