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?

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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?

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Answer 1 · 2015-03-25T18:59:36

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

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

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

So,

$P V = n R T \Rightarrow P = \frac{n R T}{V}$ $PV = nRT => P = (nRT)/V$

$P = \frac{0.3125 moles \cdot 0.082 \frac{atm \cdot L}{mol \cdot K} \cdot (273.15 + 27) K}{2.50 L}$ $P = ("0.3125"cancel("moles") * 0.082("atm" * cancel("L"))/(cancel("mol") * cancel("K")) * (273.15 + 27)cancel("K"))/(2.50cancel("L"))$

$P = 3.077 atm$ $P = "3.077 atm"$

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

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

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

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

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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?

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