# The air we breathe

The air we breathe is a mixture of nitrogen ($$\ce{N2}$$, 80%), oxygen ($$\ce{O2}$$, 20% inhaled, 16% exhaled) and carbon dioxide ($$\ce{CO2}$$, 0% inhaled, 4% exhaled) all of which may be assumed to behave as ideal gases.

(a) Calculate, in liters ($$1\; \mathrm{L} = 10^{-3} \; \mathrm{m^3}$$), the volume occupied by 1 mole of air at ambient conditions (1 atm pressure, room temperature $$\vartheta = 21\;\mathrm{°C}$$). The average human inhales only about 2% of that volume in a single breath.

I found $$V= 24.4~\mathrm{L}$$

(b) The air is warmed to body temperature ($$\vartheta = 37~\mathrm{°C}$$) as it is inhaled. Assuming the volume inhaled is constant, what is the pressure of air in the lungs when inhaled at standard pressure (1 atm) in a room with ambient temperature $$\vartheta=21~\mathrm{°C}$$?

(I am having trouble with this question.)

• Can someone help me with question b) ? – Carpediem Feb 9 '14 at 14:00
• Try writing out the ideal gas law and seeing what remains constant and what changes. – CTKlein Feb 9 '14 at 14:58
• @CTKlein We have $\frac{nRT_1}{P_1}=\frac{nRT_2}{P_2}$ But what is n in this case ? – Carpediem Feb 9 '14 at 15:01
• n is the number of moles of gas. What do you think will happen to this as you inhale? – CTKlein Feb 9 '14 at 15:07
• @CTKlein It remains unchanged. Therefore it will cancel out in the equation. Thank you – Carpediem Feb 9 '14 at 15:08

As mentioned in the comments, the amount of gas inhaled (expressed in numbers of mol $n$) stays the same. So, assuming constant volume the ideal gas equation reforms to give: $$p_2 = \frac{T_2}{T_1}\times p_1$$
When inputting all the values we get $p_2 = 1.054~\mathrm{atm}$.