# Question #d1b5d

##### 1 Answer

It will be a long answer regardless, so just scroll through and read carefully.

The difference here we'll have to probe is the mols of each gas in each flask. Having all the same masses, to some extent,

Assuming ideality, each gas from the

ideal gas lawwill exert a pressure#P_q# on the walls of flask#q# :

#P_A = (n_ART)/V = ((4 cancel("g CH"_4) xx "1 mol"/(16.0426 cancel"g"))("0.083145 L"cdot"atm/mol"cdot"K")("265 K"))/("9 L") = "0.6104 bars"# You can see that the molar mass comes in as an

inverserelationship.Therefore, the flask containing the gas with the

lowestmolar mass will contain themostmols of that gas and thus the pressure will behighestthere (i.e. in flask#bb(C)# ).

The

average molar kinetic energyis given by the equipartition theorem for ideal monatomic gases:

#barK_(avg) -= K_(avg)/(n_(gas)) = N/2 RT# ,where

#N# is the number of degrees of freedom for a certain motion, and#R# and#T# are known from the ideal gas law.

#"CH"_4# needsthreeangles to describe its rotational motion, but#"H"_2# only needstwoangles, and#"He"# hasnorotational motion to consider (under the hard sphere approximation).Furthermore, all three gases will have three translational dimensions to move in (

#x,y,z# ), and we ignore vibrations (which are only significant for methane anyway).As a result,

#"CH"_4# will have#N = 3 + 3 = 6# from translation and rotation.#"H"_2# will have#N = 3 + 2 = 5# from translation and rotation.#"He"# will have only#N = 3 + 0 = 3# from translation and rotation.Therefore, we expect that

methane will have the largest average molar kinetic energy, as multiplying by the largest number of degrees of freedomincreases#barK_(avg)# :

#color(blue)(overbrace(barK_(avg))^("Rotational + Translational only") = overbrace(6/2 RT)^(CH_4) > overbrace(5/2 RT)^(H_2) > overbrace(3/2 RT)^(He))#

It doesn't matter how many mols of each there are, because we've divided that out and made the answer intensive.

Diffusion rateis again dependent on molar mass. The lower the molar mass, the lighter the gas and thus the faster it diffuses, simply because it moves faster.Hence,

#bb("H"_2)# diffuses fastest.

The

total kinetic energyof the molecules combined, as each gas is treated ideally, will depend only on the mass and speed:

#K = 1/2 mv^2# A good choice of speed is the root-mean-square speed:

#v_(RMS) = sqrt((3RT)/M)# .We know that the molar mass,

#M# (in#"kg/mol"# ), is smallest for#"H"_2# , and is inversely proportional to the speed (small and light = fast).Therefore,

#"H"_2# is fastest, and the sample of#bb("H"_2)# will contain thelargest total kinetic energy.

We can calculate this easily without pulling out any complicated formula. You are already given the mass and volume.

#color(blue)(D) = m/V = ("4 g")/("9 L") = color(blue)("0.4444 g/L")# The

densitiesare all the same,by construction, because the pressures vary among the flasks in such a way that the densities in each flask are the same, despite their different molar masses.

Collision frequencydepends on thecollision cross-section#sigma# (and thus the size of the gas), and thereduced mass#mu# of the two colliding gases.

It is "the average number of collisions between reacting molecules per unit of time

per moles of reactant", so it isnot dependenton the mols of gas there are (as they have already been divided out).Here, the same identity gases will collide with each other:

#Z_(A A) = N_A sigma_(A A) sqrt((8k_B T)/(pimu_(A A))# ,where

#N_A# ,#k_B# , and#T# are Avogadro's number (#6.022 xx 10^(23) "mol"^(-1)# ), Boltzmann's constant (#1.38065 xx 10^(-23) "J/K"# ), and temperature in#"K"# , respectively.The only numbers that matter to us are

#sigma_(A A)# and#mu_(A A)# , thecollision cross-sectionof the gas and thereduced massof the colliding gases.

The reduced mass will involveeachgas's mass asonebody in atwo-body collision.

#sigma_(CH_4) = "0.46 nm"^2# #sigma_(He) = "0.21 nm"^2#

#sigma_(H_2) = "0.27 nm"^2#

#mu_(CH_4//CH_4) = (16.0426^2)/(16.0426 + 16.0426) = 8.0213# #mu_(He//He) = 4.0026^2/(4.0026 + 4.0026) = 2.0013# #mu_(H_2//H_2) = (2.0158^2)/(2.0158 + 2.0158) = 1.0079# All other variables the same, the ratio of the collision frequencies are:

#Z_(CH_4)/(Z_(He)) = sigma_(CH_4)/(sigma_(He)) cdot sqrt(mu_(He)/(mu_(CH_4)))#

#= (0.46)/(0.21) sqrt(2.0013/8.0213) = 1.0941#

#Z_(CH_4)/(Z_(H_2)) = sigma_(CH_4)/(sigma_(H_2)) cdot sqrt(mu_(H_2)/(mu_(CH_4)))#

#= (0.46)/(0.27) sqrt(1.0079/8.0213) = 0.6039# So, we've shown that at the same temperature and number of particles:

#color(blue)(Z_(H_2) > Z_(CH_4) > Z_(He))#