# How do you calculate the average kinetic energy of the #CH_4# molecules in a sample of #CH_4# gas at 253 K and at 545 K?

The average kinetic energy

Suppose *degrees of freedom* it has are due to linear motion, that is,

*For methane this is a decent approximation that omits vibrational degrees of freedom; I'll talk about it at the bottom of the answer.
*

*For the equipartition theorem, we have:*

#\mathbf(K_"avg" = N/2nRT)# where:

#N# is the number of degrees of freedom.#n# is the number of#\mathbf("mol")# s.#R# is the universal gas constant,#"8.314472 J/mol"cdot"K"# .#T# is the temperature in#"K"# .

NOTE: If you include the vibrational contribution to the degrees of freedom according to the equipartition theorem (

Given

Thus, we have

#color(blue)(K_"1,avg") = 6/2RT_1#

#= 3*8.314472*253#

#= "6310.68 J" ~~ color(blue)(6.31xx10^3 "J")#

#color(blue)(K_"2,avg") = 6/2RT_2#

#= 6/2*8.314472*545#

#= "13594.2 J" ~~ color(blue)(1.36xx10^4 "J")# rounded to three sig figs.

NOTE: This corresponds to an estimated constant-volume molar heat capacity of

#barC_"V,tot" ~~ 3R ~~ "24.9 J/mol"cdot"K"# , which is about#8.9%# below the true value.

This is a decent approximation but could be better because methane has four vibrational modes in its IR spectrum (

#A_1 + E + 2T_2# ),.These contribute a total of

#0.2823R# to its constant-volume molar heat capacity#barC_"V,tot"# as follows:

#barC_"V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib"#

#= 3/2R + 3/2R + 0.2823R#

#~~ color(blue)(3.2823R)# whereas the equipartition theorem predicts:

#barC_"V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib"#

#= 3/2R + 3/2R + stackrel("omit this for ideality")(overbrace(9R)#

#= color(red)(stackrel("'real'")(overbrace(12R))# , or#color(red)(stackrel("ideal")(overbrace("3R"))# which is fairly off from the true

#barC_("V,tot")# !

And just so you know I'm not making these numbers up, I got this information while doing a speed-of-sound in methane lab last year:

#barC_(V_"vib") = Rsum_(i)^(3N-6) [g_i((theta_i)/T)^2 e^(-theta_i"/"T)/(1-e^(-theta_i"/"T))^2]#

(from my lab handout)where:

#R# is the universal gas constant.#g_i# is the degeneracy of mode#i# .#E# is doubly degenerate and#T# is triply degenerate (#A# is unique, so#g_A = 1# ).#N# here is the number of atoms on#"CH"_4# :#5# .#theta_i# is the vibrational temperature in#"K"# of vibrational mode#i# . Those are given in Table 1 below.#T# is temperature in#"K"# as usual.

#barC_("V,vib")# and#barC_("V,tot")# were calculated in Excel below at#"295.15 K"# (#2.3469/8.314472 ~~ 0.2823# ).

And a summary of

#barC_("V,vib")# from various approaches is shown below:

The takeaway is that the statistical mechanics approach was the most accurate approach.

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To calculate the average kinetic energy of CH4 molecules in a sample of CH4 gas at 253 K and 545 K, you can use the equation:

Average kinetic energy = (3/2) * (kB) * (T)

Where:

- kB is the Boltzmann constant (1.38 x 10^-23 J/K)
- T is the temperature in Kelvin

For CH4 gas at 253 K: Average kinetic energy = (3/2) * (1.38 x 10^-23 J/K) * (253 K)

For CH4 gas at 545 K: Average kinetic energy = (3/2) * (1.38 x 10^-23 J/K) * (545 K)

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*Answer from HIX Tutor*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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