download the exercise with solution: entropy production gas mixing

Two well-insulated rigid tanks A and B have the same volume (V_{A}=V_{B}=1 m^{3 }) and both are filled with air with the following thermodynamic states:

Tank A: p_{A} = 1 bar ; T_{A} = 500 K

Tank B: p_{B} = 18 bar ; T_{B} = 500 K

Now these two tanks are connected with each other through a pipe, so that air from tank A and tank B can mix with each other and reach the same pressure in the end. During this mixing process, no heat is transferred.

Determine the entropy production after this mixing process.

Assume that in this case, air is an ideal gas and its specific gas constant is R=287 J/kg∙K

Solution:

Index 1: state before mixing

Index 2: state after mixing

At first, we calculate the mass of gas in both tanks before mixing:

In order to determine the entropy production, pressure and temperature p_{2} and T_{2} after mixing should also be obtained.

The boundary of the control volume chosen in this case encloses both tanks, tank A and tank B. Since both tanks are well-insulated and rigid, there are no heat transfer and work for volume change. Hence we get the simplified equation of the 1^{st} Law of thermodynamics for closed system and the final temperature T_{2}:

ΔU=ΔU_{A}+ΔU_{B}=0

→ m_{A}·c_{v}·(T_{2}-T_{A})+ m_{B}·c_{v}·(T_{2}-T_{B})=0

→ T_{2}=T_{A}= T_{B}=500K

So the final pressure p_{2} is:

Then the entropy production is:

We can obtain an important conclusion:

The entropy after gas mixing and expansion will be produced (entropy production >0 ) and therefore this process is irreversible.