<u>Answer:</u>
<u>For a:</u> The number of moles of air present in the RV is 0.047 moles
<u>For b:</u> The number of molecules of gas is ![2.83\times 10^{22}](https://tex.z-dn.net/?f=2.83%5Ctimes%2010%5E%7B22%7D)
<u>Explanation:</u>
To calculate the number of moles, we use the equation given by ideal gas follows:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
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](https://tex.z-dn.net/?f=37%5EoC%3D%5B37%2B273%5DK%3D310K)
R = Gas constant = ![0.0821\text{ L. atm }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=0.0821%5Ctext%7B%20L.%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
n = number of moles of air = ?
Putting values in above equation, we get:
![1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol](https://tex.z-dn.net/?f=1.00atm%5Ctimes%201.2L%3Dn_%7Bair%7D%5Ctimes%200.0821%5Ctext%7B%20L%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20310K%5C%5Cn_%7Bair%7D%3D%5Cfrac%7B1.00%5Ctimes%201.2%7D%7B0.0821%5Ctimes%20310%7D%3D0.047mol)
Hence, the number of moles of air present in the RV is 0.047 moles
According to mole concept:
1 mole of a compound contains
number of molecules.
So, 0.047 moles of air will contain
number of gas molecules.
Hence, the number of molecules of gas is ![2.83\times 10^{22}](https://tex.z-dn.net/?f=2.83%5Ctimes%2010%5E%7B22%7D)