Assuming ideal gas behaviour, how many atoms of Ar are contained in a typical human breath of 0.5 litre at 1.0 bar and 37oC? Air consists of 1% Ar atoms. Assuming that the argon atoms from the last breath of Plato have been distributed randomly throughout the atmosphere (5×1018m3), how long would it take to breathe one of these atoms? A typical adult breath rate is 10min−1.

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Filo · Accepted Answer

According to ideal gas equation,PV=nRTP = pressure = 1 bar = 0.987atmV = volume = 0.5 LT = temperature = 370 C    = 273+37     = 310 Kn=RTPV​    = 0.082∗3100.987∗0.5​ = 0.020 mol1 mol of air = 6.023*1023 molecules0.020 mol of air = 0.020* 6.023*1023 molecules                              = 0.12* 1023 moleculesAr atoms = 1% of air                  = 0.01* 0.12* 1023 atoms                  = 1.2*1020 atomsThese atoms are distributed uniformly in 5*1018 m3. So, number of atoms in 1 m3 = 1.2*1020/5*1018                                                 = 24 atomsit takes 1 min to breath 10 atoms.So, time taken to breath 24 atoms = 1024​                                                           = 2.4 minTime taken = 2.4 min

Question Text	Assuming ideal gas behaviour, how many atoms of Ar are contained in a typical human breath of $0.5$ litre at $1.0$ bar and $3 7^{o}$ C? Air consists of $1%$ Ar atoms. Assuming that the argon atoms from the last breath of Plato have been distributed randomly throughout the atmosphere $(5 \times 1 0^{18} m^{3})$ , how long would it take to breathe one of these atoms? A typical adult breath rate is $10 mi n^{- 1}$ .
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