when 1 moles of an ideal gas expands from 5 dm 3 to 25 dm 3 isothermally irrversibly at 2 7 ∘ C . Find work in process -

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To calculate the work done during an irreversible isothermal expansion, we use the formula $W = - P_{e x t} Δ V$ .

Given $n = 1$ mol, $V_{1} = 5$ dm³, $V_{2} = 25$ dm³, $T = 2 7^{\circ} C = 300$ K.

Since it's an expansion, the external pressure $P_{e x t}$ is the final pressure, $P_{2}$ .

Using the ideal gas law $P_{2} V_{2} = n RT$ , we find $P_{2} = \frac{n RT}{V _{2}} = \frac{1 \times 8.314 J m o l ^{- 1} K ^{- 1} \times 300 K}{25 \times 1 0 ^{- 3} m ^{3}} = 99768 P a$ .

$W = - P_{e x t} (V_{2} - V_{1}) = - 99768 P a \times (25 - 5) \times 1 0^{- 3} m^{3}$

$= - 99768 \times 20 \times 1 0^{- 3} J = - 1995.36 J \approx - 1.99 k J$

The correct answer is A.

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