Question

How many moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C?

Answer 1

PV = nRT --> n = PV/RT
n = (2×3)/(.0821×298) = 6/24.47
n = 0.245 moles O2

Answer 2

0.246 moles of oxygen

Step-by-step explanation:

The formula to use here is the ideal gas equation:

PV = nRT

where P = pressure (2atm)

V = volume (3L)

n = number of moles (unknown)

R = ideal gas constant (its either 0.082L.atm/mol.K or 8.314J/mol.K depending on the values/units provided)

T = temperature in kelvin (25°C ⇒ 25 + 273 = 298K)

For the ideal gas constant (R), 0.082 shall be used instead of 8.314 because we do not have any value having the unit of joules (J) which will be needed if we are to use 8.314

Our unknown is number of moles (n), so from the ideal gas equation

n = PV ÷ RT

n = (2 × 3) ÷ (0.082 × 298)

n = 0.246 moles of oxygen