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
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
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.
Edit:
P = lb x (ft lb(force)) / lb(mass) °R) °R / (cu in x (1ft/12in)
If you were converting cubic inches to cubic feet, you would multiply by 1/1728, not 1/12.
To find the original mass of gas in the tank, you can use the ideal gas law equation: PV = mRT, where P is the pressure, V is the volume, m is the mass, R is the ideal gas constant, and T is the temperature.
In this case, you have the following information:
R = 45.2 (given in the question)
V = 12 cu.ft = 12 * 1728 cu.in = 20736 cu.in (conversion from cubic feet to cubic inches)
T = 80°F = 540°R (conversion from Fahrenheit to Rankine)
Now, let's calculate the original mass inside the tank:
m = (PV) / (RT)
= (150 * 20736 * (1/12)) / (45.2 * 540)
= 10.6195 lb
Therefore, the original mass inside the tank is approximately 10.6195 pounds.
To find the original pressure, you can rearrange the ideal gas law equation:
P = (mRT) / V
Using the calculated mass (m = 10.6195 lb), we can substitute the values into the equation:
P = (10.6195 * 45.2 * 540) / (20736 * (1/12))
= 216.5398 psi
Therefore, the original pressure inside the tank is approximately 216.5398 psi.