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(C) The work done by the system during expansion against a constant external pressure is given by the formula: $W = -P_{ext} 	imes (V_2 - V_1)$.Given

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$15$

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(C) The work done by the system during expansion against a constant external pressure is given by the formula: $W = -P_{ext} 	imes (V_2 - V_1)$.Given: $P_{ext} = 2 \ atm$,$V_1 = 5 \ L$,$V_2 = X \ L$,and $W = -2,026.4 \ J$.First,convert the work from Joules to $L \cdot atm$ using the conversion factor $1 \ L \cdot atm = 101.32 \ J$:$W = \frac{-2,026.4 \ J}{101.32 \ J \cdot L^{-1} \cdot atm^{-1}} = -20 \ L \cdot atm$.Now,substitute the values into the work formula:$-20 \ L \cdot atm = -2 \ atm 	imes (X - 5) \ L$.Dividing both sides by $-2 \ atm$:$10 = X - 5$.Therefore,$X = 15 \ L$.

If $5 \ L$ of an ideal gas at a constant external pressure of $2 \ atm$ expands isothermally to a final volume of '$X$' $L$,the system does a work of $-2,026.4 \ J$. '$X$' (in $L$) is $(1 \ L \cdot atm = 101.32 \ J)$

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