Question

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 .

Answer 1

$PV=nRT$
$n= \frac{PV}{RT}$
$n= \frac{740 torr*2.15 L}{62.4 \frac{torr*L}{mol*K} *(37celsius+273)}$
n = 0.0822 mol
$0.0822 mol*6.022*10 ^{23} = 4.95*10 ^{22} molecules$

Answer 2

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