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 10.0 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 Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 26.4 atm. Assume the temperature of the tank remains constant.

since PV=nRT and RT and V are to remain constant,

P/n = RT/V must be constant.

n is usually in moles, but we can just as easily call it grams, since the ratio of moles to grams is constant for O2.

We have reduced P by a factor of 26.4/43.0, so we must reduce the quantity of O2 by the same factor.

10.0 * 26.4/43.0 = 6.14 kg

the answer, of course, is 10.0-6.14 = 3.86 kg have been removed.

To determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 26.4 atm, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature

Since the temperature remains constant, we can simplify the equation to:

PV = constant

Now, let's find the initial volume of the tank. The initial gauge pressure is 43.0 atm, which means the interior pressure is (atm + atm) = 1 atm + 43 atm = 44 atm.

Since the pressure outside the tank is typically around 1 atm, the gauge pressure is the difference between the interior and exterior pressure. Therefore, the initial interior pressure is 44 atm - 1 atm = 43 atm.

We know that PV = constant, so we can write:

P1V1 = P2V2

Now, let's substitute the values we have:

P1 = 43 atm
V1 = volume of the tank (unknown)
P2 = 26.4 atm
V2 = volume of the tank when the pressure is 26.4 atm (unknown)

We can rearrange the equation to solve for V2:

V2 = (P1 * V1) / P2

To find the mass of oxygen that has been withdrawn, we need to find the change in number of moles (Δn). We can use the ideal gas law equation again:

PV = nRT

Since the temperature remains constant, the equation simplifies to:

P1V1 = n1RT
P2V2 = n2RT

Divide both equations by RT to get:

P1V1 / RT = n1
P2V2 / RT = n2

Now, we can find the change in the number of moles (Δn) using:

Δn = n1 - n2
= (P1V1 / RT) - (P2V2 / RT)
= (P1V1 - P2V2) / RT

The mass of oxygen can be calculated using the molar mass of oxygen (32 g/mol) and the change in number of moles (Δn):

Mass of oxygen = Δn * molar mass of oxygen
= Δn * 32 g/mol

Therefore, the mass of oxygen that has been withdrawn from the tank when the pressure reading is 26.4 atm is equal to Δn * 32 g/mol.