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The vapour pressure of water at {80}^{{}}{C} is 355 mm of Hg.

Question

The vapour pressure of water at $80^{\circ} C$ is 355 mm of Hg. 1 L vessel contains $O_{2}$ at $80^{\circ} C$ , which is saturated with water and the total pressure being 760 mm of Hg. The contents of the vessel were pumped into 0.3 L vessel at the same temperature. What is the partial presure of $O_{2}$ ?

1350 Hg

2178.3 Hg

121.5 Hg

355 Hg

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Alexander Discussed

The vapour pressure of water at

80^{\circ} C

is 355 mm of Hg. 1 L vessel contains

O_{2}

80^{\circ} C

, which is saturated with water and the total pressure being 760 mm of Hg. The contents of the vessel were pumped into 0.3 L vessel at the same temperature. What is the partial presure of

O_{2}

14 mins ago

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14 mins ago

Text solutionVerified

Answer: aSolution: (a) Volume of

O_{2}

=1 L
and

P_{O_{2}}

=760-355=405 mm Hg
Since, the temperature is constant, using Boyle's law we get

1 L \times 405 m m = 0.3 L \times P^{'} o_{2}

P'o_(2) =1350 mmHg

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Question Text	The vapour pressure of water at $80^{\circ} C$ is 355 mm of Hg. 1 L vessel contains $O_{2}$ at $80^{\circ} C$ , which is saturated with water and the total pressure being 760 mm of Hg. The contents of the vessel were pumped into 0.3 L vessel at the same temperature. What is the partial presure of $O_{2}$ ?
Answer Type	Text solution:1
Upvotes	150

The vapour pressure of water at 80∘C is 355 mm of Hg. 1 L vessel contains O2​ at 80∘C, which is saturated with water and the total pressure being 760 mm of Hg. The contents of the vessel were pumped into 0.3 L vessel at the same temperature. What is the partial presure of O2​?

Text solutionVerified

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