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?

1 answer:

5 0

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

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