The pressure gauge on a tank registers the gauge pressure, which is the difference between the interior and exterior pressure. When the tank is full of oxygen (O2), it contains 9.00 kg of the gas at a gauge pressure of 40.0 atm. Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 25.0 atm. Assume the temperature of the tank remains constant.
I set up a ratio using the ideal gas law.
Then I found oxygen in moles.
n= (mass of oxygen)/(molar mass)
PV/nRT (initial) = PV/nRT (final)
I just assumed that V and T are constant, and R is a constant.
so I come out with
P1/n1 = P2/n2
I solved for n2=n1*(P2/P1)
Then I converted back to kg. Then I subtracted that value from the initial mass value. However, I'm not getting the correct answer. Can someone let me know what I'm doing wrong?
3 answers
I assume you used P1 = 41 atm and P2 = 26 atm because they are gauge?
P1 = 41 atm
P2 = 26 atm
P V = n R T
V is constant
R is constant
T is constant
so
P1 /n1 = p2/n2
n is proportional to kg, no need for moles here
so
41/9 = 26/x
x = 9*26/41 = 5.7 kg left
or 9-5.7 = 3.3 removed
P2 = 26 atm
P V = n R T
V is constant
R is constant
T is constant
so
P1 /n1 = p2/n2
n is proportional to kg, no need for moles here
so
41/9 = 26/x
x = 9*26/41 = 5.7 kg left
or 9-5.7 = 3.3 removed
I actually used P1=40 atm and P2=25 atm
I got a very similar answer (3.375 kg removed), but I'm still getting it wrong. Am I not taking something into account?
I got a very similar answer (3.375 kg removed), but I'm still getting it wrong. Am I not taking something into account?
1 answer