# You have a 1 L container of a gas at 20°C and 1 atm. Without opening the container, how could you tell whether the gas is chlorine or fluorine?

Jul 15, 2017

The ideal gas equation is $p V = n R T$, where:
- $p$ = pressure, $1 a t m \approx 100000 P a$
- $V$ = volume, $1 L = 0.001 {m}^{3}$
- $T$ = temperature, ${20}^{\circ} C \approx 293 K$
- $R$ = gas constant.
- $n$ = number of moles

$n = \frac{p V}{R T} = \frac{100000 \cdot 0.001}{8.31 \cdot 293} = 0.0410706292 m o l$

Asuming the mass of the container is negligible, $n = \frac{m}{M}$, where:
- $m$ = mass ($g$)
- $M$ = molar mass ($g$ $m o {l}^{- 1}$)

$m = n M$, is you know the mass of the container+gass, and substitute it in, you will get a value for molar mass, and therefore find the gas.h

Oct 3, 2017

Determine the mass of the container plus gas.

#### Explanation:

This assumes that you know the mass of the container.

We can use the Ideal Gas Law to calculate the mass of the gas.

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

$m = \frac{p V M}{R T}$

If the gas is chlorine,

m = (1 color(red)(cancel(color(black)("atm"))) × 1 color(red)(cancel(color(black)("L"))) × 70.91 "g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1""mol"^"-1"))) × 293.15 color(red)(cancel(color(black)("K")))) = "2.9 g"

If the gas is fluorine,

m = (1 color(red)(cancel(color(black)("atm"))) × 1 color(red)(cancel(color(black)("L"))) × 38.00 "g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1""mol"^"-1"))) × 293.15 color(red)(cancel(color(black)("K")))) = "1.6 g"

Thus, if the mass of container plus gas is 2.9 g more than the mass of the empty container, the gas is chlorine.

If the mass difference is only 1.6 g, the gas is fluorine.