The oxygen tank with a volume of 3.33 L that contained 7.3 moles of molecular oxygen at 298K was in a building during a fire in an adjacent room. It was subjected to a temperature of 1400 K. What was the internal pressure in atm at this time?
volume = V = 3.33 L
temperature = T= 298K
number of moles = n = 7.3 moles
R = 0.0821 L-atm/mol-K
Ideal gas equation
PV=nRT
P = nRT/V = 7.3 x 0.0821 x 298 / 3.33 = 53.63 atm
Now, the gas subjected to a temperture of 1400K
Now
P1 = 53.63 atm P2 = ?
T1= 298K T2 = 1400K
according to Gay-Lussacs law
P1/T1 = P2/T2 at constant volume
53.63 / 298 = P2 / 1400
P2 = 251.95 atm
pressure at the temperature at 1400K = 251.95 atm
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