At what temperature will 0.731 moles of neon gas occupy 10.30 L at 2.50 atm? Use the equation: MM= ρRTP
Solution:
n = 0.731 mol
P = 2.50 atm
V = 10.30 L
R = universal gas constant = 0.0821 L atm K−1 mol−1
T = ???
Ideal gas law can be used:
PV = nRT
To find the temperature, solve the equation for T:
T = PV / nR
T = (2.50 atm × 10.30 L) / (0.731 mol × 0.0821 L atm K−1 mol−1) = 429 K
T = 429 K = 156∘C
Answer: at 429 K
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