Question

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after a forced exhalation, is 1,200. mL.

(a) How many moles of air are present in the RV at 1.00 atm and 37.0°C? mol air
(b) How many molecules of gas are present under these conditions? × 10 molecules air

Answer 1

For a: The number of moles of air present in the RV is 0.047 moles

For b: The number of molecules of gas is
`2.83* 10^(22)`

Step-by-step explanation:

• For a:

To calculate the number of moles, we use the equation given by ideal gas follows:

`PV=nRT`

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air =
`37^oC=[37+273]K=310K`

R = Gas constant =
`0.0821\text{ L. atm }mol^(-1)K^(-1)`

n = number of moles of air = ?

Putting values in above equation, we get:

`1.00atm* 1.2L=n_(air)* 0.0821\text{ L atm }mol^(-1)K^(-1)* 310K\\n_(air)=(1.00* 1.2)/(0.0821* 310)=0.047mol`

Hence, the number of moles of air present in the RV is 0.047 moles

• For b:

According to mole concept:

1 mole of a compound contains
`6.022* 10^(23)` number of molecules.

So, 0.047 moles of air will contain
`(0.047* 6.022* 10^(23))=2.83* 10^(22)` number of gas molecules.

Hence, the number of molecules of gas is
`2.83* 10^(22)`