Question

How many grams of oxygen gas must be in an 8.50 L container to exert a pressure of 0.900 atm at a temperature of 20.00°C?

Answer 1

The general gas law states that the product of the pressure and the volume of the gas, divided by the thermodynamic temperature, in kelvin, is equal to a constant. This constant, in turn, is described by the Clapeyron equation, note:

`PV\text{ = nRT}`

Where:

P = pressure exerted by the gas

V = volume occupied by this gas

T = temperature in kelvin

R = universal ideal gas constant (0.082 atm.L/mol.K)

n = number of moles

Step 1 - Transform 20 °C into kelvin using the formula: Tk = 273 + Tc

Tk = 273 + 20

Tk = 293 K

Step 2 - Replace the values on the formula:

P = 0.900 atm

V = 8.50L

n = ???

R = 0.082 atm.L/mol.K

T = 293

PV = nRT

0.900 x 8.50 = n x 0.082 x 293

n = 0.106/24.026

n = 0.004407 moles of oxygen

Step 3 - We transform moles into gram using the formula: mass = moles x molar mass

molar mass of O2 = (2x16) = 32 g/mol

mass = 0.004407 x 32

mass = 0.141 grams of O2

Answer: 0.141 grams of O2