Quantity of work (in joules) done by the gas if it expands against a constant pressure of 0.980 atm and the change in volume ($∆\mathrm{V}$) is 25.0 L, is

Work done - p$∆$V

$\begin{array}{l}=0.980 \mathrm{atm}\times 25.0 \mathrm{L}=24.5 L \mathrm{atm}\\ =\frac{24.5 L \mathrm{atm}\times 8.314 \mathrm{J} {\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}}{0.0821 L \mathrm{atm} {\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}}=2481\mathrm{J}=2.481\times {10}^3\mathrm{J}\end{array}$

Note Use value of R (gas constant to convert L atm to J

$\mathrm{R}=0.0821 L \mathrm{atm} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}=8.314 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$