Respuesta :
Answer:
a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]
[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]
We have;
[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]
[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]
[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]
p₂ = 100000/0.941 = 106.265 kPa
[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
