What is the pressure, in atmospheres, of 6.0 grams of h2 gas if it has a volume of 73 liters when the temperature is 25 c?

PV = nRT   (where P = pressure; V = volume; n = number of moles; R = gas constant; T = Temperature)

Moles of H₂ = mass of H₂ ÷ molar mass of H₂
                    =  6.0 g ÷ 2 g/mol
                    =  3.0 mol

To get the answer in atmospheres, let's convert C to K 
      25 °C ≡ 298 K

Now, based on the formula above
     P = (nRT) ÷ V

    P = (3.0 mol × (0.082 L · atm/mol · K) × 298 K) ÷  73 L
       ≈  1 .004 atm

oxygen

Explanation: