If 5.0 moles of O2 and 3.0 moles of N2 are placed in a 30.0 L tank at a temperature of 250 °C...?

Consider the resulting gaseous mixture:
Volume, V = 30.0 L
Total no. of moles, n = (5.0 + 3.0) mol = 8.0 mol
Temperature, T = (273 + 250) K = 523 K
Gas constant, R = 0.08206 L atm / (mol K)

Gas law: PV = nRT
Then, P = nRT/V

Pressure, P = 8.0 × 0.08206 × 523 / 30.0 atm = 11.4 atm

 nO2 = 5.0mol

nN2 = 3.0mol

Volume, V = 30L

Temp., T = 250oC = 523K

Total number of moles, n = nO2 + nN2

= 5.0mol + 3.0mol

n = 8.0mol

PV = nRT

P = (8.0mol x 0.0821LatmK-1mol-1 x 523K) / 30L

= 11.45atm

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