Question

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

Answer 1

Answer:

174 K

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.