10. A mouse is placed in a sealed chamber with air at 761.0 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 2.40 L, and the temperature is held constant at 306K. After two hours the pressure inside the chamber falls to 712.9 torr. What mass of oxygen has the mouse consumed?

To answer this question you can use gas law (PV=nRT) solving for moles of air that are proportional to moles of oxygen. These moles most be converted to grams with molecular weight:

n = PV/RT

Where, at initial conditions:

P is pressure (761,0torr×[tex]\frac{1 atm}{760 torr}[/tex]) = 1,0013 atm

V is volume 2,40 L

R is gas constant 0,082 atmL/molK

T is temperature 306 K

Solving, moles of aire are: 0,09577 moles

21% of air is oxygen, thus:

0,09577×0,21 = 0,02011 initial moles of oxygen

The final moles of oxygen are:

n = PV/RT

With the same initial conditions just changing pressure:

P is 712,9torr×[tex]\frac{1 atm}{760 torr}[/tex] = 0,9380 atm

Final moles are: 0,08972 moles of air

The moles of oxygen are:

0,08972×0,21 = 0,01884 final moles of oxygen

Thus, moles of oxygen consumed by the mouse are:

0,02011 - 0,01884 = 1,27x10⁻³ moles

In grams:

1,27x10⁻³ moles ×[tex]\frac{32 g}{1moleO_{2}}[/tex] = 0,04064 g of O₂