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February 1, 2015

February 1, 2015

Posted by **Janus** on Monday, March 1, 2010 at 8:01am.

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

- Thermodynamics -
**drwls**, Monday, March 1, 2010 at 9:45amYour 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

- Thermodynamics -
**Janus**, Monday, March 1, 2010 at 10:17amMy 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.

- Thermodynamics -
**Janus**, Monday, March 1, 2010 at 10:21amEdit:

P = lb x (ft lb(force)) / lb(mass) °R) °R / (cu in x (1ft/12in)

- Thermodynamics -
**drwls**, Monday, March 1, 2010 at 11:57amIf you were converting cubic inches to cubic feet, you would multiply by 1/1728, not 1/12.

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