A sample of oxygen collected over water at
32C and 1038 torr has a volume of 0.741 L.
What volume would the dry oxygen occupy
under standard conditions?The vapor pres-
sure of water at 32C is 35.7 torr.
Answer in units of L
This problem can be solved using the combined gas law equation.
P1V1/T1 = P2V2/T2
***The key to this type of problem is to let the units be the same for the 1's and 2's.
*** at standard condition, T is 273.15 and P is 1atm or 760 torr
P1 = 1038 torr P2 = 760 torr
T1 = 32 + 273.15 = 305.15K T2 = 273.15 K
V1 = 0.741L V2 = ?
applying the formula above, substitue the values given.
(1038)(0.741)/ 305.15 = (760)(V2)/(273.15)
V2 = 0.906 L
hope this helps :)