Question

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

Answer 1

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 moles
• R is gas constant
• T is temperature (in K)

Step-by-step 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

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

Topic: AP Chemistry

Unit: Gas Laws