# Question #5ea04

##### 1 Answer

#### Explanation:

Your tool of choice here will be the **ideal gas law equation**, which looks like this

#color(blue)(ul(color(black)(PV = nRT)))#

Here

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

Notice that the equation uses *moles* of gas, but that you are given the **mass** of the sample. The first thing to do here will be to convert the sample from grams to moles by using the **molar mass** of oxygen gas.

#3.8 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.119 moles O"_2#

Now, room temperature is usually taken to mean a temperature of

#T["K"] = 20^@"C" + 273.15 = "293.15 K"#

No mention of the pressure at which the gas is being kept was made, but you can assume that you're dealing with normal pressure, i.e. with the pressure *at sea level*.

#P = "1.0 atm"#

Rearrange the ideal gas law equation to solve for

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

Plug in your values to find

#V = (0.119 color(red)(cancel(color(black)("moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 293.15color(red)(cancel(color(black)("K"))))/(1.0color(red)(cancel(color(black)("atm"))))#

#color(darkgreen)(ul(color(black)(V = "2.9 L")))#

The answer is rounded to two **sig figs**.