Question

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 1

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

Answer 2

Answer: Heat transfer (Q) is 521 kJ.

Explanation: In the piston-cilinder assembly, we can suppose that the oxygen act as an ideal gas, so, it can be used the General Gas Equation:

PV=$\frac{m}{M}$RT, where:

P is pressure;

V is volume;

m is mass;

M is molar mass;

R is a constant: R = 8.314.$10^{-5}$ m³bar.K⁻¹.mol⁻¹;

T is temperature;

Using this equation, find the intial temperature:

PV = $\frac{m}{M}$RT

1.2 = $\frac{2}{16}$.8.314.$10^{-5}$.T

T = 1.924.$10^{5}$ K

To determine the final temperature, use Combined Gas Law:

$\frac{P_{i} . V_{i} }{T_{i} }$ = $\frac{P.V }{T}$, in which, the left side of the equality is related to the initial values and the right side, to the final values.

As pressure is constant:

$\frac{V_{i} }{T_{i} } =\frac{V}{T}$

T = $\frac{V.T_{i} }{V_{i} }$

T = $\frac{4.1.924.10^{-5} }{2}$

T = 3.85.$10^{5}$ K

With the temperatures, calculate the heat transfer of the process:

Q = m.k.ΔT, where:

k is heat constant

ΔT = T - $T_{i}$

Q = m.k.ΔT

Q = 2.1.35.(3.85 - 1.92).$10^{5}$

Q = 521 kJ

The heat transfer in the process is 521 kJ.