A 0.5 mol sample of N2 is in a 6L container at 2 atm. what is the temperature of the gas in K

Using the gas equation
PV=nRT
T=PV/nR
T= 2 atm * 6L/0.5 mol* 0.082
T= 292.68 kelvin

300 K

General Formulas and Concepts:

Atomic Structure

Diatomic ElementsMoles

Gas Laws

Ideal Gas Law: PV = nRT

P is pressureV is volumen is molesR is gas constantT is temperature

Explanation:

Step 1: Define

Identify variables

[Given] n = 0.5 mol N₂

[Given] V = 6 L

[Given] P = 2 atm

[Given] R = 0.0821 L · atm · mol⁻¹ · K⁻¹

[Solve] T

Step 2: Solve for T

Substitute in variables [Ideal Gas Law]:                                                          (2 atm)(6 L) = (0.5 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)TMultiply [Cancel out units]:                                                                               12 atm · L = (0.04105 L · atm · K⁻¹)TIsolate T [Cancel out units]:                                                                             292.326 K = TRewrite:                                                                                                             T = 292.326 K

Step 3: Check

Follow sig fig rules and round. We are given 1 sig fig as our lowest.

292.326 K ≈ 300 K

What do you mean by that? like the contents? the thc level?