Mass of Oxygen Withdrawn from Tank at 28.0 atm

[hot2moli]

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 15.0 kg of the gas at a gauge pressure of 41.0 atm. Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 28.0 atm. Assume the temperature of the tank remains constant.

I though you could use the Ideal gas law, converting 15kg into moles, but would the volume be considered constant here? Because if not, then I do not see how you could apply it to PV= nRT.

 

[G01]

The volume is the volume of the tank itself. So, yes. The volume is constant. What isn't constant here (other than the pressure of course) is the number of mols of oxygen.

 

[hot2moli]

so I did P/n = P2/n2
and I am not getting the right answer.
P= 41atm
n= (15000g)/(16g/mol)
P2= 28
and then I solve for n2 multiplying by 16g/mol to get mass of oxygen?

WHERE I HAVE I GONE WRONG? :( PLEASE HELP!

 

[Ygggdrasil]

Remember that 41atm is the gauge pressure, not the actual pressure.

 

[hot2moli]

I'm sorry.. I still am not understanding...25atm wouldn't be the actual than either, but can't you make them proportional still because it's the same system?...and even if it's not the actual P, wouldn't you just add 1atm to it to account for air pressure?

 

[G01]

I'm sorry.. I still am not understanding...25atm wouldn't be the actual than either, but can't you make them proportional still because it's the same system?...and even if it's not the actual P, wouldn't you just add 1atm to it to account for air pressure?

No, the proportionality will only hole for the actual pressures, since the addition of 1 atm to the numerator and denominator of P_1/P_2 will not give the same fraction. i.e.:

$$\frac{P_{gauge 1}}{P_{guage2}}=/=\frac{P_1}{P_2}$$

Work with the actual pressures.