If 0.600 moles of Nitrogen gas is collected at a pressure of 234 kPa in a
container which has a volume of 12.0 L, what is the temperature of the
nitrogen gas in the container? Express the temperature in Kelvin. (use the R
value for kPa)
563.18 K
O 4680 K
O 52 atm
O 337.91K

563.18 K

General Formulas and Concepts:

Gas Laws

Ideal Gas Law: PV = nRT

P is pressure (in kPa)L is volume (in L)n is molesR is gas constantT is temperature (in K)

Explanation:

Step 1: Define

[Given] n = 0.600 mol N₂

[Given] P = 234 kPa

[Given] L = 12.0 L

[Given] R = 8.314 L · kPa · mol⁻¹ · K⁻¹

[Solve] T

Step 2: Solve for T

Substitute in variables [Ideal Gas Law]:                                                          (234 kPa)(12.0 L) = (0.600 mol)(8.314 L · kPa · mol⁻¹ · K⁻¹)TMultiply:                                                                                                             2808 kPa · L = (4.9884 L · kPa · K⁻¹)T[Division Property of Equality] Isolate temp:                                                   T = 562.906 K

Topic: AP Chemistry

Unit: Gas Laws

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