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

Answer:

0.0979 kg/m³

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

Answer:

$\rho = 0.098 kg/m^3$

Explanation:

As we know that for ideal gas we will have

$PV = nRT$

here we can convert it into the form of density

$PV = \frac{m}{M}RT$

now we have

$PM = \rho RT$

now if the temperature will remain constant then in that case

$\frac{P}{\rho} = constant$

so we will have

$\frac{P_1}{P_2} = \frac{\rho_1}{\rho_2}$

here we can plug in all values in it

$\frac{7720}{600} = \frac{1.26}{\rho}$

now we have

$\rho = \frac{600}{7720}(1.26)$

$\rho = 0.098 kg/m^3$