Question

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.

Answer 1

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

Step-by-step 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