Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1

To find the total pressure in a mixture of dioxygen (O₂) and dihydrogen (H₂), we can use the ideal gas law, which is expressed as PV = nRT. Here, P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. Let's break this down step by step.

Step 1: Convert the Temperature to Kelvin

The temperature given is 27°C. To convert this to Kelvin, we add 273.15:

T = 27 + 273.15 = 300.15 K

Step 2: Calculate the Number of Moles of Each Gas

Next, we need to determine the number of moles of dioxygen and dihydrogen. We can use the formula:

n = mass (g) / molar mass (g/mol)

• Dioxygen (O₂): The molar mass of O₂ is approximately 32 g/mol.
• Dihydrogen (H₂): The molar mass of H₂ is approximately 2 g/mol.

Now, let's calculate the moles:

• For dioxygen: n(O₂) = 8 g / 32 g/mol = 0.25 mol
• For dihydrogen: n(H₂) = 4 g / 2 g/mol = 2 mol

Step 3: Calculate the Total Number of Moles

The total number of moles in the mixture is the sum of the moles of dioxygen and dihydrogen:

n(total) = n(O₂) + n(H₂) = 0.25 mol + 2 mol = 2.25 mol

Step 4: Apply the Ideal Gas Law

Now that we have the total number of moles, we can use the ideal gas law to find the total pressure:

PV = nRT

Rearranging for P gives us:

P = nRT / V

Step 5: Substitute the Values

We have:

• n = 2.25 mol
• R = 0.083 bar dm³ K⁻¹ mol⁻¹
• T = 300.15 K
• V = 1 dm³

Plugging in these values:

P = (2.25 mol) × (0.083 bar dm³ K⁻¹ mol⁻¹) × (300.15 K) / (1 dm³)

Step 6: Calculate the Pressure

Now, let's do the math:

P = (2.25 × 0.083 × 300.15) / 1

P ≈ 56.1 bar

Final Result

The total pressure in the mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a 1 dm³ vessel at 27°C is approximately 56.1 bar.