Respuesta :
Answer:
Atmospheric density, [tex]\rho_{atm} = 0.0462 kg/m^{3}[/tex]
Given:
Gas constant for dry air, R = 287[tex]m^{2}s^{- 2}[/tex]
Temperature of gas, [tex]T = 280 K[/tex]
Pressure of gas, [tex]P = 130 kPa[/tex]
Solution:
Now, to calculate the atmospheric density, we follow:
The eqn for an ideal gas is given by:
[tex]P_{g}V = nRT_{g}[/tex]
where
n = no. of moles of gas
n =[tex]\frac{mass of gas, m_{g}}{Molecular mass of gas, M_{g}}[/tex]
Also, we can write,
[tex]n = \frac{P_{g}V}{RT_{g}}[/tex]
Comparing both the values of n, we get:
[tex]\frac{m_{g}}{M_{g}} = \frac{P_{g}V}{RT_{g}}[/tex]
[tex]\rho_{atm} = \frac{m_{g}}{V} = \frac{P_{g}M_{g}}{RT_{g}}[/tex] (1)
Now, the molecular mass of air can be calculated as:
[tex]M_{g} = molar fraction of O_{2}\times molar mass of O_{2} + molar fraction of N_{2}\times molar mass of N_{2}[/tex]
[tex]M_{g} = 0.21\times 32 + 0.78\times 28 = 28.56 g/mol[/tex]
(Since, Oxygen and nitrogen constitutes to about 21% and 78% in the atmosphere)
Using the value above in eqn (1):
[tex]\rho_{atm} = \frac{130\times 10^{3}\times 28.56}{287\times 280} = 0.0462 kg/m^{3}[/tex]
