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What is the volume occupied by 14.3 g of argon gas at a pressure of 1.35 atm and a temperature of 436 K?

1 Answer

8 votes

Answer:

9.487 L

Step-by-step explanation:

argon mole wt = 39.948 gm/mole

14.3 gm / 39.948 gm/mole = .357965 moles of argon

Use Ideal Gas Law:

PV = n RT where R = .082057 L-atm / K-mol

V = .357965 * .082057 * 436 / 1.35 = 9.487 L