Question

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?

Answer 1

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