Engineering Thermal Fluids Question?

First determine the volume in tank 1 after the bleed off:

Pi Vi / Ti = Pf Vf / Tf

(600 kpa * 300 L) / 333K = (300 kpa * X liter) / 323K

solve for X = 582 L .......in tank 1 after bleed off

Since tank 1 can only hold 300L, then that means 282 L moved into Tank 2

Now compare tank 1 with 300L known volume moving 282L over to tank 2 with known temp and pressure. But the 282L will change because of the effect of the temp and pressure so volume is still unknown variable in tank 2:

Pi Vi / Ti = Pf Vf / Tf

(300 kpa * 300L) / 323K = (190 kpa * Volume2) / 300K

Solve for tank2 volume = 440 L

let ,p1=600kpa

t1=333k

v1=0.3m^3

let p2=300kpa

t2=323k & v1=v2=0.3m^3

again let p3=190kpa

t3=300k

v3=?(to be find)

R for CO2=0.189kj/kgk

first mass of the co2 is to be found out...

using ideal gas eq.......p1v1=m1Rt1

we get m1=2.86kg

(since it is given that tanks are rigid volume of the gas will be not change)

let x amout of gas escapes from tank 1....

therefore appling ideal gas eq for final condition for tabk1...we get

p2v2=(m1-x)Rt2 (v2=v1)

x will come out to be 1.3857kg........

since x amout of gas escapes to tank 2

therfore appling ideal gas eq for tank3 we get

p3v3=xRt3

on solving v3 =0.4135m3...........

(note ; tanks are rigid volume of the gas in the tank is not going to change

& R for co2 needs to be calculated..........& t should be in k and pressure should be in kpa)