Question

One mole of ideal gas is confined in a piston-cylender device, which is one foot in diameter. The piston can be assumed weightless and frictionless. The internal and external pressures are both initially 1 atm. An additional weight of 10 lbm is placed on top of the pistion and the piston drops until the gas balances the force pushing the piston downward. The temperature stays at a constant 80 degrees F.

What is the Final Pressure?

What is the Final Volume?

Answer 1

n = 1 mol

D = 1 ft

P = 1 atm

W =10 lbm

T = 80 F constant

a)

final P

Pfinal = Pinitial + Padded

Padded = W/A

A = pi*(D/2)^2

A = 3.1416*(0.3048/2)^2 = 0.072966 m^2

W = m*g

m = 10*0.454 = 4.54 kg

g = 9.8 m/s2

W = 4.54*9.8 = 44.492 N

Padded = W/A

PAdded = 44.492/0.072966 = 609.76 Pa

Ptotal = 101325 pa + 609.76 Pa = 101934.76 Pa or 1.006 atm

b) final volume

Apply ideal gas law ratios

P1V1 = P2V2

Vf = V1*(P1/P2) = V1*1/1.006 = 0.994 of the original volume

V = from ideal gas

PV = nRT

V = nRT/P = 1*0.0082*(26+273)/(1) = 2.4518 L

Then

Vf = V1*0.994 = 0.994 *2.4518 = 2.4371 L

c)

work done

W = f*d

W = P*V = 1.006*(Vf-Vi) = 1.006*(2.4371 -2.4518 ) = 0.0147882 L*atm

Since 101.3 L*atm = 1 J then

W = 1.4984 J