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Two kilograms of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m3 and 1 bar, respectively. Heat transfer to the oxygen occurs at constant pressure until the volume is doubled. Determine the heat transfer for the process, in kJ, assuming the specific heat ratio is constant, k = 1.35. Kinetic and potential energy effects can be ignored.

Answer :

gbenewaeternity

Answer:

Explanation:

Given Data;

Initial voulume V₁ = 2m³

Initial pressure  P₁ = 1 bar

Heat ratio k - 1.35

Mass of oxygen = 2kg

n = mole of oxygen = 32

Step 1: For an ideal gas situation,

P₁V₁ = mRT₁/M----------------------------------------1

Where;

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

M = mole of oxygen

m = mass

R = 8.314. m³bar.K⁻¹.mol⁻¹; Constant value

Substituting into the equation to find T₁, we have

1 * 2 = (2 * 8.314 * T₁)/32

2 = 16.628T₁/32

T₁ = (2*32)/16.628

T₁ = 64/16.628

T₁ = 3.8489 K

Step 2:

Calculating T₂, We use the formula

V₁/T₁ = V₂/T₂-------------------------------------2

Since the initial volume is doubled, V₂ = 4m³

Substituting, we have

2/3.8489 = 8/T₂

0.5196 = 4/T₂

T₂ = 4/0.5196

T₂ = 7.6982 K

Step 3:

For an ideal gas, specific heat is a function of temperature

Therefore,

Δu = u₂-u₁ = cv(T₂-T₁)---------------------------------------3

But cv = (R/M/)(k-1)

cv= (8.314/32)/(1.35 -1)

cv = 0.2598/0.35

cv = 0.7423

From Equation 3, Heat transfer for the process is calculated as;

Δu =  cv(T₂-T₁)

Δu = 0.7423(7.6982 - 3.8489)

Δu = 0.7423 * 3.8493

Δu  = 2.8573

lhmarianateixeira

In this exercise we have to use our knowledge of volume to calculate the heat transferred, so we have:

[tex]\Delta u = 2.8573[/tex]

Organizing the information given in the statement we have that:

  • Initial voulume V₁ = 2m³
  • Initial pressure  P₁ = 1 bar
  • Heat ratio k - 1.35
  • Mass of oxygen = 2kg
  • n = mole of oxygen = 32

They are using the formula of an ideal gas as:

[tex]P_1V_1 = mRT_1/M[/tex]

Where;

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₁ = Initial temperature
  • M = mole of oxygen
  • m = mass
  • R = Constant

Finding the value of T in the given equation we have:

[tex]1 * 2 = (2 * 8.314 * T_1)/32\\2 = 16.628T_1/32\\T_1 = (2*32)/16.628\\T_1 = 64/16.628\\T_1 = 3.8489 K[/tex]

Finding the value of  T₂, We have:

[tex]V_1/T_1 = V_2/T_2\\2/3.8489 = 8/T_2\\0.5196 = 4/T_2\\T_2 = 4/0.5196\\T_2 = 7.6982 K[/tex]

Then the heat transferred can be written as:

[tex]\Delta u = u_2-u_1 = cv(T_2-T_1)\\cv= (8.314/32)/(1.35 -1)\\cv = 0.2598/0.35\\cv = 0.7423\\\Delta u = cv(T_2-T_1)\\\Delta u = 0.7423(7.6982 - 3.8489)\\\Delta u = 0.7423 * 3.8493\\\Delta u = 2.8573[/tex]

See more about heat at brainly.com/question/1429452

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