Since the piston is in equilibrium at all times the process is quasi static.
Considering the forces acting on the piston,
PgasS=PoS+Mg
Pgas=Po+MgS⇒ constant
Hence it is an isobaric process.
If heat is provided to this system both volume and temperature increases.
Rate of heat released to the system
dQdt=q J/s
Using first law of thermodynamics,
ΔQ=ΔU+W
Writing this in differential form w.r.t time,
dQdt=dUdt+dWdt
q=nCvdTdt+dWdt
Since the piston is in equilibrium at all times, the velocity with which the piston moves is constant
⇒ KE is constant.
∴W=PdV
Using this,
q=nCvdTdt+dPdVdt...(i)
Using ideal gas equation,
PdVdt=nRdTdt...(ii)
For monoatomic gas,
Cv=32R
Let at some time
t the piston be at a distance
x from the initial position and it moves further by a distance
dx in time
dt
At this time, change in volume,
ΔV=Sxand rate of change of volume,
dVdt=Sdxdt=Sv
So we convert all the terms involving temperature into volume in equation (i) using equation (ii)
q=32PdVdt+PdVdt=52PdVdt=52PSv
v=2q5PS
v=2q5S(Po+MgS)