A 0.5 mol sample of N2 is in a 6L container at 2 atm. what is the temperature of the gas in K
1 answer:
Answer:
300 K
General Formulas and Concepts:
<u>Atomic Structure</u>
<u>Gas Laws</u>
Ideal Gas Law: PV = nRT
<em>P</em> is pressure <em>V</em> is volume <em>n</em> is moles <em>R</em> is gas constant <em>T</em> is temperature Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>n</em> = 0.5 mol N₂
[Given] <em>V</em> = 6 L
[Given] <em>P</em> = 2 atm
[Given] <em>R</em> = 0.0821 L · atm · mol⁻¹ · K⁻¹
[Solve] <em>T</em>
<em />
<u>Step 2: Solve for </u><em><u>T</u></em>
Substitute in variables [Ideal Gas Law]: (2 atm)(6 L) = (0.5 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)T Multiply [Cancel out units]: 12 atm · L = (0.04105 L · atm · K⁻¹)T Isolate <em>T</em> [Cancel out units]: 292.326 K = T Rewrite: T = 292.326 K
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig as our lowest.</em>
292.326 K ≈ 300 K
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