At what temperature will a 1.00 g sample of neon gas exert a pressure of 500. torr in a 5.00 L container?

Answer 1

PV = nRT

P = pressure = 500 torr x 1 atm/760 torr = 0.6579 atm

V = volume in L = 5.00 L

n = moles of gas = 1.00 g Ne x 1 mol Ne/20.18 g = 0.04955 mols

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = ?

Solving for T...

T = PV/nR = (0.6579 atm)(5.00 L) / (0.4955 mols)(0.0821 Latm/Kmol)

T = 80.9 K (-192ºC)