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5 votes
At what temperature will 1.50 moles of nitrogen exert a pressure of 1.072 atm in a 20.0 L cylinder?

238 K
0.00574 K
337 K
174 K

1 Answer

5 votes

Answer:

174 K

Step-by-step explanation:

To answer this question we will use the PV=nRT formula, where:

  • P = 1.072 atm
  • V = 20.0 L
  • n = 1.50 mol
  • R = 0.082atm·L·mol⁻¹·K⁻¹
  • T = ?

We input the data given by the problem:

  • 1.072 atm * 20.0 L = 1.50 mol * 0.082atm·L·mol⁻¹·K⁻¹ * T

And solve for T:

  • T = 174 K

Meaning the correct answer is the fourth option, 174 K.

User JuicyFruit
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