dq=du+dw---1

at the final and the initial time,v=0 (since sytem is at equliibrium)

.: for the piston=change in KE =0

.:Wgas +Wgr+Watm=0

.:Wgas=Mgx+P₀A

on substituting we get Wgas=300

Also,if u write the equllibrium equations of piston at initial and final postions(or by observation),we get

initial pressure=final pressure=>process is isobaric

for isobaric process =dw=nRdT=300

also U=f/2nRT

f=5(.: oxygen diatomic)

.:U=5/2*300=750

from 1 we get ,

Q=300+750=1050J