The vapor pressure of water at 80,^oC is 355. | vedclass

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(B) The total pressure is given by Dalton's Law: $P_{Total} = P_{H_2O} + P_{O_2}$.Given $P_{Total} = 760\, mm$ and $P_{H_2O} = 355.5\, mm$ at $80\,^oC

Vedclass · Accepted Answer

$809$

Vedclass · Answer

(B) The total pressure is given by Dalton's Law: $P_{Total} = P_{H_2O} + P_{O_2}$.Given $P_{Total} = 760\, mm$ and $P_{H_2O} = 355.5\, mm$ at $80\,^oC$.Therefore,the initial partial pressure of $O_2$ is $P_{O_2(1)} = 760 - 355.5 = 404.5\, mm$.When the gas is transferred to a $50\, mL$ vessel at constant temperature,we apply Boyle's Law: $P_1V_1 = P_2V_2$.Here,$P_1 = 404.5\, mm$,$V_1 = 100\, mL$,$V_2 = 50\, mL$.$P_2 = \frac{404.5 	imes 100}{50} = 809\, mm$.Thus,the partial pressure of $O_2$ in the $50\, mL$ vessel is $809\, mm$.

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