The pressure gauge on a tank registers the gauge pressure, which is the difference between the interior pressure and exterior pressure. When the tank is full of oxygen (O2), it contains 14.0 kg of the gas at a gauge pressure of 43.0 atm. Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 29.4 atm. Assume the temperature of the tank remains constant.

9.67 kg

Explanation:

If the gauge pressure is 43 atm, the absolute pressure is 44 atm

p1 = 44 atm = 4.46 MPa

If the gauge pressure is 29.4 atm, the absolute pressure is 30.4 atm

p2 = 30.4 atm = 3.08 MPa

The ideal gas state equation is

p * V = m * R * T

We can set two eqaution for these two times

p1 * V1 = m1 * R * T1

p2 * V2 = m2 * R * T2

SInce V1 = V2 and T1 = T2

p1 = m1 * R

p2 = m2 * R

Rearranging

R = p1 / m1

R = p2 / m2

Equating:

p1 / m1 = p2 / m2

m2 = m1 * p2 / p1

m2 = 14 * 3.08 / 4.46 = 9.67 kg