# 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"?

Mar 18, 2016

${K}_{E} = \text{4.7 kJ}$

#### Explanation:

As you know, kinetic energy is given by the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {K}_{E} = \frac{1}{2} \cdot m \cdot {v}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$m$ - the mass of the object
$v$ - its speed

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, ${v}_{\text{rms}}$, which is defined as the square root of the square of the average velocities of the gas particles

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

$R$ - the universal gas constant, equal to ${\text{8.314472 J mol"^(-1)"K}}^{- 1}$
$T$ - the absolute temperature of the gas
${M}_{M}$ - the molar mass of the gas

Plug this into the equation for kinetic energy to get

${K}_{E} = \frac{1}{2} \cdot m \cdot {v}_{\text{rms}}^{2}$

${K}_{E} = \frac{1}{2} \cdot m \cdot {\left[\sqrt{\frac{3 R T}{M} _ M}\right]}^{2} = \frac{3}{2} \cdot \frac{m}{M} _ M \cdot R T$

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

$n = \frac{m}{M} _ M$

Plug this into the equation to get

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{E} = \frac{3}{2} \cdot n R T} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $\text{380 K}$.

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

${K}_{E} = \frac{3}{2} \cdot 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{mole"))) * "8.314472 J" color(red)(cancel(color(black)("mol"^(-1)))) * color(red)(cancel(color(black)("K"^(-1)))) * 380color(red)(cancel(color(black)("K}}^{- 1}}}}$

${K}_{E} = \text{4739.2 J}$

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

${K}_{E} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{4.7 kJ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Mar 18, 2016

I also got ${K}_{\text{avg" = "4.7 kJ}}$!

Below, I show how I would get ${K}_{\text{avg}} = \frac{3}{2} n R T$ from the statistical mechanics definition of average kinetic energy for a monatomic ideal gas.

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

$\setminus m a t h b f \left(\left\langleE\right\rangle = {K}_{\text{avg}} = - {\left(\frac{\partial \ln Q}{\partial \beta}\right)}_{N , V}\right)$

where:

• Q = q^N/(N!) is known as the partition function.
• q_"linear" = sum_i g_ie^(-betaepsilon) = ((2pimk_BT)/(h^2))^(3"/"2)V = ((2pim)/(h^2beta))^(3"/"2)V is a measure of thermally accessible energy states for translational motion, with degeneracies $g$ and energy states $\epsilon$. The exponent is $\frac{3}{2}$ because each degree of freedom from each direction of motion ($x , y , z$ give a total of $3$ directions) contributes $\frac{1}{2}$.
• $\beta = \frac{1}{{k}_{B} T}$ is a "thermodynamic beta" to make equations look nicer because ${k}_{B} T$ shows up a lot. :)
• $m$ is the mass of the gas in grams. This will not matter in the end!
• $h \approx 6.626 \times {10}^{- 34} \text{J"*"s}$ is Planck's constant, but we will not need to worry about it.
• $N$ is the number of gas particles in the "ensemble", which is just the technical term for a group of gas particles in the context of statistical mechanics.
• $V$ is the volume, which came from the fact that the gas is able to travel in three dimensions. We won't need to worry about this later.
• $T$ is temperature in $\text{K}$, while ${k}_{B} \approx 1.3806 \times {10}^{- 23} \text{J/K}$ is the Boltzmann constant.

According to the first equation, we will take the derivative of $\ln Q$ with respect to $\beta$ and keep $N$ and $V$ constant.

This looks hard, but using $\ln$ makes it much easier in the end. First, let's simplify this using the properties of logarithms. (The hard part is simplifying it.)

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 $\beta$, we can simply ignore all the non-$\setminus m a t h b f \left(\beta\right)$ 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 ${k}_{B}$ or $T$ gives $0$. Nice!

So, we get:

$\textcolor{b l u e}{\left\langleE\right\rangle} = - {\left(\frac{\partial \ln Q}{\partial \beta}\right)}_{V , N}$

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

$= \frac{3 N}{2} \cdot \frac{1}{\beta}$

$= \textcolor{b l u e}{\frac{3}{2} N {k}_{B} T}$

Now, note that $N {k}_{B} = n R$, because $\frac{N}{n} = {N}_{A}$ and ${N}_{A} {k}_{B} = R$, where $n$ is the number of $\text{mol}$s and ${N}_{A} = 6.0221413 \times {10}^{23}$. Therefore, we have:

$\setminus m a t h b f \left(\left\langleE\right\rangle = {K}_{\text{avg}} = \frac{3}{2} n R T\right)$

So, after all that, we get the average kinetic energy for $\text{1 mol}$ of monatomic ideal gas to be:

$\textcolor{b l u e}{\left\langleE\right\rangle} = \frac{3}{2} \cdot \cancel{\text{1 mol"cdot"8.314472 J/"cancel"mol"cdotcancel"K" * 380 cancel"K}}$

$= 4739.24904$ $\text{J}$

$\implies \textcolor{b l u e}{4.7}$ $\textcolor{b l u e}{\text{kJ}}$