The formula for calculating maximum work done (reversible work) is

W = -2.303*n*R*T*log(V2/V1)

RT= 8.314*300

Calculating n , given 16g of oxygen

n= (Given amount of oxygen)/(Molecular mass of Oxygen, O2)

n= 16/32 = 0.5

and, according to the question

V2 = 25 dm3

V1 = 5 dm3

Putting all the values in work done equation we get,

W = 2.303*(0.5)*8.314*300*log(25/5)

W = 2.01 x 10^3 Joule