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Question

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross sectional area 4×103m2. Initially the gas is at 27C and volume of gas is 1.2×103m3, system is in equilibrium and spring is in natural state. The sample is heated by a heater until piston moves out slowly by 5 cm. Calculate final temperature of the gas and heat supplied by heater. Take k (spring constant) = 8000 N/m, mass of piston = 20 kg (Assume there is no loss of heat through coils and walls, take Patm=105Pa )

Final temperature of the gas (in K):
Heat supplied by heater (in J):

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Solution



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 K583 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=nRTΔ(PV)=nRΔT
nΔT=PfVfPiViR
For monoatomic gas, Cv=32R
Putting these in equation (i)
ΔU=32RPfVfPiViR=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+105×4×1030.05
Wgas=40 J

ΔQ=ΔU+W=255+40=295 J

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