Asample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample is saturated with water vapor. if the total pressure of the mixture at 26.4 °c is 765 torr, what is the partial pressure of oxygen? the vapor pressure of water at 26.4 °c is 25.81 mm hg.

To calculate the partial pressure of oxygen we need to make reference to this formula from Dalton's law of partial pressures: P(total) = P(gas1) + P(gas2). In our question, P(total) = 765 torr which is equivalent to 765 mm hg. Also P(gas1) is vapour pressure which is 25.81. To obtain the partial pressure of oxygen , we subtract P(gas1) from P(total) i.e P(gas2)= P(total)- P(gas2). Therefore: P(gas2)= 765mm hg-25.81mm hg P(gas2)=739.19 mm hg. This implies that partial pressure of oxygen = 739.19 mm hg

Work done = force ×distancebut force = mass ×gravityand gravity = 10m\s force = 2×10 = 20 work done = 20×3 = 60 joules