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

The atmospheric density given a temperature of 280 K, pressure of 130 kPa, and the gas constant for dry air of 287 m2/s2.K, is approximately 1.623 kg/m3.

Step-by-step explanation:

The question asks us to calculate the atmospheric density given the temperature, pressure, and specific gas constant for dry air. To do this, we can use the Ideal Gas Law rearranged to solve for density (ρ):

ρ = P / (Rair × T)

where P is the pressure, Rair is the specific gas constant for dry air, and T is the temperature. Substituting the given values, P = 130 kPa, Rair = 287 m2/s2.K, and T = 280 K, we can calculate the density as:

ρ = 130,000 Pa / (287 m2/s2.K × 280 K)

ρ ≈ 1.623 kg/m3

Answer 2

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 =
`(mass of gas, m_(g))/(Molecular mass of gas, M_(g))`

Also, we can write,

`n = (P_(g)V)/(RT_(g))`

Comparing both the values of n, we get:

`(m_(g))/(M_(g)) = (P_(g)V)/(RT_(g))`

`\rho_(atm) = (m_(g))/(V) = (P_(g)M_(g))/(RT_(g))` (1)

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

`M_(g) = molar fraction of O_(2)* molar mass of O_(2) +  molar fraction of N_(2)* molar mass of N_(2)`

`M_(g) = 0.21* 32 +  0.78* 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) = (130* 10^(3)* 28.56)/(287* 280) = 0.0462 kg/m^(3)`