Questions
A vessel of volume 1660 contains 0.1 mole of oxygen and 0.2 mole of nitrogen. If the temperature of the mixture is 300 K, find its pressure.
detailed solution
Correct option is C
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 PaTalk to our academic expert!
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