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.
2 answers:
<h2>Answer:</h2>
(a) 14.72 kg
(b) 414 kJ
(c) 414 kJ
<h2>Explanation:</h2>(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
T = temperature of the air = 142K
<em>Substitute these values into equation (i) as follows;</em>
3 x 10⁵ x 2 = m x 287.05 x 142
6 x 10⁵ = m x 40761.1
600000 = 40761.1m
m =
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
<em>Substitute these values into equation (ii) as follows;</em>
3 x 10⁵ x 2 = 1.5 x 10⁵ x V₂
6 x 10⁵ = 1.5 x 10⁵V₂
6 = 1.5V₂
V₂ =
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 = ------------(iv)
Also, remember, in thermodynamics the work done, W, when a gas expands or compresses in volume is given by the following;
W =
<em>Substitute the value of P in equation (iv) into the above equation as follows;</em>
W =
W =
W =
<em>Integrating gives;</em>
W = k ln [V] -----------------(v)
<em>Putting the values of the integral limits V₁ and V₂ </em>into equation (v)
W = k ln [V₂ - V₁]
<em>Substitute the values of k, V₂ and V₁ into equation above as follows;</em>
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
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