Question

What pressure (in torr) is exerted by 10.0 g of O2 in a 2.50 L container at a temperature of 27 degree C?

Answer 1

The pressure exercited by the gas will be `"2300 torr"`.

So, you know that you have a certain amount of gas, 10.0 g to be exact, in a 2.5-L container at `27^@"C"`. You can use the ideal gas law equation to solve for the pressure in atm, then use a simple conversion factor to go from atm to torr.

The number of moles of oxygen present in the container is

`"10.0"cancel("g") * "1 mole O"_2/("32.0"cancel("g")) = "0.3125 moles O"_2`

So,

`PV = nRT => P = (nRT)/V`

`P = ("0.3125"cancel("moles") * 0.082("atm" * cancel("L"))/(cancel("mol") * cancel("K")) * (273.15 + 27)cancel("K"))/(2.50cancel("L"))`

`P = "3.077 atm"`

Since 1 atm is defined as being equal to 760 torr, you'll get

`"3.077"cancel("atm") * "760 torr"/(cancel("1 atm")) = "2338.5 torr"`

Rounded to two sig figs, the number of sig figs given for 27 degrees Celsius, the answer will be

`P = color(red)("2300 torr")`