One-litre of oxygen at a pressure of 1 atm and two-litre of nitrogen at a pressure of 0.5 atm , are introduced into a vessel of volume one-litre. If there is no change in temperature, the final pressure of the mixture of the gases [ in atm ] is

Question

One-litre of oxygen at a pressure of 1 atm and two-litre of nitrogen at a pressure of 0.5 atm , are introduced into a vessel of volume one-litre. If there is no change in temperature, the final pressure of the mixture of the gases [ in atm ] is, 1 . 5, 2 . 5, 2, 4

MARKS App · Accepted Answer

Ideal gas equation is given by p V = n R T ...(i) For oxygen, p = 1 atm, V = 1 L, n = n O 2 Therefore, Eq. (i) becomes ∴ 1 × 1 = n O 2 R T ⇒ n O 2 = 1 R T For nitrogen p = 0.5 atm, V = 2 L, n = n N 2 ∴ 0.5 × 2 = n N 2 R T ⇒ n N 2 = 1 R T For mixture of gas p m i x V m i x = n m i x R T Here, n m i x = n O 2 + n N 2 ∴ p m i x V m i x R T = 1 R T + 1 R T ⇒ p m i x V m i x = 2 ( V m i x = 1 )

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