If #"H"_2# and #"He"# are at #"1 atm"# and #"2 atm"# respectively, find the ratio of speeds for #"H"_2# vs. #"He"#?

A priori, you should be able to recognize that #H_2# is lighter than #He#, so it should be faster.

But the first method shown is more robust, because it does not rely on the gas being ideal, only isotropic.

We can consider their RMS speed:

#u_(RMS) = sqrt((3RT)/M)#

For ideal gases, we know that #PV = nRT#, so we also have:

#u_(RMS) = sqrt((3PV)/(nM))#

where #P#, #V#, #n#, #R#, and #T# are known from the ideal gas law, while #M# is the molar mass in #"kg/mol"#.

METHOD 1: SAME TEMPERATURE

In the first method, we would find that at the SAME temperature:

#(u_(RMS)(H_2))/(u_(RMS)(He)) = sqrt(M_(He)/M_(H_2))#

#= sqrt(("0.0040026 kg/mol")/("0.0020158 kg/mol"))#

#~~ sqrt2#

So,

#color(blue)(u_(RMS)(H_2) ~~ sqrt2u_(RMS)(He))#

METHOD 2: DIFFERENT PRESSURES

In the second method, we would find that #"H"_2# at #"1 atm"# and #"He"# at #"2 atm"# would involve the following assumptions:

• #He# and #H_2# are assumed to have the same molar volume ONLY if they are at the same conditions, given that they are PRESUMED to be ideal gases.
• They have the same volume. Therefore, since the container pressure is twice as much for #He#, the molar volume is half as large for #He#, and there is twice the mols of #He# to leave the volume the same as for #H_2#.

#(u_(RMS)(H_2))/(u_(RMS)(He)) = sqrt((P_(H_2))/(n_(H_2)M_(H_2)))/(sqrt((P_(He))/(n_(He)M_(He))))#

We have just mentioned:

• #P_(He) = 2P_(H_2)#
• #n_(He) = 2n_(H_2)#

Therefore, we get:

#(u_(RMS)(H_2))/(u_(RMS)(He)) = sqrt((P_(H_2))/(n_(H_2)M_(H_2)))/(sqrt((2P_(H_2))/(2n_(H_2)M_(He))))#

#= sqrt((1)/(M_(H_2)))/(sqrt((2)/(2M_(He))))#

#= sqrt(M_(He)/M_(H_2))#

And so, as before,

#color(blue)(u_(RMS)(H_2) ~~ sqrt2u_(RMS)(He))#