Question

Oxygen initially is at 35°F and 16.8 lbf/in2. It fills a closed, rigid 5 ft3 tank filled with a paddle wheel. During the process, a paddle wheel provides 4 Btu of energy transfer by work to the gas. During the process, the gas temperature increases up to 90°F. Assuming ideal gas behavior and ignoring K.E and P.E effects, determine the mass of the oxygen in the tank, the amount of heat transfer in Btu, and the final pressure of the tank.

Answer 1

m = 228 gm

ΔQ = 4.35 Btu

P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa

Step-by-step explanation:

FOR MASS OF OXYGEN:

We will use ideal gas equation for initial conditions:

P₁V₁ = nRT₁

P₁V₁ = mRT₁/M

where,

P₁ = Initial Pressure = (16.8 lbf/in²)(6894.76 Pa/1 lbf/in²) = 1.15 x 10⁵ Pa

V₁ = Initial Volume = (5 ft³)(0.028316 m³/1 ft³) = 0.1415 m³

m = mass of oxygen = ?

R = Universal Gas Constant = 8.314 J/mol.k

T₁ = Initial Temperature = 35°F = 274.67 k

M = Molecular Mass of Oxygen = 32 gm

Therefore,

(1.15 x 10⁵ Pa)(0.1415 m³) = m(8.314 J/mol.k)(274.67 k)/(32 gm)

m = (1.15 x 10⁵ Pa)(0.1415 m³)(32 gm)/(8.314 J/mol.k)(274.67 k)

m = 228 gm

FOR AMOUNT OF HEAT TRANDFER:

Using First Law of Thermodynamics:

ΔQ = ΔU +W

Since, this is a constant volume process. Therefore, the work done in this process must be equal to zero:

ΔQ = ΔU + 0

ΔQ = ΔU

and,

ΔU = m Cv ΔT

Therefore,

ΔQ = m Cv ΔT

where,

ΔQ = Heat Transfer Amount = ?

m = mass of oxygen = 228 g = 0.228 kg

Cv = Molar Specific Heat of Oxygen at Constant Volume = 0.659 KJ/kg.k

ΔT = Change in temperature = 305.22 k - 274.67 k = 30.55 k

Therefore,

ΔQ = (0.228 kg)(0.659 KJ/kg.k)(30.55 k)

ΔQ = (4.6 KJ)(0.9478 Btu/1 KJ)

ΔQ = 4.35 Btu

FOR FINAL PRESSURE IN TANK:

Using equation of state:

P₁V₁/T₁ = P₂V₂/T₂

Here, the volume will be constant due to rigid tank. Hence,

V₁ = V₂ = V

Therefore,

P₁V/T₁ = P₂V/T₂

P₁/T₁ = P₂/T₂

(16.8 lbf/in²)/(35°F) = P₂/(90°F)

P₂ = (16.8 lbf/in²)(90°F)/(35°F)

P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa