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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.)

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  • $\begingroup$ Can someone help me with question b) ? $\endgroup$ – Carpediem Feb 9 '14 at 14:00
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    $\begingroup$ Try writing out the ideal gas law and seeing what remains constant and what changes. $\endgroup$ – CTKlein Feb 9 '14 at 14:58
  • $\begingroup$ @CTKlein We have $\frac{nRT_1}{P_1}=\frac{nRT_2}{P_2}$ But what is n in this case ? $\endgroup$ – Carpediem Feb 9 '14 at 15:01
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    $\begingroup$ n is the number of moles of gas. What do you think will happen to this as you inhale? $\endgroup$ – CTKlein Feb 9 '14 at 15:07
  • $\begingroup$ @CTKlein It remains unchanged. Therefore it will cancel out in the equation. Thank you $\endgroup$ – Carpediem Feb 9 '14 at 15:08
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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 $p_2 = 1.054~\mathrm{atm}$.

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