download the exercise with solution: Air Mixing

A well-insulated rigid tank is divided into part A and part B through a partition, as shown in figure. Initially, side A of the tank contains 0.5m^{3} air at 6bar and 400K, and the other side B is filled with 0.2 m^{3} air at 12bar and 500K. The partition is then removed, and the air mixes within the entire tank. Determine the final temperature of air.

Here the air is assumed to be ideal gas.

The specific gas constant for air is R_{air}=287 J/kg∙K and the specific heat capacity at constant pressure c_{p}=1 J/g∙K

Solution:

We first calculate the mass of air in each side of tank:

Since tank is rigid and its volume remains constant, there is no work for volume change. In addition, there is no other kind of work such as electrical work involves. → W=0

As the tank is well-insulated (adiabatic), no heat transfer occurs. → Q=0

According to the 1.Law of thermodynamics for closed system, the equation reduces to ∆U_{sys}=0, that is, ∆U_{A}+∆U_{B}=0.

Because the air is considered as ideal gas and we assume the final temperature T_{2}:

∆U_{A}=m_{A}∙c_{v}∙(T_{2} – T_{A}) and ∆U_{B}=m_{B}∙c_{v}∙(T_{2} – T_{B})

Therefore: ∆U_{A}+∆U_{B}= m_{A}∙c_{v}∙(T_{2} – T_{A})+ m_{B}∙c_{v}∙(T_{2} – T_{B})=0

Now we calculate the final pressure p_{2} after removing the partition:

Since entropy is the property of state, we can calculate the entropy change of this system separately. That means,

∆S_{sys}=∆S_{A}+∆S_{B}

where

- ∆S
_{A}is the entropy change of air in side A - ∆S
_{B}is the entropy change of air in side B

We notice that ∆S >0. That means, this process is irreversible which follows the 2.Law of thermodynamics.