Question

$4.0$mol of monatomic gas $A$interacts with $3.0$mol of monatomic gas $B$. Gas $A$initially has $9000J$of thermal energy, but in the process of coming to thermal equilibrium it transfers $1000J$of heat energy to gas $B$. How much thermal energy did gas $B$have initially?

Short answer

The Initial Thermal engergy gas $B$value is,$E_{\mathrm{B},\mathrm{i}}=5000\mathrm{J}.$

Step-by-step solution

Step: 1 Thermal Energy:

Heat is transferred through collisions between high-energy molecules and low-energy molecules during a thermal interaction, and both systems finally achieve the same final temperature, which is the equilibrium state. By using an equation, the number of moles is connected to the ultimate thermal energy.

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

where $E_{tot}$is the total energy

$E_{\mathrm{A},\mathrm{i}}+E_{\mathrm{B},\mathrm{i}}$

For gas $A$, we solve $E_{tot}$in the form

$E_{\mathrm{tot}}=\frac{n_A+n_B}{n_A}{E}_{\mathrm{A},\mathrm{f}}$.

Step: 2 Finding total energy:

The initial thermal energy of gas $A$is $E_{\mathrm{A},\mathrm{i}}=9000\mathrm{J}$and it transfers $\mathrm{\Delta }{E}_{\mathrm{A}}=1000\mathrm{J}$to gas $B$.so, the final energy o gas $A$is

$E_{\mathrm{A},\mathrm{f}}=E_{\mathrm{A},\mathrm{i}}-\mathrm{\Delta }{E}_{\mathrm{A}}=9000\mathrm{J}-1000\mathrm{J}=8000\mathrm{J}$.

Now,we put the values of $n_A,n_B$and $E_{A,f}$to get $E_{tot}$by

$\begin{array}{r}{E}_{\text{tot}}=\frac{n_A+n_B}{n_A}{E}_{\mathrm{A},\mathrm{f}}\\ =\frac{4.0\mathrm{mol}+3.0\mathrm{mol}}{4.0\mathrm{mol}}\left(8000\mathrm{J}\right)\\ =14000\mathrm{J}.\end{array}$

Step: 3 Initial thermal energy of gas B:

The total energy of the system equals the original thermal energy of both gases since the system's whole energy is preserved. The ga's initial thermal energy is now available $B$by

$\begin{array}{r}{E}_{\mathrm{B},\mathrm{i}}=E_{\mathrm{tot}}-E_{\mathrm{A},\mathrm{i}}\\ =14000\mathrm{J}-9000\mathrm{J}\\ =5000\mathrm{J}.\end{array}$