# What is the average kinetic energy of a gas in a closed system at #"380 K"#, if the universal gas constant is #R = "8.314472 J/K"cdot"mol"#?

##### 2 Answers

#### Answer:

#### Explanation:

As you know, **kinetic energy** is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)K_E = 1/2 * m * v^2color(white)(a/a)|)))" "# , where

Now, you are asked to find the *average kinetic energy* of the gas, and not its kinetic energy, because a gas is made up of a large number of particles, each with its *specific* speed and direction of movement, i.e .**velocity**.

Some of these gas molecules will move slower, some will move faster, which is why an **average speed** is needed in order to be able to calculate the average kinetic energy.

This tells you that you have to use the **root-mean-square speed**, **average velocities** of the gas particles

#color(blue)(|bar(ul(v_"rms" = sqrt((3RT)/M_M)color(white)(a/a)|)))" "# , where

**absolute temperature** of the gas

**molar mass** of the gas

Plug this into the equation for kinetic energy to get

#K_E = 1/2 * m * v_"rms"^2#

#K_E = 1/2 * m * [sqrt((3RT)/M_M)]^2 = 3/2 * m/M_M * RT#

Now, mass divided by molar mass will given **number of moles**

#n = m/M_M#

Plug this into the equation to get

#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_E = 3/2 * nRT)color(white)(a/a)|)))#

Now, since no mention of number of moles was made, you can assume that you're dealing with **one mole** of gas kept at a temperature of

This means that the average kinetic energy of the gas will be

#K_E = 3/2 * 1color(red)(cancel(color(black)("mole"))) * "8.314472 J" color(red)(cancel(color(black)("mol"^(-1)))) * color(red)(cancel(color(black)("K"^(-1)))) * 380color(red)(cancel(color(black)("K"^(-1))))#

#K_E = "4739.2 J"#

Expressed in *kilojoules* and rounded to two **sig figs**, the answer will be

#K_E = color(green)(|bar(ul(color(white)(a/a)"4.7 kJ"color(white)(a/a)|)))#

I also got

Below, I show how I would get **monatomic ideal gas**.

An alternative approach to this (albeit harder) is to start from the **statistical mechanics** definition of the *average kinetic energy*.

#\mathbf(<< E >> = K_"avg" = -((dellnQ)/(delbeta))_(N,V))#

where:

#Q = q^N/(N!)# is known as thepartition function.#q_"linear" = sum_i g_ie^(-betaepsilon) = ((2pimk_BT)/(h^2))^(3"/"2)V = ((2pim)/(h^2beta))^(3"/"2)V# is a measure ofthermally accessible energy statesfortranslational motion, with degeneracies#g# and energy states#epsilon# . The exponent is#3/2# because each degree of freedom from each direction of motion (#x,y,z# give a total of#3# directions) contributes#1/2# .#beta = 1/(k_BT)# is a"thermodynamic beta"to make equations look nicer because#k_BT# shows up a lot. :)#m# is themass of the gasin grams. This will not matter in the end!#h ~~ 6.626xx10^(-34) "J"*"s"# isPlanck's constant, but we will not need to worry about it.#N# is thenumber of gas particlesin the "ensemble", which is just the technical term for a group of gas particles in the context of statistical mechanics.#V# is thevolume, which came from the fact that the gas is able to travel in three dimensions. We won't need to worry about this later.#T# istemperaturein#"K"# , while#k_B ~~ 1.3806xx10^(-23) "J/K"# is theBoltzmann constant.

According to the first equation, we will take the derivative of

This looks hard, but using

#lnQ = ln(q^N/(N!))#

#= ln((((2pim)/(h^2beta))^(3"/"2)V)^N/(N!))#

#= ln((((2pim)/(h^2beta))^(3N"/"2)V^N)/(N!))#

#= ln(((2pim)/(h^2beta))^(3N"/"2)V^N) - lnN!#

#= ln((2pim)/(h^2beta))^(3N"/"2) + lnV^N - lnN!#

#= (3N)/2ln((2pim)/(h^2)) - (3N)/2lnbeta + NlnV - lnN!#

Well now, this looks much nicer.

When we take the partial derivative with respect to **ignore all the non-****terms** (that's how the partial derivative works; you focus on the variable in the denominator of the derivative only). That means anything that doesn't contain

So, we get:

#color(blue)(<< E >>) = -((dellnQ)/(delbeta))_(V,N)#

#= -[cancel((3N)/2ln((2pim)/(h^2))) color(green)(stackrel("keep")(overbrace(-(3N)/2d/(dbeta)[lnbeta]))) + cancel(NlnV) - cancel(lnN!)]#

#= (3N)/2 *1/beta#

#= color(blue)(3/2Nk_BT)#

Now, note that

#\mathbf(<< E >> = K_"avg" = 3/2 nRT)#

So, after all that, we get the **average kinetic energy** for

#color(blue)(<< E >>) = 3/2cdotcancel"1 mol"cdot"8.314472 J/"cancel"mol"cdotcancel"K" * 380 cancel"K"#

#= 4739.24904# #"J"#

#=> color(blue)(4.7)# #color(blue)("kJ")#