Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm8 at 27 ∘C. (R = 0.083 bar dm8 K−1 mol−1

Total pressure =  56.77 bar

Explanation:

Given data:

Mass of dioxygen = 8 g

Mass of dihydrogen = 4 g

Volume of vessel = 1 dm³

Temperature = 27°C (27+273 = 300 K)

R = 0.083 bar.dm³ / mol.K

Total pressure = ?

Solution:

Number of moles of dioxygen:

Number of moles = mass/molar mass

Number of moles = 8 g/ 32 g/mol

Number of moles = 0.25 mol

Pressure of dioxygen:

PV = nRT

P = nRT/V

P = 0.25 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 6.97 bar.dm³  /1 dm³

P = 6.97 bar

Number of moles of dihydrogen:

Number of moles = mass/molar mass

Number of moles = 4 g/ 2 g/mol

Number of moles = 2 mol

Pressure of dihydrogen:

PV = nRT

P = nRT/V

P = 2 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 49.8 bar.dm³  /1 dm³

P = 49.8 bar

Total pressure of mixture in a vessel:

Total pressure = P (O₂) + P(H₂)

Total pressure =  6.97 bar + 49.8 bar

Total pressure =  56.77 bar