Container divided in two parts with ideal gas

Question

I wanted to ask your help with the following problem:

Consider a closed container, that is, with insulated walls impervious, and rigid. The container is divided into two parts by a wall having the same properties. Each part contains a portion of the same ideal gas. initially on the left has a volume $V_{1}$, $N_{1}$ atoms and a temperature $T_{1}$. On the right is $V_{2}$, $N_{2}$ and $T_{2}$.

The wall that separates the gas loses its rigidity property, but still is insulated and waterproof. Suppose that $T_{1}\neq T_{2}$, $V_{1}\neq V_{2}$ and $N_1 \neq N_2$, initially. Find the final thermodynamic state.

I understand that by thermodynamic end state refers to temperatures, volumes, entropies, number of atoms, change energy, work and heat.

However, I could not get the volumes as a function of baseline variables and neither obtaining the entropies.

Answer 1

$pV = NR_mT$ gas equation

$pv=R_mT/M=p/\rho$ gas equation using specific volume

$pv^n=const$ polytropic

$M$ is the molar mass and $R_m$ the universal gas constant

$n=\kappa=c_p/c_v$ for the isentropic case (loss of rigidity is assumed to mean frictionless here). We know that the change in volume for both volumes as well as the pressure after the change of state must be the same. If we lable the variables after the change of state with a prime, we can write the following equations:

$v_1'=v_1(p_1/p_1')^{1/\kappa}$

$v_2'=v_2(p_2/p_2')^{1/\kappa}$

$v_1'-v_1=v_2-v_2'$

$v_1'-v_1=R_m/M(T_1'/p_1'-T_1/p_1)$

$v_2'-v_2=R_m/M(T_2'/p_2'-T_2/p_2)$

$p_2'=p_1'$

That's 6 equations for 6 primed unknowns.

Answer 2

Work and heat are not state functions. In this problem, there is no heat or work done by the system as a whole (considering both parts of the container), due to insulation and no change in total volume. The work done on one part equals the work done by the other part. There is no heat by either part due to insulation. This is an adiabatic process. There is a relationship between pressure and volume in terms of heat capacity for ideal gas adiabatic processes.

You have to decide whether or not to assume reversibility. If you assume reversiblity, there is no change in entropy. If you do not assume reversibility, you have no way of finding entropy.

That's all for now since it's a homework question.