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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 $1.5 L$ 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 $3 L$ vessel at the same temperature. What is the partial pressure of $O_{2}$ ?

202.50 mm of Hg

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217.40 mm of Hg

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115.50 mm of Hg

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150.40 mm of Hg

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Solution

The correct option is A $202.50 mm of Hg$
Volume of $O_{2} = 1.5 L$
$P_{O_{2}} = 760 - 355 = 405 mm of Hg$
Since the temperature is constant,
Using Boyle's law,
$P_{O_{2}} V_{O_{2}} = P_{O_{2}}^{'} V_{O_{2}}^{'}$
$1.5 L \times 405 mm = 3 L \times P_{O_{2}}^{'}$
$∴ P_{O_{2}}^{'} = 202.5 mm of Hg$

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The vapour pressure of water at 80∘C is 355 mm of Hg. The total pressure of a 1.5 L vessel containing O2 saturated with water at 80∘C is 760 mm of Hg. The contents of the vessel are pumped into a 3 L vessel at the same temperature. What is the partial pressure of O2?

The correct option is A 202.50 mm of HgVolume of O2=1.5 L PO2=760−355=405 mm of Hg Since the temperature is constant, Using Boyle's law, PO2VO2=P′O2V′O2 1.5 L×405 mm=3 L×P′O2 ∴ P′O2=202.5 mm of Hg

The correct option is A $202.50 mm of Hg$
Volume of $O_{2} = 1.5 L$
$P_{O_{2}} = 760 - 355 = 405 mm of Hg$
Since the temperature is constant,
Using Boyle's law,
$P_{O_{2}} V_{O_{2}} = P_{O_{2}}^{'} V_{O_{2}}^{'}$
$1.5 L \times 405 mm = 3 L \times P_{O_{2}}^{'}$
$∴ P_{O_{2}}^{'} = 202.5 mm of Hg$