• # question_answer The workdone during the expansion of a gas from a volume of $4\text{ }d{{m}^{3}}$ to $6\text{ }d{{m}^{3}}$against a constant external pressure of 3 atm is A)  $-6\,J$                       B)  $-608\,J$C)  $+304\,J$                 D)  $-304\,J$

$W=-{{P}_{ext}}\,({{V}_{2}}-{{V}_{1}})$ $=-3(6-4)$ $=-6L-atm$ $=-6\times 101.32\,J$ $(\because \,\,1\,\,L-atm=101.32\,J)$                 $=-608\,\,J$