Asked by Janus
A 12 cu.ft tank contains hydrogen sulfide gas at 150psia & 80°F after 5 pounds of the gas have been drawn out. Before any gas left the tank the temperature was 70°F.
What mass of gas was in the tank originally and what pressure?
just to check:
R=45.2
V=12cu.ft = 20736cu.in
t=80°f = 540°R
m=(PV)/(RT)
m=(150 x 20736 x (1/12)) / (45.5 x 540)
m=10.6195
original mass inside the tank
10.6195+5=15.6195lb
original pressure
P=(mRT)/V
P=(15.6195 x 45.2 x 530) / (20736 x (1/12))
P= 216.5398
What mass of gas was in the tank originally and what pressure?
just to check:
R=45.2
V=12cu.ft = 20736cu.in
t=80°f = 540°R
m=(PV)/(RT)
m=(150 x 20736 x (1/12)) / (45.5 x 540)
m=10.6195
original mass inside the tank
10.6195+5=15.6195lb
original pressure
P=(mRT)/V
P=(15.6195 x 45.2 x 530) / (20736 x (1/12))
P= 216.5398
Answers
Answered by
drwls
Your method looks OK but I would have to check the value of R in such unconventional units. I don't understand whay you multiplied V by 1/12 instead of 12
Answered by
Janus
My mistake I forgot to add the units
Unit of R = (ft lb(force)) / lb(mass) °R
P = lb x (ft lb(force)) / lb(mass) °R) / (cu in x (1ft/12in)
the (1ft/12in) is suppose to make the P into lb / sq in.
Unit of R = (ft lb(force)) / lb(mass) °R
P = lb x (ft lb(force)) / lb(mass) °R) / (cu in x (1ft/12in)
the (1ft/12in) is suppose to make the P into lb / sq in.
Answered by
Janus
Edit:
P = lb x (ft lb(force)) / lb(mass) °R) °R / (cu in x (1ft/12in)
P = lb x (ft lb(force)) / lb(mass) °R) °R / (cu in x (1ft/12in)
Answered by
drwls
If you were converting cubic inches to cubic feet, you would multiply by 1/1728, not 1/12.
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