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(D) The work done during expansion is given by the formula $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.Given: $W = -800 \ J$,$P_{ext} = 1.6 \ bar$,$

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(D) The work done during expansion is given by the formula $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.Given: $W = -800 \ J$,$P_{ext} = 1.6 \ bar$,$V_1 = 3 \ dm^3$.Since $1 \ dm^3 \ bar = 100 \ J$,we convert the work done to $dm^3 \ bar$: $W = -800 \ J = -8 \ dm^3 \ bar$.Substituting the values into the equation: $-8 \ dm^3 \ bar = -1.6 \ bar 	imes (V_2 - 3 \ dm^3)$.$V_2 - 3 = \frac{-8}{-1.6} = 5$.$V_2 = 5 + 3 = 8 \ dm^3$.

Calculate the final volume when $2$ moles of an ideal gas expand from $3 \ dm^3$ at constant external pressure $1.6 \ bar$ and the work done in the process is $800 \ J$. (in $dm^3$)

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