Aballoon contains 5l of oxygen at a pressure of 90 kpa. how much oxygen will the ballon contain if the pressure is lowered to 67 kpa.

1- According to this formula:
P1V1 = P2V2

2-By converting P from KPa to atm:
1 KPa = 0.00986923267 atm
So, 
P1= 0.888231 atm
P2 = 0.661239 atm

by substitution on the previous formula:
P1V1 = P2V2
0.888231 x 5 = 0.661239 x V2
So,
V2(for the oxygen will the ballon contain) = 6.7 L

Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol