Question

$2.0$mol of monatomic gas $A$initially has role="math" localid="1648440865095" $5000J$of thermal energy. It interacts with $3.0$mol of monatomic gas $B$, which initially has $8000J$ of thermal energy.
a. Which gas has the higher initial temperature?
b. What is the final thermal energy of each gas?

Short answer

$\left(a\right)$The higher initial temperature is, $T_{\mathrm{B},\mathrm{i}}>T_{\mathrm{A},\mathrm{i}.}$

$\left(b\right)$The Final Thermal energy of gas is,${\mathrm{E}}_{\mathrm{A},\mathrm{f}}=5000\mathrm{J}\text{and}{\mathrm{E}}_{\mathrm{B},\mathrm{f}}=7800\mathrm{J}.$

Step-by-step solution

Step: 1 a Find the initial temperature: (part a)

Once we have two systems, each contains thermal energy at a different temperature at the start. When the two systems are coupled, they interact thermally until they attain thermal equilibrium. Each system has its own set of starting thermal energy, which is determined by

$E_{1\mathrm{i}}=\frac{3}{2}{n}_{1}R{T}_{1\mathrm{i}}$

Our target is to find the initial temperature of each gas, so we solve equation for $T_i$to be

$T_{1\mathrm{i}}=\frac{2}{3}\frac{E_{1\mathrm{i}}}{n_{1}R}$

For gas $A$,the number of moles is localid="1648393481785" $n_A=2mol$and the thermal energy is $E_{A\mathrm{i}}=5000\mathrm{J}$,so the values in $T_{\mathrm{A},\mathrm{i}}$is

$\begin{array}{r}{T}_{\mathrm{A},\mathrm{i}}=\frac{2}{3}\frac{E_{\mathrm{A},\mathrm{i}}}{n_{A}R}\\ =\frac{2}{3}\frac{5000\mathrm{J}}{\left(2.0\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})}\\ =200\mathrm{K}.\end{array}$

For gas $B$,the number of moles is $n_B=3mol$and the thermal energy is $E_{B\mathrm{i}}=8000\mathrm{J}$,so the values in ${\mathit{T}}_{\mathbf{B}\mathbf{,}\mathbf{i}}$is

$\begin{array}{r}{T}_{\mathrm{B},\mathrm{i}}=\frac{2}{3}\frac{E_{\mathrm{A},\mathrm{i}}}{n_{B}R}\\ =\frac{2}{3}\frac{8000\mathrm{J}}{\left(3.0\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})}\\ =214\mathrm{K}.\end{array}$

As shown by the two results, gas $B$has a higher initial temperature than gas $A$.

$T_{\mathrm{B},\mathrm{i}}>T_{\mathrm{A},\mathrm{i}}$.

Step: 2 Thermal energy related to number of moles:

During a thermal interaction, heat is transmitted through collisions between high-energy molecules and low-energy molecules, and both systems eventually reach the same final temperature, which is the equilibrium state. The amount of moles in the form determines the final thermal energy.

$E_{1,\mathrm{f}}=\frac{n_1}{n_1+n_2}{E}_{\mathrm{tot}}$.

where $E_{tot}$is the total initial energy

$E_{\mathrm{A},\mathrm{i}}+E_{\mathrm{B},\mathrm{i}}=13000\mathrm{J}.$

Step: 3 b Final Thermal Energy: (part b)

For gas $A$the number of moles is $2$moles and for $B$the number of moles is $3$moles,the final thermal energy for each gas is

$\begin{array}{r}{E}_{\mathrm{A},\mathrm{f}}=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}{E}_{\mathrm{tot}}\\ =\frac{2.0\mathrm{mol}}{2.0\mathrm{mol}+3.0\mathrm{mol}}\left(13000\mathrm{J}\right)\\ =5200\mathrm{J}\\ E_{\mathrm{B},\mathrm{f}}=\frac{n_{\mathrm{B}}}{n_{\mathrm{B}}+n_{\mathrm{A}}}{E}_{\mathrm{tot}}\\ =\frac{3.0\mathrm{mol}}{3.0\mathrm{mol}+2.0\mathrm{mol}}\left(13000\mathrm{J}\right)\\ =7800\mathrm{J}.\end{array}$