Question
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?
a) What are the partial pressures of argon and nitrogen? b) What will be the total tank pressure?
Answers
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%
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%
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?
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