# What must the temperature, pressure, and volume units be in any gas law?

##### 2 Answers

The combination of all these laws can be derived from

#\mathbf(PV = nRT),#

by choosing two or three variables to keep constant, so whatever applies to the ideal gas law applies to the rest.

*For all of these laws, the volume, temperature, and pressure must match the units of the universal gas constant you choose so you get the answer in the right units.*

The universal gas constant in gas law problems is often used as:

#"R" = "0.082057 L"cdot"atm/mol"cdot"K"# (1)#"R" = "0.083145 L"cdot"bar/mol"cdot"K"# (2)

**(1)**

#R = (PV)/(nT)#

Looking at only the units, we see:

#("L"cdot"atm")/("mol"cdot"K") = (stackrel(V)overbrace("L")cdot stackrel(P)overbrace("atm"))/(stackrel(n)overbrace("mol")cdot stackrel(T)overbrace("K"))#

So these are clearly equivalent units. That means if you use **(1)**, then **volume** is in **temperature** is in **pressure** is in

**(2)**

#R = (PV)/(nT)#

Looking at only the units, we see:

#("L"cdot"bar")/("mol"cdot"K") = (stackrel(V)overbrace("L")cdot stackrel(P)overbrace("bar"))/(stackrel(n)overbrace("mol")cdot stackrel(T)overbrace("K"))#

As before, these are clearly equivalent units. That means if you use **(2)**, then **volume** is in **temperature** is in **pressure** is in

#### Answer:

The temperature must be in 'K for the ratios to work out. The pressure and volume units are arbitrary, as long as they are consistent.

#### Explanation:

Be aware however that using them to calculate actual molar values requires the correct use of units and conversion factors. You cannot mix the units and obtain the correct answers.