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statuscvo [17]
4 years ago
6

A 75.0-liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. What is the temperature of the canister?

Chemistry
1 answer:
alexandr1967 [171]4 years ago
8 0

Answer:

  • <u>312 K</u>

Explanation:

<u>1) Data:</u>

<em />

a) V = 75.0 liter

b) n = 15.82 mol

c) p = 546.8 kPa

d) T = ?

<u>2) Formula:</u>

  • Ideal gas equation: pV = nRT

Where:

  • n = number of moles
  • V = volume
  • p = absolute pressure
  • T = absolute temperature
  • R = Universal Gas constat: 8.314 kPa - liter / K-mol

<u>3) Solution:</u>

a) <u>Solve the equation for T</u>:

  • T = pV / (nR)

b) <u>Substitute and compute</u>:

  • T = 546.8 kPa × 75.0 l iter / (15.82 mol × 8.314 kPa-liter/K-mol) = 312 K

(since the volume is expressed with 3 significant figures, the answer must show also 3 significant figures)

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