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When 0.655 L of Ar at 1.20 atm and 227°C is mixed with 0.250 L of O2 at 501 torr and 127°C in a 400 mL flask at 27°C, what is the pressure in the flask?

2007-10-13 17:55:18 · 2 answers · asked by raleighzia 1 in Science & Mathematics Chemistry

2 answers

Please use ideal gas law to figure out the partial presure of both Ar and O2 (at the final volume and temperature):
PV/T = pv/t or P = pvT/(Vt)
Ar: 0.655*1.20*300/(0.400*500) = 1.179 (atm)
O2: 0.250*(501/760)*300/(0.400*400) = 0.309 (atm)
So the total pressure = 1.179atm + 0.309atm = 1.49 atm

2007-10-13 19:04:59 · answer #1 · answered by Hahaha 7 · 0 0

Pressure (symbol: p) is the force per unit area applied on a surface in a direction perpendicular to that surface.

2007-10-13 19:38:09 · answer #2 · answered by Anonymous · 0 0

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