Question

An insulated rigid tank is divided into two compartments by a partition. One compartment contains 7 kg of oxygen gas at 40°C and 100 kPa, and the other compartment contains 4 kg of nitrogen gas at 20°C and 150 kPa. Now the partition is removed, and the two gases are allowed to mix. Determine
(a) the mixture temperature and

(b) the mixture pressure after equilibrium has been established.

Answer 1

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

$Number \ of\ moles \ of \ oxygen, n_1 = \frac{Mass \ of \ oxygen}{Molar \ mass \ of \ oxygen} = \frac{7000}{32} = 218.75 \ moles$

From the universal gas equation, we have;

P·V = n·R·T

$V_1 = \frac{n_1RT_1}{P_1} =\frac{218.75 \times 8.3145 \times 313.15 }{100000 } = 5.696 m^3$

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

$Number \ of\ moles \ of \ nitrogen, n_2 = \frac{Mass \ of \ nitrogen}{Molar \ mass \ of \ nitrogen} = \frac{4000}{28.0134} = 142.79 \ moles$

From the universal gas equation, we have;

P·V = n·R·T

$V_2 = \frac{n_2RT_2}{P_2} =\frac{142.79\times 8.3145 \times 293.15 }{150000 } = 2.32 m^3$

Therefore

We have for nitrogen

$\frac{c_p}{c_v} = 1.4$

$c_p - c_v = 296.8 J/KgK$

Therefore;

$Nitrogen, \ c_p = 1036 \ J/(kg \cdot K)$

$Nitrogen, \ c_v = 740\ J/(kg \cdot K)$

The molar heat capacities of Nitrogen are therefore as follows;

$Nitrogen, \ \tilde{c_p} = 29.134 \ kJ/(kmol \cdot K)$

$Nitrogen, \ \tilde{c_v} = 20.819 \ kJ/(kmol \cdot K)$

For oxygen we have

$Oxygen, \ \tilde{c_p} = 29.382 \ kJ/(kmol \cdot K)$

$Nitrogen, \ \tilde{c_v} = 21.068 \ kJ/(kmol \cdot K)$

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 = $\sum \left (n_i\tilde{c_v}_i T_i \right )$ = 0.21875 × 21.068 × 313.15 + 0.14279×20.819×293.15 = 2,314.65 kJ

After mixing, we have

U₂ = $T_3 \sum \left (n_i\tilde{c_v}_i \right )$ = T (0.21875 × 21.068  + 0.14279×20.819) = T×7.58137001

Therefore the final temperature, T is then

$T_3 = \frac{\sum \left (n_i\tilde{c_v}_i T_i \right )}{\sum \left (n_i\tilde{c_v}_i \right )} =\frac{2,314.65 }{7.58137001} =305.30761550 \ K$

The mixture temperature, T₃ = 305.31 K

(b) The mixture pressure, P₃ after equilibrium has been established is given as

$P_3 = \frac{n_3 \tilde{R}T_3}{V_3}$

Where:

n₃ = n₁ + n₂ = 0.21875 + 0.14279 = 0.36154 kmol = 361.54 moles

$\tilde{R}$ = 8.3145 J/(gmol·K)

Therefore ,

$P_3 = \frac{361.54 \times 8.3145 \times 305.31 }{8.016 } = 114,492.1766706961 Pa$

P₃ ≈ 114.5 kPa.