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?

6.3 atm

General Formulas and Concepts:

Atomic Structure

Reading a Periodic TableMoles

Ideal Gas Law

Ideal Gas Law Formula: PV = nRT

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

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

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

Ithink the answer is 56.