The correct option is D −271.17 kcal
At STP, 16 g of O2 or 12 mole of O2 will occupy 11.2 L volume.
Now, if the volume is doubled, it means V2 = 22.4 L
So, (V2−V1) = 22.4 - 11.2 = 11.2 L
At STP = P = 1 atm
Now, w=−P×(V2−V1)=−1×11.2 L. atm = -11.2 L.atm = −11.2×101.3 J =
w=−1134.56 J
Since, 1 cal = 4.184 J
w = 1134.564.184 = −271.17 kcal