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?

Mar 25, 2015

The pressure exercited by the gas will be $\text{2300 torr}$.

So, you know that you have a certain amount of gas, 10.0 g to be exact, in a 2.5-L container at ${27}^{\circ} \text{C}$. 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

${\text{10.0"cancel("g") * "1 mole O"_2/("32.0"cancel("g")) = "0.3125 moles O}}_{2}$

So,

$P V = n R T \implies P = \frac{n R T}{V}$

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

$P = \text{3.077 atm}$

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

$\text{3.077"cancel("atm") * "760 torr"/(cancel("1 atm")) = "2338.5 torr}$

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

$P = \textcolor{red}{\text{2300 torr}}$