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## Homework Statement

A horizontal container with adiabatic walls has a vertical wall inside it which divides the container in two. The wall is free to move back and forth and is not adiabatic. Initially, 1 kg of air is in the compartment to the left of the wall, at 5 bar and 350 K, and to the right we have 3 kg of CO2, at 2 bar and 450 K. They are both to be considered ideal gases, with constant heat capacity. Determine the heat exchanged between the two gases till they reach an equilibrium configuration. M(air)= 28.96 kg/kmol, C_v(air) = 0.726 kJ/kg K. M(CO2)= 44 kg/kmol, C_v(CO2) = 0.750 kJ/kg K.

## Homework Equations

M = m/n

Ideal gas law: PV = nRT

First law of Thermodynamics: ΔU = Q - W

Work in PVT system, assuming quasistatic process, and following sign convention used in First law above: W = ∫pdV, between two volumes.

## The Attempt at a Solution

First of all, we need the equilibrium configuration. Applying the Frst law of Thermodynamics to all of the system (the container), we find that ΔU = 0 for it. So,

ΔU_air = - ΔU_CO2

But ΔU = m C_v ΔT for an ideal gas, assuming constant heat capacity. So, we have, knowing that both gases will eventually reach a final equilibrium temperature T_f,

ΔU_air = - ΔU_CO2 → m_air C_v(air) (T_f - 350) = m_CO2 C_v(CO2) (450 - T_f)

from where we can easily solve for T_f, the equilibrium temperature.

To get the equilibrium pressure, we know that it'll be the same value as well in both gases, because they'll be in mechanical equilibrium. For this, we use the ideal gas law, having obtained the amount of moles for both gases using the data provided for the mass and molar mass for each gas. Hence,

P_f V_f(air) = n_air R T_f

P_f V_f(CO2) = n_CO2 R T_f

P_f V_f(CO2) = n_CO2 R T_f

and dividing both equations term by term, we arrive at an equation relating V_f(air) and V_f(CO2). But we can already obtain another equation relating V_f(air) and V_f(CO2), since we can can calculate the initial volumes of both gases, and we know that the sum is the same, so V_f(air) + V_f(CO2) = known value, is another equation, from which we solve for V_f(air) and V_f(CO2). Then, using either value, we plug back into either equation of state for the final equilibrium state and solve for P_f.

Now, to get the heat exchanged between both gases, I've thought of using the First law. We already know ΔU for both gases, and we also know that ΔU = Q - W by the First law. So,

Q_air = ΔU_air + W_air

Q_CO2 = ΔU_CO2 + W_CO2

Q_CO2 = ΔU_CO2 + W_CO2

and my question is evaluating the W work term. We can use W = ∫pdV for

*quasistatic processes*, but this process (one of the gases quickly expands, while the other is compressed, till both reach equilibrium) doesn't seem quasistatic, so I am not sure how to evaluate the W term, nor if this is the correct approach.

Any help will be appreciated!

Thanks.