Question

The gas constant for dry air Rais 287 m2/s2 . K (here K stands for degrees Kelvin) Assuming the temperature is 280 K and the pressure is 130 kPa, what is the atmospheric density?

Answer 1

Answer:

Atmospheric density, $\rho_{atm} = 0.0462 kg/m^{3}$

Given:

Gas constant for dry air, R = 287$m^{2}s^{- 2}$

Temperature of gas, $T = 280 K$

Pressure of gas, $P = 130 kPa$

Solution:

Now, to calculate the atmospheric density, we follow:

The eqn for an ideal gas is given by:

$P_{g}V = nRT_{g}$

where

n = no. of moles of gas

n =$\frac{mass of gas, m_{g}}{Molecular mass of gas, M_{g}}$

Also, we can write,

$n = \frac{P_{g}V}{RT_{g}}$

Comparing both the values of n, we get:

$\frac{m_{g}}{M_{g}} = \frac{P_{g}V}{RT_{g}}$

$\rho_{atm} = \frac{m_{g}}{V} = \frac{P_{g}M_{g}}{RT_{g}}$         (1)

Now, the molecular mass of air can be calculated as:

$M_{g} = molar fraction of O_{2}\times molar mass of O_{2} +  molar fraction of N_{2}\times molar mass of N_{2}$

$M_{g} = 0.21\times 32 +  0.78\times 28 = 28.56 g/mol$

(Since, Oxygen and nitrogen constitutes to about 21% and 78% in the atmosphere)

Using the value above in eqn (1):

$\rho_{atm} = \frac{130\times 10^{3}\times 28.56}{287\times 280} = 0.0462 kg/m^{3}$