Question

A rigid tank that contains 2 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 4 kg of O2 at 25°C and 150 kPa. The valve connecting the two tanks is opened, and the two gases are allowed to mix. If the final mixture temperature is 25°C, determine the volume of each tank and the final mixture pressure.

Answer 1

Answer:

The volume in the first tank = 0.32 $m^{3}$

The volume in the second tank = 2.066 $m^{3}$

The  final pressure of the mixture = 203.64 K pa

Explanation:

First Tank

Mass = 2 kg

Pressure = 550 k pa

Temperature = 25 °c = 298 K

Gas constant for nitrogen = 0.297 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

550 × V = 2 × 0.297 × 298

V = 0.32 $m^{3}$

This is the volume in the first tank.

Second tank

Mass =  4 kg

Pressure = 150 K pa

Temperature = 25 °c = 298 K

Gas constant for oxygen = 0.26  $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

150 × V = 4 × 0.26 × 298

V = 2.066 $m^{3}$

This is the volume in the second tank.

This is the iso thermal mixing. i.e.

$P_{3} V_{3} = P_{1} V_{1} + P_{2} V_{2}$ ----- (1)

$V_{3} = V_{1} + V_{2}$

$V_{3} = 0.32 + 2.066$

$V_{3} = 2.386 \ m^{3}$

Put this value in equation (1)

$P_{3}$ × 2.386 =  550 × 0.32 + 150 × 2.066

$P_{3}$ = 203.64 K pa

Therefore the  final pressure of the mixture = 203.64 K pa

Answer 2

The volume of the first tank has been $\rm 0.32\;m^3$. The volume of the second tank has been $\rm 2.066\;m^3$. The final pressure of the mixture has been 203.646 kPa.

The final volume of the first tank has been given by ideal gas equation as:

$PV=mRT$

Where the pressure of the tank, $P=550 \;\rm {kPa}$

The mass of gas, $m=2\;\rm kg$

The gas constant for nitrogen, $R=0.297 \;\rm kJ/kg.K$

The temperature of the gas, $T=25\;^\circ \rm C;\;298\;K$

Substituting the values for volume (V) of nitrogen tank:

$\rm 550\;\times\;\textit V=2\;\times\;0.297\;\times\;298\\\textit V=0.32\;m^3$

The volume of the first tank has been $\rm 0.32\;m^3$.

The volume of the second tank of gas has been given as:

The pressure of the tank, $P=150 \;\rm {kPa}$

The mass of gas, $m=4\;\rm kg$

The gas constant for nitrogen, $R=0.26 \;\rm kJ/kg.K$

The temperature of the gas, $T=25\;^\circ \rm C;\;298\;K$

Substituting the values for volume (V) of oxygen tank:

$\rm 150\;\times\;\textit V=2\;\times\;0.26\;\times\;298\\\textit V=2.066\;m^3$

The volume of the second tank has been $\rm 2.066\;m^3$.

The final pressure of the mixture has been given by the equation of isothermal mixing as:

$P_3V_3=P_1V_1\;+\;P_2V_2$

Where $P_3V_3$ has been the pressure and volume of the final solution.

$P_1V_1$ has been the pressure and volume of the nitrogen tank.

$P_2V_2$ has been the pressure and volume of the oxygen tank.

The final volume of solution, $V_3=V_1\;+\;V_2$

$V_3=0.32\;+\;0/266\;\text m^3\\V_3=2.386\;\text m^3$

Substituting the values:

$P_3\;\times\;2.386=(550\;\times\;0.32\;)+\;(150\;\times\;2.066)\\P_3\;\times\;2.386=176\;+\;309.9\\P_3=\dfrac{485.9}{2.386} \\P_3=203.646\;\rm kPa$

The final pressure of the mixture has been 203.646 kPa.

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