What volume will 12.0 g of oxygen gas (#O_2#) occupy at 25°C and a pressure of 52.7 kPa?

1 Answer
Mar 5, 2016

Answer:

#"17.6 L"#

Explanation:

This is a classic example of how an ideal gas law equation practice problem looks like.

The ideal gas law equation looks like this

#color(blue)(| bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)))|)" "#, where

#P# - the pressure of the gas
#V# - the volume occupied by the gas
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.0821("atm" * "L")/("mol" * "K")#
#T# - the absolute temperature of the gas

Now, notice that the problem does not provide you with the number of moles of gas present in the sample. However, it does provide with the mass of the sample and the identity of the gas.

This means that you can use its molar mass as a conversion factor to go from grams to moles

#12.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of molecular oxygen")) = "0.375 moles O"_2#

Your biggest challenge when it comes to this type of problems will often be to make sure that the units given to you match those used in the expression of the ideal gas constant ,#R#.

In this case, you have to make sure that you convert the temperature from degrees Celsius to Kelvin and the pressure from kilopascals to atm by using the conversion factors

#T["K"] = t[""^@"C"] + 273.15#

#"1 atm " = " 101.325 kPa"#

Rearrange the ideal gas law equation to solve for #V#

#PV = nRT implies V = (nRT)/P#

Plug in your values to get

#V = (0.375color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25.0)color(red)(cancel(color(black)("K"))))/(52.7/101.325color(red)(cancel(color(black)("atm"))))#

#V = color(green)(|bar(ul(color(green)("17.6 L"))|) -># rounded to three sig figs