# How do you solve gas laws problems with the given "at STP" and with mass?

Jan 15, 2017

I like to use the Ideal Gas Law.

#### Explanation:

The Ideal Gas Law is:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

• $P$ is the pressure
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the universal gas constant
• $T$ is the temperature

We can rearrange the Ideal Gas Law to get

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

Also,

$\text{Moles" = "mass"/"molar mass} = \frac{m}{M}$

$V = \left(\frac{m}{M}\right) \frac{R T}{P} = \frac{m R T}{P M}$

EXAMPLE

What is the volume of 10.0 g of hydrogen at STP?

Solution

Remember that STP is defined as 0 °C and 1 bar.

$m = \text{10.0 g}$
$R = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T = \text{(0 + 273.15) K" = "273.15 K}$
$P = \text{1 bar}$
$M = \text{2.016 g·mol"^"-1}$

V = (mRT)/(PM) = (10.0 color(red)(cancel(color(black)("g"))) × "0.083 14" color(red)(cancel(color(black)("bar"))) ·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 2.016 color(red)(cancel(color(black)("g·mol"^"-1")))) = "113 L"
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