Respuesta :
Answer:
(a) The mixture temperature, T₃ is 305.31 K
(b) The mixture pressure, P₃ after establishing equilibrium is 114.5 kPa
Explanation:
Here we have the initial conditions as
Oxygen compartment
Mass of oxygen = 7 kg
Molar mass of oxygen = 32.00 g/mol
Pressure in compartment, P₁ = 100 kPa
Temperature of oxygen, T₁ = 40 °C = 313.15 K
Number of moles of oxygen, n₁ is given by
[tex]Number \ of\ moles \ of \ oxygen, n_1 = \frac{Mass \ of \ oxygen}{Molar \ mass \ of \ oxygen} = \frac{7000}{32} = 218.75 \ moles[/tex]
From the universal gas equation, we have;
P·V = n·R·T
[tex]V_1 = \frac{n_1RT_1}{P_1} =\frac{218.75 \times 8.3145 \times 313.15 }{100000 } = 5.696 m^3[/tex]
For the Nitrogen compartment, we have
Mass of nitrogen = 4 kg
Molar mass of oxygen = 28.0134 g/mol
Pressure in compartment, P₂ = 150 kPa
Temperature of oxygen, T₂ = 20 °C = 293.15 K
Number of moles of nitrogen, n₂ is given by
[tex]Number \ of\ moles \ of \ nitrogen, n_2 = \frac{Mass \ of \ nitrogen}{Molar \ mass \ of \ nitrogen} = \frac{4000}{28.0134} = 142.79 \ moles[/tex]
From the universal gas equation, we have;
P·V = n·R·T
[tex]V_2 = \frac{n_2RT_2}{P_2} =\frac{142.79\times 8.3145 \times 293.15 }{150000 } = 2.32 m^3[/tex]
Therefore
We have for nitrogen
[tex]\frac{c_p}{c_v} = 1.4[/tex]
[tex]c_p - c_v = 296.8 J/KgK[/tex]
Therefore;
[tex]Nitrogen, \ c_p = 1036 \ J/(kg \cdot K)[/tex]
[tex]Nitrogen, \ c_v = 740\ J/(kg \cdot K)[/tex]
The molar heat capacities of Nitrogen are therefore as follows;
[tex]Nitrogen, \ \tilde{c_p} = 29.134 \ kJ/(kmol \cdot K)[/tex]
[tex]Nitrogen, \ \tilde{c_v} = 20.819 \ kJ/(kmol \cdot K)[/tex]
For oxygen we have
[tex]Oxygen, \ \tilde{c_p} = 29.382 \ kJ/(kmol \cdot K)[/tex]
[tex]Nitrogen, \ \tilde{c_v} = 21.068 \ kJ/(kmol \cdot K)[/tex]
The final volume, V₃ then becomes
V₃ = V₁ + V₂ = 5.696 m³ + 2.32 m³ = 8.016 m³
(a) For adiabatic mixing of gases the final temperature of the mixture is then found as follows
Therefore before mixing
U₁ = [tex]\sum \left (n_i\tilde{c_v}_i T_i \right )[/tex] = 0.21875 × 21.068 × 313.15 + 0.14279×20.819×293.15 = 2,314.65 kJ
After mixing, we have
U₂ = [tex]T_3 \sum \left (n_i\tilde{c_v}_i \right )[/tex] = T (0.21875 × 21.068 + 0.14279×20.819) = T×7.58137001
Therefore the final temperature, T is then
[tex]T_3 = \frac{\sum \left (n_i\tilde{c_v}_i T_i \right )}{\sum \left (n_i\tilde{c_v}_i \right )} =\frac{2,314.65 }{7.58137001} =305.30761550 \ K[/tex]
The mixture temperature, T₃ = 305.31 K
(b) The mixture pressure, P₃ after equilibrium has been established is given as
[tex]P_3 = \frac{n_3 \tilde{R}T_3}{V_3}[/tex]
Where:
n₃ = n₁ + n₂ = 0.21875 + 0.14279 = 0.36154 kmol = 361.54 moles
[tex]\tilde{R}[/tex] = 8.3145 J/(gmol·K)
Therefore ,
[tex]P_3 = \frac{361.54 \times 8.3145 \times 305.31 }{8.016 } = 114,492.1766706961 Pa[/tex]
P₃ ≈ 114.5 kPa.
