Question 1. Air at initial temp of 300K & initial volume of 0.002 m^3 is contain in a piston cylinder arrangement. Initially a spring is attached to the piston having a spring constant of 10 k N/m is touching the piston but no force exerting on it. Now heat is added to the air & it expended slowly to occupy a final volume of 0.03 m^3 and the area of piston is 0.02 m^2. The atmospheric pressure assume to be 100KPa. Then determine the final pressure inside the cylinder.

THERMODYNAMICS

CHAPTER: FIRST LAW OF THERMODYNAMICS

Question 1. Air at initial temp of 300K & initial volume of 0.002 m^3 is contain in a piston cylinder arrangement. Initially a spring is attached to the piston having a spring constant of 10 kN/m is touching the piston but no force exerting on it. Now heat is added to the air & it expended slowly to occupy a final volume of 0.03 m^3 and the area of piston is 0.02 m^2. The atmospheric pressure assume to be 100KPa. Then determine the final pressure inside the cylinder.

Solution  T1 = 300K

V1 = 0.002 m^3 

V2  = 0.03 m^3

Spring constant (K) = 10 KN/m

A = 0.02 m^2

Patm = 100 KPa 

Psp = pressure applied by spring

Patm = pressure applied by atmosphere

P1 = inside pressure of piston at position one 

P2 = inside pressure of piston at position two

 

P1 =  Patm + Psp1   ……………..............................(0) 

                                                 ( Psp1 = pressure of  spring at position one  is zero =0)

P1 = Patm = 100 KPa ………………………………………..(1)

P2 = Patm + Psp2

P2 = 100 + Psp2 ………………………………………………..(2)

We need to find displacement “x” and pressure “Psp2”

V2 – V1 = A2x

0.003 – 0.002 = 0.02x

 x = 0.05 m   ………………………………………......…..…….(3)

Now

Psp2 = F/A2  = kx/A2                    (F= spring force, A2 = area of piston at position 2)

Put value of "x" in equation (4)

Psp2 = 10x/0.02 = 25 KPa     …………………………..(4)

Put “Psp2” value in equation (2)

P2 = 100 + Psp2

P2 = 100 +  25 = 125 KPa    ……………………………..(5)

Final pressure inside the cylinder is P2 =125 KPa