# 22.4 L of hydrogen gas contains how many molecules?

Aug 25, 2016

At 0°"C" temperature and $1$ atmosphere pressure, this volume is one mole. This is Avogrado's number of molecules, $\textcolor{b l u e}{6.02 \times {10}^{23}}$.

#### Explanation:

The temperature and pressure chosen for this answer, 0°"C" and $1$ atmosphere, were intended to be "standard" temperature and pressure, and the question was apparently formulated with that standard in mind. But the standard pressure was changed to $100$ kPa (one atmosphete is $101.3$ kPa) in 1982, and confusingly some references still quote the old standard (e.g., http://www.kentchemistry.com/links/GasLaws/STP.htm). One solution to this dilemma that fits with the question is to avoid the "standard" label and refer to the temperature and pressure specifically.

The connection between $22.4 \text{ L}$ and one mole, at 0°"C" and $1$ atmosphere, comes from the Ideal Gas Law

$P V = n R T$

$P$ = pressure, $1$ atm
$V$ = volume, see below

$n$ = moles of gas, put in $1$ mole

$R$ = gas constant, in pressure-volume unots it's $\textcolor{b l u e}{0.08206 \left(\text{L atm")/("mol K}\right)}$

$T$ = tempeature, 0°"C" = 273.15 "K"

So

$\left(1 \text{ atm")(V)=(1" mol")(.08206 ("L atm")/("mol K"))(273.15" K}\right)$

Solve for $V$ and to three significant figures you get $\textcolor{b l u e}{22.4 \text{ L}}$. Thus at 0°"C" temperature and $1$ atmosphere pressure, $22.4 \text{ L}$ of an ideal gas is one mole.

Aug 26, 2016

At a pressure of 1 bar and a temperature of 0 °C, 22.4 L of hydrogen gas contains 5.94 × 10^23color(white)(l) "molecules".

#### Explanation:

The number of molecules depends on the temperature and pressure of the gas.

I shall arbitrarily assume that the gas is at STP (1 bar and 0 °C).

We can use the Ideal Gas Law to calculate the moles of hydrogen:

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

We can rearrange this to get

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

$P = \text{1 bar}$
$V = \text{22.4 L}$
$R = \text{0.0.08 314 bar·L·K"^"-1""mol"^"-1}$
$T = \text{(0 + 273.15) K" = "273.15 K}$

n = (1 color(red)(cancel(color(black)("bar"))) × 22.4 color(red)(cancel(color(black)("L"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K")))) = "0.9864 mol"

We know that 1 mol of a gas contains 6.022 × 10^23 color(white)(l)"molecules".

$\text{No. of molecules" = 0.9864 color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23color(white)(l) "molecules")/(1 color(red)(cancel(color(black)("mol")))) = 5.94 × 10^23color(white)(l) "molecules}$