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Q.
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 $(\Delta V )$ is $25.0 \,L ,$ is
Thermodynamics
Solution:
Work done $= p \Delta V$
$=0.980 \,atm \times 25.0\, L =24.5 \,L \,atm$
$=\frac{24.5\, L \,atm \times 8.314 \,J \,mol ^{-1} K ^{-1}}{0.0821\, L \,atm\, mol ^{-1} K ^{-1}}=2481\, J =2.481 \times 10^{3} \,J$
Note Use value of $R$ (gas constant to convert $L$ atm to $J$
$R =0.0821 \,L$ atm $mol ^{-1} K ^{-1}=8.314 \,J\, mol ^{-1} K ^{-1}$