A rigid tank that contains 4.0 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 6.0 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.

the volume of the oxygen tank is

V ox = 0.844 m³

the volume of the nitrogen tank is

V ni = 2.359 m³

the final pressure is

P = 255. 534 kPa

Explanation:

taking into account that

n = m/M

where

n = number of moles, m = mass, M = molecular weight

then

n oxigen = m ox / M ox = 6.0 kg / (32 gr/mol) * 1000gr/kg = 187.5 moles

n nitrogen = m ni / M ni = 4.0 kg / (28 gr/mol) * 1000gr/kg = 142.857 moles

from the ideal gas law

P*V=n*R*T

where P = absolute pressure, V = volume occupied by the gas, R = ideal gas constant = 8.314 J / (mol K), T = absolute temperature

V=n*R*T/P

replacing values

for the oxygen tank, T ox = 25°C = 298 K, P = 550 kPa = 550000 Pa,

V ox = n*R*T/P = 187.5 mol * 8.314 J / (mol K) * 298 K / 550000 Pa = 0.844 m³

V ox = 0.844 m³

for the nitrogen tank, T ni = 298 K, P = 150000 Pa

V ni = n*R*T/P = 142.857 mol * 8.314 J / (mol K) * 298 K / 150000 Pa = 2.359 m³

V ni = 2.359 m³

when the gases mix, they occupy a volume of

V = V ox + V ni = 0.844 m³ + 2.359 m³ = 3.203 m³

and total number of moles of gas of the mixture is

n = n oxigen + n nitrogen = 187.5 moles + 142.857 moles = 330.357 moles

therefore

P*V=n*R*T

P = n*R*T/V

replacing values

P = n*R*T/V = 330.357 mol*8.314 J / (mol K) * 298 K / 3.203 m³ * 1 kPa/1000Pa = 255. 534 kPa

P = 255. 534 kPa