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 11.0 kg of the gas at a gauge pressure of 35.0 atm. Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 19.4 atm. Assume the temperature of the tank remains constant.

mass of oxygen that has been withdrawn from the tank = 4.77 kg

Explanation:

We are given;

Initial mass of oxygen gas;m1 = 11 kg

Initial gauge pressure;p'1 = 35 atm

Final gauge pressue;p'2 = 19.4 atm

Now, for thia question we cannot use gauge pressure but absolute pressures.

Thus, formula for absolute pressure = gauge pressure + external atmospheric pressure.

External atmospheric pressure = 1 atm.

Thus;

P1 = 35 + 1 = 36 atm

P2 = 19.4 + 1 = 20.4 atm

Now, formula for ideal gas equation is;

PV = nRT = (m/M)RT

Where;

m is mass of gas

M is molar mass of gas

R is gas constant

T is temperature

Thus, for state 1,we have;

P1 × V1 = (m1/M)RT1 - - - (eq 1)

For state 2,we have;

P2 × V2 = (m2/M)RT2 - - - - - (eq 2)

Divide equation 1 by equation 2 to give;

(P1 × V1)/(P2 × V2) = (m1 × T1)/(m2 × T2)

The gas remains confined in the tank, thus, V1 = V2

Also, the temperature is constant, thus T1 = T2.

Thus, V1 & V2 and also T1 & T2 will cancel out to give;

P1/P2 = m1/m2

Since we are looking for the mass of oxygen( Δm) that has been with withdrawn from the tank, we have to rewrite the equation to accommodate Δm. Now, Δm is simply m1 - m2

Thus,

From P1/P2 = m1/m2, we have;

P1/P2 = m1/(-(m1 - m2) + m1)

This gives;

P1/P2 = m1/(m1 - Δm)

Cross multiply to get;

(m1 - Δm)(P1/P2) = m1

Cross multiply to get;

m1 - Δm = (P2/P1)m1

Δm = m1 - (P2/P1)m1

Δm = m1(1 - (P2/P1))

Plugging in the relevant values to get;

Δm = 11(1 - (20.4/36))

Δm = 4.77 kg

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