Question

What would be the pressure, in atm, exerted by 3.2 moles of oxygen with a volume of 12.3 liters and a temperature of 294 K?

Answer 1

6.3 atm

General Formulas and Concepts:

Atomic Structure

• Reading a Periodic Table
• Moles

Ideal Gas Law

Ideal Gas Law Formula: PV = nRT

• P is pressure (in atm)
• V is volume (in L)
• n is number of moles
• R is gas constant (0.0821 L · atm · mol⁻¹ · K⁻¹)
• T is temperature (in K)

Step-by-step explanation:

Step 1: Define

Identify variables

[Given] n = 3.2 mol O₂

[Given] V = 12.3 L

[Given] T = 294 K

[Solve] P

Step 2: Solve for P

1. Substitute in variables [Ideal Gas Law Formula]: P(12.3 L) = (3.2 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)(294 K)
2. Multiply [Cancel out units]: P(12.3 L) = 77.2397 L · atm
3. Isolate P [Cancel out units]: P = 6.27965 atm

Step 3: Check

Follow sig figs and round. We are given 2 sig figs as our lowest.

6.27965 atm ≈ 6.3 atm