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HIGVICIOUS HIGVICIOUS
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10 years ago
A 345 mL sample of 0.5454 g of He gas is at a pressure of 900 torr. What is the temperature of this gas in Kelvin?
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Subject Expert
10 years ago
Hi HIGVICIOUS,

For this we need to use the gas law equation: PV = nRT

P = 900/760 = 1.18atm
V = .345L
n = .5454/4 = .13635 mol
R = .08206 atm*L/(mol*K)

1.18*.345 = .13635* .08206*T
.4071 = .0112T
T = 36 Kelvin

Hope this helps,
Laser
padre,  ruffus
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