(a)
V=2.68 L = 0.00268 m³,
t=37°C => T=310 K,
p=743 torr =99058.246 Pa
R=8.31 J/mol
pV=nRT
n= pV/RT = 99058.246•0.00268/8.31•310 =0.1 mol
N= n•N₀=0.1•6.02•10²³=6.02•10²² molecules
(b)
V=5•10³ L=5 m³
μ=28.98 g/mol=28.98•10⁻³ kg/mol
t=0℃ =273 K
p =1 atm = 101325 Pa
pV=mRT/μ
m=pVμ/RT=101325•5• 28.98•10⁻³/8.31•273 =6.47 kg
Calculate the number of molecules in a deep breath of air whose volume is 2.68 L at body temperature, 37°C, and a pressure of 743 torr
(b) The adult blue whale has a lung capacity of 5.0 103 L. Calculate the mass of air (assume an average molar mass 28.98 g/mol) contained in an adult blue whale's lungs at 0.0°C and 1.00 atm, assuming the air behaves ideally.
how do i do these?
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