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

##### 1 Answer

#### 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

*universal gas constant*, usually given as

**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* ,

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

#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 threesig figs