Amouse is placed in a sealed chamber with air at 755.0 torr. this chamber is equipped with enough solid koh to absorb any co2 and h2o produced by the mouse. the gas volume in this chamber is measured to be exactly 2.20 l, the temperature is held constant at 292 k. after two hours the pressure inside the bottle falls to 710.2 torr. what mass of oxygen has the mouse consumed?

mass O2 = 3,456 10⁻² gr

Explanation:

To solve this problem we use the equation of state of ideal gases

     PV = nRT

With the initial data we calculate the initial amount of gas

     n = PV / RT

Let's reduce the magnitudes to SI units

     P1 = 755.0 torr (1 10⁵Pa / 760 torr) = 0.99 10⁵ Pa

     V1 = 2.20 L (m3 / 10³ L) = 2.20 10⁻³ m3

     P2 = 710.2 = 0.93 10⁵ Pa

     n = 0.99 105 2.20 10⁻³ / (8.314 292)

     n = 8.97 10⁻² mol

Now we can calculate the moles consumed

     P1 V = n1 RT

     P2 V = n2 RT

     P1 / P2 = n1 / n2

     n2 = n1  P2/P1

     n2 = 8.97 10⁻² 0.93 10⁵ / 0.99 10⁵

     n2 = 8.43 10⁻² mol

These are the moles of air left in the chamber, the moles consumed is the difference with the initials

     Δn = n1-n2

     Δn = 8.97 10⁻² - 8.43 10⁻²  

     Δn = 0.54 10⁻² mol = 5.4 10⁻³ mol

These are the moles of air consumed, in the normal atmosphere only 20% is oxygen, therefore, the mole of oxygen is

     n O2 = 0.20 5.4 10⁻³

     n O2 = 1.08 10⁻³ mol

The oxygen molecule has two atoms and the molecular weight of oxygen is 15,999 gr / mol

      PM = 2 PA = 32 gr / mol

     mass O2 = Pm mol

     mass O2 = 32 1.08 10⁻³

     mass O2 = 3,456 10⁻² gr

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