Question

Consider a container like that shown in Figure $20.12$, with $n_1$moles of a monatomic gas on one side and $n_2$moles of a diatomic gas on the other. The monatomic gas has initial temperature $T_{1i}$. The diatomic gas has initial temperature$T_{2i}$ .
a. Show that the equilibrium thermal energies are

$\begin{array}{r}{E}_{1\mathrm{f}}=\frac{3n_1}{3n_1+5n_2}\left(E_{1\mathrm{i}}+E_{2\mathrm{i}}\right)\\ E_{2\mathrm{f}}=\frac{5n_2}{3n_1+5n_2}\left(E_{1\mathrm{i}}+E_{2\mathrm{i}}\right)\end{array}$

b. Show that the equilibrium temperature is

$T_{\mathrm{f}}=\frac{3n_1{T}_{1\mathrm{i}}+5n_2{T}_{2\mathrm{i}}}{3n_1+5n_2}$

c.$2.0g$ of helium at an initial temperature of role="math" localid="1648474536876" $300K$interacts thermally with $8.0g$of oxygen at an initial temperature of$600K$ . What is the final temperature? How much heat energy is transferred, and in which direction?

Short answer

$\left(a\right)$The thermal energies equilibrium,

$\begin{array}{r}{E}_{1f}=\frac{3n_1}{3n_1+5n_2}\left(E_{1i}+E_{2i}\right),\\ E_{2f}=\frac{5n_2}{3n_1+5n_2}\left(E_{1i}+E_{2i}\right).\end{array}$

$\left(b\right)$The equilibrium temperature is $T=\frac{3n_1{T}_1+5n_2{T}_2}{3n_1+5n_2}.$

$\left(c\right)$Heat energy transfers at final temperature,localid="1648536524385" $\text{T=}488\mathrm{K}\text{.}$

Step-by-step solution

Step: 1  Finding equilibrium themal energies: (part a)

Taking the original energies into account $E_{1i}$and $E_{2i}$, the amount of energy, which will remain constant due to the law of conservation of energy, will be

$E=E_{1i}+E_{2i}$

We know that the degrees of freedom vary. $i$, the number of moles $n$and temperature $T$the energy is given as

$E=\frac{i}{2}nRT$

This means that the final energy will be

$E=\frac{i_1}{2}{n}_{1}R{T}_1+\frac{i_2}{2}{n}_{2}R{T}_2$

Let us note, however, that the temperature at the end remains constant; so, we can factorise, resulting in

$E=\frac{1}{2}\left(i_1{n}_1+i_2{n}_2\right)RT$

$i_1{n}_1$and $i_2{n}_2$are the ones who are in charge of the energy generated by the first and second gases, respectively.

We can get a ratio if we design one.

$\frac{E_{xf}}{E}=\frac{i_x{n}_x}{i_1{n}_1+i_2{n}_2}$

where $x=\{1,2\}$implies the presence of two gases. As a result, the ultimate energies will be.

$E_{xf}=\frac{i_x{n}_x}{i_1{n}_1+i_2{n}_2}E$

Swapping the basic energies and degrees of freedom for the final energy $i_1=3$and $i_2=5$, we get

$\begin{array}{r}{E}_{1f}=\frac{3n_1}{3n_1+5n_2}\left(E_{1i}+E_{2i}\right)\\ E_{2f}=\frac{5n_2}{3n_1+5n_2}\left(E_{1i}+E_{2i}\right)\end{array}.$

Step: 2  Finding Equilibrium temperature: (part b)

Then let return to the initial and final expressions of total energy as the sum of the energies of the two gases. We have them on the left and right, respectively, and we know they are equal.

$\frac{i}{2}\left(i_1{n}_1+i_2{n}_2\right)RT=\frac{i_1}{2}{n}_{1}R{T}_1+\frac{i_2}{2}{n}_{2}R{T}_2$

Simplifying, we get

$\left(i_1{n}_1+i_2{n}_2\right)T=i_1{n}_1{T}_1+i_2{n}_2{T}_2$

This means that the final temperature will, after substituting the degrees of freedom, be:

$T=\frac{3n_1{T}_1+5n_2{T}_2}{3n_1+5n_2}$.

Step: 3  Finding Final temperature: (part c)

We have $n_1=0.5$and $n_2=0.5$.

The Final temperature will be

$$\begin{gathered} T=\frac{3\cdot 0.5\cdot 300+5\cdot 0.5\cdot 600}{3\cdot 0.5+5\cdot 0.5} \\ T=487.5\mathrm{K} \\ \mathrm{T}\simeq 488\mathrm{K}. \end{gathered}$$