A vessel of volume 1660 cm3 contains 0.1 mole of oxygen and 0.2 mole of nitrogen. If the temperature of the mixture is 300 K, find its pressure.

We have pV = nRT   ⇒ p = nRTVThe partial pressure due to oxygen isp1 = (0.1 mol)(8.3 J/mol-K)(300 K)(1660×10-6m3) = 1.5×105 PaSimilarly, the partial pressure due to nitrogen isp2 = 3.0 × 105 PaThe total pressure in the vessel is p = p1+p2 = (1.5+3.0)×105 Pa = 4.5 × 105 Pa