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

May 11, 2016

The combination of all these laws can be derived from

$\setminus m a t h b f \left(P V = n R T\right) ,$

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:

• $\text{R" = "0.082057 L"cdot"atm/mol"cdot"K}$ (1)
• $\text{R" = "0.083145 L"cdot"bar/mol"cdot"K}$ (2)

(1)

$R = \frac{P V}{n T}$

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 $\setminus m a t h b f \left(\text{L}\right)$, temperature is in $\setminus m a t h b f \left(\text{K}\right)$, and pressure is in $\setminus m a t h b f \left(\text{atm}\right)$.

(2)

$R = \frac{P V}{n T}$

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 $\setminus m a t h b f \left(\text{L}\right)$, temperature is in $\setminus m a t h b f \left(\text{K}\right)$, and pressure is in $\setminus m a t h b f \left(\text{bar}\right)$.

May 11, 2016