Question

The density of some air at a pressure of 7720mmhg is 1.26kgm^-³. Determine its density at a pressure at 600mmHg?​

Answer 1

0.0979 kg/m³

Step-by-step explanation:

Treating air as an ideal gas:

PV = nRT

n/V = P/RT

Density depends on both pressure and temperature. If we assume the temperature is constant, we can say that density is directly proportional to pressure.

ρ = kP

ρ/P = k

Writing a proportion:

(1.26 kg/m³) / (7720 mmHg) = ρ / (600 mmHg)

ρ = 0.0979 kg/m³

Answer 2

`\rho = 0.098 kg/m^3`

Step-by-step explanation:

As we know that for ideal gas we will have

`PV = nRT`

here we can convert it into the form of density

`PV = (m)/(M)RT`

now we have

`PM = \rho RT`

now if the temperature will remain constant then in that case

`(P)/(\rho) = constant`

so we will have

`(P_1)/(P_2) = (\rho_1)/(\rho_2)`

here we can plug in all values in it

`(7720)/(600) = (1.26)/(\rho)`

now we have

`\rho = (600)/(7720)(1.26)`

`\rho = 0.098 kg/m^3`