the Ar will effectively be "squeezed" into 80% of its previous volume
this will increase the tank pressure to 125% of the original
the pp of Ar is 80% of the total
the pp of nitrogen is 20%
A 24 L welding tank contains of 285 bar of argon at 28 °C. Nitrogen is added to obtain an 80:20 argon:nitrogen mixture.
a) What are the partial pressures of argon and nitrogen? b) What will be the total tank pressure?
3 answers
To find total pressure since you have such a high pressure you have to treat the gases as non-ideal. this means that you have to find the pseudocritical temperature and pressure of the gas mixture which is
T'c= Ya(Tca)+Yb(Tcb)
and
P'c= Ya(Pca)+Yb(Pcb)
where Y is the mole fraction of the gas.
next you find the Pseudoreduced Temp and Pressure :
T'r= T/T'c
P'r=P/P'c
then look at a compressibility chart and find Zm- the compessibility factorof the gas mixture.
Next use equation V/n=Zm*R*T/P
and solve for P
This is called Kay's rule if you want to look it up in more detail.
T'c= Ya(Tca)+Yb(Tcb)
and
P'c= Ya(Pca)+Yb(Pcb)
where Y is the mole fraction of the gas.
next you find the Pseudoreduced Temp and Pressure :
T'r= T/T'c
P'r=P/P'c
then look at a compressibility chart and find Zm- the compessibility factorof the gas mixture.
Next use equation V/n=Zm*R*T/P
and solve for P
This is called Kay's rule if you want to look it up in more detail.
I think it's important that you define what you mean by 80:20 mixture. Is that 80:20 by mass, by moles, or by volume?
1 answer