Question

Air in a piston–cylinder assembly, initially at 3 bar, 142 K, and a volume of 2 m3. The air undergoes a process to a state where the pressure is 1.5 bar, during which the pressure–volume relationship is pV = constant. Assuming ideal gas behavior for the air, determine: the mass of the air, in kg and the work and heat transfer, each in KJ.

Answer 1

Answer:

(a) 14.72 kg

(b) 414 kJ

(c) 414 kJ

Explanation:

(a) To get the mass of the air, the ideal gas equation will be used and is given as follows;

PV = mRT            ----------------(i)

Where;

P = pressure of the air = 3 bar = 3 x 10⁵Pa

V = volume of the air = 2m³

m = mass of air

R = specific gas constant of air = 287.05 $\frac{J}{kgK}$

T = temperature of the air = 142K

Substitute these values into equation (i) as follows;

3 x 10⁵ x 2 = m x 287.05 x 142

6 x 10⁵ = m x 40761.1

600000 = 40761.1m

m = $\frac{600000}{40761.1}$

m = 14.72kg

Therefore, the mass of the air is 14.72 kg

(b) According to the question, the relationship between the pressure and the volume is given by

PV = k         (where k = constant)

This implies that;

P₁V₁ = P₂V₂ = ... = k           -------------------(ii)

Where;

P₁ = initial pressure of air = 3 bar = 3 x 10⁵Pa

V₁ = initial volume of air = 2m³

P₂ = final pressure of air = 1.5 bar = 1.5 x 10⁵Pa

V₂ = final volume of air

Substitute these values into equation (ii) as follows;

3 x 10⁵ x 2 = 1.5 x 10⁵ x V₂

6 x 10⁵ = 1.5 x 10⁵V₂

6 = 1.5V₂

V₂ = $\frac{6}{1.5}$

V₂ = 4m³

Also, from equation (ii)

P₁V₁ = k               --------------(iii)

Substitute the values of P₁ and V₁ into equation (iii) as follows;

3 x 10⁵ x 2 = k

k = 6 x 10⁵ m³Pa

Recall that PV = k

This implies that;

P = $\frac{k}{V}$            ------------(iv)

Also, remember, in thermodynamics the work done, W, when a gas expands or compresses in volume is given by the following;

W = $\int\limits^{V_2}_{V_1} {P} \, dV$

Substitute the value of P in equation (iv) into the above equation as follows;

W = $\int\limits^{V_2}_{V_1} {\frac{k}{V} } \, dV$

W = $k \int\limits^{V_2}_{V_1} {\frac{1}{V} } \, dV$

W = $k \int\limits^{V_2}_{V_1} {V^{-1} } \, dV$

Integrating gives;

W = k ln [V]                -----------------(v)

Putting the values of the integral limits V₁ and V₂ into equation (v)

W = k ln [V₂ - V₁]

Substitute the values of k, V₂ and V₁ into equation above as follows;

W = 6 x 10⁵ ln [4 - 2]

W = 6 x 10⁵ ln [2]

W = 6 x 10⁵ (0.69)

W = 4.14 x 10⁵

W = 414 kJ

Therefore, the work done is 414 kJ

(c) The heat transfer Q, the work done, W, and the change in internal energy, ΔU, in a thermodynamic system are related by the following relation;

Q - W = ΔU                  ----------(vi)

If the values of P₂, V₂ are substituted into equation (i) to find the value of T₂, it will be found that T₁ and T₂ are the same. i.e T₁ = T₂ = 142K. Therefore, the change in internal energy, ΔU = 0.

Equation (vi) the becomes

Q - W = 0

Q = W            [Substitute the value of W = 414 kJ]

Q = 414 kJ

Therefore, the heat transfer is 414 kJ

Answer 2

Answer:

mass of the air = 14.62kg

Workdone = 415.88 kJ

Heat transfer = 415.88 kJ

Explanation:

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