Calculate the number of molecules in a deep breath of air whose volume is 2.15 L at body temperature, 37 ∘C, and a pressure of 740 torr .
\(PV=nRT\)
\(n= \frac{PV}{RT}\)
\(n= \frac{740 torr*2.15 L}{62.4 \frac{torr*L}{mol*K} *(37celsius+273)}\)
n = 0.0822 mol
\(0.0822 mol*6.022*10 ^{23} = 4.95*10 ^{22} molecules\)
PV = nRT
R = 0.0821 L atm / mol K
(ideal gas constant)
First, convert 735 torr to atm. Divide by 760.
(1 atm = 760 torr)
735 torr 1 atm / 760 torr = 0.967 atm
Then, convert 37 C to Kelvin. Just add 273.
37 C = 310K
n = PV / RT
= (0.967)(2.07) / (0.0821)(310)
= 0.0786 mol
0.0786 mol 6.02 10^23 molecules / 1 mol = 4.73 10^22 molecules