2.1 Molecular model of an ideal gas  (Page 5/19)

Density of air at stp and in a hot air balloon

Strategy and solution

1. We are asked to find the density, or mass per cubic meter. We can begin by finding the molar mass. If we have a hundred molecules, of which 78 are nitrogen, 21 are oxygen, and 1 is argon, the average molecular mass is $\frac{78m_{{\text{N}}_2}+21m_{{\text{O}}_2}+m_{\text{Ar}}}{100}$ , or the mass of each constituent multiplied by its percentage. The same applies to the molar mass, which therefore is
   
   $M=0.78M_{{\text{N}}_2}+0.21M_{{\text{O}}_2}+0.01M_{\text{Ar}}=29.0\text{g/mol}\text{.}$
   
   Now we can find the number of moles per cubic meter. We use the ideal gas law in terms of moles, $pV=nRT,$ with $p=1.00\text{atm}$ , $T=273\text{K}$ , $V=1{\text{m}}^3$ , and $R=8.31\text{J/mol}·\text{K}$ . The most convenient choice for R in this case is $R=8.31\text{J/mol}·\text{K}$ because the known quantities are in SI units:
   
   $n=\frac{pV}{RT}=\frac{(1.01\times {10}^5\text{Pa})(1{\text{m}}^3)}{(8.31\text{J/mol}·\text{K})(273\text{K})}=44.5\text{mol}\text{.}$
   
   Then, the mass $m_{\text{s}}$ of that air is
   
   $m_{\text{s}}=nM=(44.5\text{mol})(29.0\text{g/mol})=1290\text{g}=1.29\text{kg}.$
   
   Finally the density of air at STP is
   
   $\rho =\frac{m_{\text{s}}}{V}=\frac{1.29\text{kg}}{1{\text{m}}^3}=1.29{\text{kg/m}}^3.$
2. The air pressure inside the balloon is still 1 atm because the bottom of the balloon is open to the atmosphere. The calculation is the same except that we use a temperature of $120\text{ºC}$ , which is 393 K. We can repeat the calculation in (a), or simply observe that the density is proportional to the number of moles, which is inversely proportional to the temperature. Then using the subscripts 1 for air at STP and 2 for the hot air, we have
   
   ${\rho }_2=\frac{T_1}{T_2}{\rho }_1=\frac{273\text{K}}{393\text{K}}(1.29{\text{kg/m}}^3)=0.896{\text{kg/m}}^3.$

Significance

Using the methods of Archimedes’ Principle and Buoyancy , we can find that the net force on $2200{\text{m}}^3$ of air at $120\text{ºC}$ is $F_b-F_g={\rho }_{\text{atmosphere}}Vg-{\rho }_{\text{hot air}}Vg=8.49\times {10}^3\text{N}\mathrm{,}$ or enough to lift about 867 kg. The mass density and molar density of air at STP, found above, are often useful numbers. From the molar density, we can easily determine another useful number, the volume of a mole of any ideal gas at STP, which is 22.4 L.