Container divided into 2 parts with $p, \ V, \ T$ known, what is the final temperature

Question

So, this problem was given to 10th graders in a town physics olympiad. A thermally insulated container is divided by a thin thermoconductor wall in 2 compartments with volumes $V_1 = 3L$, respectively $V_2 = 4L$ which contain the same kind of gas at pressures $p_1 = 3 \ atm$, respectively $p_2 = 2 \ atm$ and temperatures $T_1 = 300 \ K$, respectively $T_2 = 400 K $. Find the temperature $T$ after reaching thermal equilibrium.

First of all, I am quite confused about the thin thermoconductor wall. Does the gases exchange heat? I tried this: $$p_1V_1 = \nu_1RT_1$$ and $$p_2V_2 = \nu_2RT_2$$ give $$\nu_1 = \frac{p_1V_1}{RT_1} = \frac{3}{R}$$ and $$\nu_2 = \frac{p_2V_2}{RT_2} = \frac{2}{R}$$ Then, when, equilibrium is reached, $$p_f(V_1 + V_2) = (\nu_1 + \nu_2)RT_f$$ gives $$p_f (V_1 + V_2) = 5T_f$$ and, finally, $$T_f = \frac{p_f(V_1 + V_2)}{5}$$ Unfortunately, I could not find $p_f$. Can we say $p_f = p_1 + p_2$? Also, what's with the wall?

I am sorry if there are mistranslations, I did my best.

Answer 1

Initial Conditions

• $P_1$ = $3~atm~~~~~~~P_2 = 2~atm$
• $V_1$ = $3~L~~~~~~~~V_2 = 4~L$
• $T_1$ = $300~K ~~~~~~T_2 = 400~K$

So, $$\frac{n_1}{n_2} = \frac{3}{2}$$ $n$ - no. of moles of the gas

As the volume is constant so Work Done is Zero, but temperature changes so there is some change in the internal energy of the gases.

From $1st$ Law of Thermodynamics: $$Q = \Delta U + W$$ here $W = 0$

So $Q = \Delta U$, so exchange in energy is equal to change in the internal energy of the gas.

$$\Delta U_1 = \Delta U_2$$ $$n_1.C_v.\Delta T_1 = n_2.C_v.\Delta T_2$$ As the gases are same so the molar heat capacity of the gas at constant volume [$C_v$] is same for both

This gives, $$\frac{n_1}{n_2} = \frac{400-T}{T-300} = \frac{3}{2}$$ So, $T = 340 K$ $->$ the final temperature of both the gases at equilibrium