Question

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?

Answer 1

The ideal gas equation is `pV=nRT`, where:
- `p` = pressure, `1atm~~100000Pa`
- `V` = volume, `1L=0.001m^3`
- `T` = temperature, `20^circC~~293K`
- `R` = gas constant.
- `n` = number of moles

`n=(pV)/(RT)=(100000*0.001)/(8.31*293)=0.0410706292mol`

Asuming the mass of the container is negligible, `n=m/M`, where:
- `m` = mass (`g`)
- `M` = molar mass (`g` `mol^(-1)`)

`m=nM`, 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

Answer 2

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.

`color(blue)(bar(ul(|color(white)(a/a)pV = nRT = m/MRTcolor(white)(a/a)|)))" "`

`m = (pVM)/(RT)`

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.