The vapour pressure of water at 80-C is 355 mm of Hg- The total pressure of a 1 L vessel containing O2 saturated with water at 80-C is 760 mm of Hg- The contents of the vessel are pumped into a 0-3 L vessel at the same temperature- What is the partial pressure of O2-

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Byju's Answer

Standard XII

Chemistry

Boyle's Law

The vapour pr...

Question

The vapour pressure of water at $80^{\circ} C$ is $355 mm$ of $H g$ . The total pressure of a $1 L$ vessel containing $O_{2}$ saturated with water at $80^{\circ} C$ is $760 mm$ of $H g$ . The contents of the vessel are pumped into a $0.3 L$ vessel at the same temperature. What is the partial pressure of $O_{2}$ ?

1350 mm of Hg

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2178.3 mm of Hg

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121.5 mm of Hg

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355 mm of Hg

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Solution

The correct option is A $1350 mm of Hg$
Volume of $O_{2} = 1 L$
$P_{O_{2}} = 760 - 355 = 405 mm of Hg$
Since the tempearture is constant,
Using Boyle's law:
$P_{O_{2}} V_{O_{2}} = P_{O_{2}}^{'} V_{O_{2}}^{'}$
$1 L \times 405 mm = 0.3 L \times P_{O_{2}}^{'}$
$∴ P_{O_{2}}^{'} = 1350 mm of Hg$

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Boyle's Law

Standard XII Chemistry

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The vapour pressure of water at 80∘C is 355 mm of Hg. The total pressure of a 1 L vessel containing O2 saturated with water at 80∘C is 760 mm of Hg. The contents of the vessel are pumped into a 0.3 L vessel at the same temperature. What is the partial pressure of O2?

The correct option is A 1350 mm of HgVolume of O2=1 L PO2=760−355=405 mm of Hg Since the tempearture is constant, Using Boyle's law: PO2VO2=P′O2V′O2 1 L×405 mm=0.3 L×P′O2 ∴ P′O2=1350 mm of Hg

The correct option is A $1350 mm of Hg$
Volume of $O_{2} = 1 L$
$P_{O_{2}} = 760 - 355 = 405 mm of Hg$
Since the tempearture is constant,
Using Boyle's law:
$P_{O_{2}} V_{O_{2}} = P_{O_{2}}^{'} V_{O_{2}}^{'}$
$1 L \times 405 mm = 0.3 L \times P_{O_{2}}^{'}$
$∴ P_{O_{2}}^{'} = 1350 mm of Hg$