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?

2 Answers
Jul 15, 2017

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

Oct 3, 2017

Answer:

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.