What is the temperature of 0.545 mole of neon in a 2.00 L vessel at 4.68 atm?
Given:
P = 4.68 atm
V = 2.0 L
n = 0.545 mol
use:
P * V = n*R*T
4.68 atm * 2 L = 0.545 mol* 0.08206 atm.L/mol.K * T
T = 209 K
Answer: 209 K
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