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Calculate the density of oxygen, O2, under each of the following conditions:

a) STP
b) 1.00 atm and 35.0°C

Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma

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IthaloAbreu

Answer:

a) 1.43 g/L

b) 1.27 g/L

Explanation:

Oxygen is an ideal gas, so, using the ideal gas equation:

PV = nRT  

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature (always in Kelvin!).

n = mass (m)/molar mass (MM), so:

[tex]PV = \frac{m}{MM}RT[/tex]

PVMM = mRT

[tex]PMM = \frac{m}{V} RT[/tex]

m/V is the density (d), so:

d = [tex]\frac{PMM}{RT}[/tex]

R = 0.082 atm.L/(mol.K) and MM of O2 = 2x 16 = 32 g/mol

a) for STP, P = 1 atm and T = 0ºC = 273 K

d = [tex]\frac{1x32}{0.082x273}[/tex]

d = 1.43 g/L

b) P = 1 atm and T = 35ºC + 273 = 308 K

d = [tex]\frac{1x32}{0.082x308}[/tex]

d = 1.27 g/L

As oxygen is an ideal gas, the ideal gas equation is as follows:

[tex]PV= nRT[/tex]

P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature. n= m/MM

  • Formula: [tex]d=\frac{PMM}{RT}[/tex]
  • a) for STP, P = 1 atm and T = 0ºC = 273 K, putting values in formula gives d = 1.43 g/L.
  • b) P = 1 atm and T = 35ºC + 273 = 308 K, putting values in formula gives d = 1.27 g/L.

Thus, answers for calculated densities are 1.43g/L and 1,27g/L.

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