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A quantity of N2 occupies a volume of 1.5L at 300K and 1.2atm. The gas expands to a volume of 3.1L as the result of a change in both temperature and pressure.


Find the density of the gas at these new conditions.

Respuesta :

Eduard22sly

Answer: 0.661g/L

Explanation:

T = 300K

P = 1.2atm

V = 1.5L

R = 0.082atm.L/K /mol

n =?

From the ideal gas equation PV = nRT

n = PV /RT = (1.2x1.5)/(300x0.082)

n = 0.0732mol

Molar Mass of N2 = 28g/mol

Mass conc. = n x molar Mass

Mass conc = 0.0732 x 28

Mass conc of N2 = 2.0496g

Volume = 3.1L

Mass = 2.0496g

Density =?

Density = Mass /volume

Density = 2.0496/3.1

Density = 0.661g/L

onyebuchinnaji

The density of the nitrogen gas at the new condition is 1.35 g/L.

The given parameters;

  • volume of the nitrogen gas, N₂ = 1.5 L
  • temperature of the gas, T = 300 K
  • pressure of the gas, P = 1.2 atm
  • final volume of the gas, 3.1 L

The number of moles of the gas in the new condition is calculated as follows;

[tex]PV = nRT\\\\n = \frac{PV}{RT} \\\\n = \frac{1.2 \times 3.1 }{0.082 \times 300} \\\\n = 0.15 \ mole[/tex]

The mass of the nitrogen gas is calculated as;

[tex]m = 0.15 \times (28 \ g/mol)\\\\m = 4.2 \ g[/tex]

The density of the nitrogen gas is calculated as follows;

[tex]\rho = \frac{mass}{volume} \\\\\rho = \frac{4.2}{3.1} \\\\\rho = 1.35 \ g/L[/tex]

Thus, the density of the nitrogen gas at the new condition is 1.35 g/L.

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