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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.

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  • $\begingroup$ Your equation $p_f(V_1 + V_2) = (\nu_1 + \nu_2)RT_f$ is invalid, as there is no well-defined final pressure; the wall allows heat transfer but not pressure equalization. You need an additional equation that relates, through the heat capacity, the energy increase in one gas to the energy decrease in the other gas. $\endgroup$
    – Chemomechanics
    Apr 14, 2021 at 19:05
  • $\begingroup$ Can we use the heat capacity for compartments? I thought it can be used only for objects. $\endgroup$
    – andu eu
    Apr 14, 2021 at 19:29
  • $\begingroup$ The heat capacity is that of the gas (specifically, the molar heat capacity). You're meant to ignore the heat capacity of the actual compartment material. $\endgroup$
    – Chemomechanics
    Apr 14, 2021 at 19:53
  • $\begingroup$ Are you familiar with the concept of molar heat capacity at constant volume? $\endgroup$
    – Chet Miller
    Apr 14, 2021 at 20:00
  • $\begingroup$ @andueu Hint: to determine the final equilibrium temperature just equate the heat lost by gas 2 to the heat gained by gas 1. $\endgroup$
    – Bob D
    Apr 14, 2021 at 20:14

1 Answer 1

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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

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