The piston is moving slowly
⇒ quasistatic process
Let us assume that at any given time the piston has moved by a distance
x.
xi=0 It is given that the piston is always in the equlibrium
⇒ the net force on piston is zero at all times.
Considering the FBD of the piston,
PgasA=PoA+mg+kx Pgas=Po+mgA+kxA Since the is no addition or loss of the gas, the number of moles of the ideal gas in the cylinder remains conserved.
∴ni=nf Using ideal gas equation,
PiViTi=PfVfTf Substituting the values of pressures and volumes with corresponding value of
x,we have
(Po+mgA)VoTo=(Po+mgA+kxfA)(Vo+Axf)To Substituting all the values we get,
Tf=583.33 K≃583 K In the second part we need to find the heat supplied to the system,
ΔQ=ΔU+W where,
ΔU=nCvΔT...(i)
Since the number of moles is not speciefied in the problem, we make use of the ideal gas equation,
PV=nRT⇒Δ(PV)=nRΔT ⇒nΔT=PfVf−PiViR For monoatomic gas,
Cv=32R Putting these in equation (i)
ΔU=32RPfVf−PiViR=255 J Workdone by the gas
Wgas=∫PdV=∫x=5x=0(Po+mgA+kxA)Adx Integrating ans substing we get,
W=40 J The gas is doing work on the piston. The piston is moving slowly
⇒ KE is not changing.
Using Work-Energy theorem,
Wnet on the piston =
ΔKE=0 Again using the FBD of piston,
Wgravity+Wspring+Watm+Wgas=0 Wgravity=−mgxf,Wspring=−12kx2f, Watm=−PoAxf Substituting these in the above equation,
Wgas=mgxf+12kx2f+PoAxf Putting the values,
Wgas=20×100×0.05+128000×(0.05)2+105×4×10−30.05 Wgas=40 J ΔQ=ΔU+W=255+40=295 J