Calculate $q . w . Δ U$ and $Δ H$ for the reversible isothermal expansion of one mole of an ideal gas at $127^{o} C$ from a volume of $10 {d m}^{3}$ to $20 {d m}^{3}$ .

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Solution

Since, the process is isothermal,
$Δ U = Δ H = 0$
From the first law of thermodynamics
$Δ U = q + w = 0$
$q = - w$
$w = - 2.303 n R T log \frac{V_{2}}{V_{1}} = - 2.303 \times 1 \times 8.314 \times 400 log \frac{20}{10}$
$= - 2.303 \times 1 \times 8.314 \times 0.3010 = - 2305.3 J$ (Work done by system)
$q = - w = 2305.3 J$ (Heat absorbed by the system)

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