# Thread: Ideal gas equation, fins volume

1. ## Ideal gas equation, fins volume

Two tanks (A and B) are connected together by a small pipe which contains a valve.
Initially the valve is closed and tank A, which has a volume of 1 m3 contains air
at a pressure of 6 atm and a temperature of 60C. Tank B contains a mixture of
nitrogen and oxygen (95 mol% N2) at 12 atm pressure and 90C .
The valve is then opened and the contents of the two tanks mix. After complete
mixing has taken place, the mixed gas contains 85 mol% N2. Calculate the volume
of tank B.

Consider that air contains 79 mol% nitrogen and 21 mol% oxygen and that the
volume of any gas at 273 K and 1 atm is 22.414 m3 kmol��1 - alternatively, 1 atm
= 101.325 kPa, R = 8.3143 kJ kmol��1 K��1.

I know I have to make use of the ideal gas equation, however not sure how to start, what is my unknown?

2. Originally Posted by Tweety
I know I have to make use of the ideal gas equation, however not sure how to start, what is my unknown?
I think I have worked it out, but I am not sure if this is the correct method/answer, so if anyone could check my working, would greatly appreciate.

pv = nRT

Tank A:

$v = 1 m^{3}$

$P = 6 atm$

$n=?$

$N_{2}$ : $0.79 mol\%$

$O_{2}$ $, 0.21mol\%$

$T= 333k$

Tank B:

$n=?$

$N_{2}$ : $0.95 mol\%$

$O_{2}$ : $0.05mol\% 0_{2}$

$p= 12 atm$

$T = 363k$

$V=?$

Tank C:

$N_{2} : 0.85 mol\%$

$O_{2} : 0.15mol\%$

1 atm = 101325 pa
6 atm = 607950 pa

$moles \\\\\ in\\\\\\\\\\ tank A : 607950 \times 1 = n \times 8.3143 \times 333$

n = 219.6 moles

$N_{2}: (219.6 \times 0.79 ) + 0.95B = 0.85c$

$O_{2} : (219.6 \times 0.21) + 0.05B = 0.15 C$

solve the simultaneous equation for 'B' moles, and than use that value in the Pv= nRT to work out the volume, which I got to be 0.327.