Question

A 80.0 cm3 box contains helium at a pressure of 2.50 atm and a temperature of 110 ∘C. It is placed in thermal contact with a 210 cm3 box containing argon at a pressure of 4.20 atm and a temperature of 390 ∘C

What is the final thermal energy of each gas? Ef(He),Ef(Ar)

How much heat energy is transferred, and in which direction?

What is the final temperature?

What is the final pressure in each box?

What is the final temperature?

What is the final pressure in each box? pf(He), pf(Ar) =

Answer 1

First, Let's consider that for both gases, the final state is at the same temperature. Also, the volume remains constant, which lead us to a ischoric process. Besides, according to ideal gas law:

Constant gas for helium is 2.0771 kJ/kgK, and for argon is 0.2081 kJ/kgK.

As konwn, heat flows from high temperature to low temperature. Then, it flows from argon to helium (argon looses heat, and helium gains heat). Since it is a isochoric process, there is no work. Then, heat equals to internal energy change at constant volume. Let's consider an average specific heat at constant volume for both gases. Specific heats are 3.11 kJ/kgK and 0.3134 kJ/kgK for helium and argon respectively.

Final thermal energy is obtained using this final temperature as follow:

Heat transfered can be obtained using helium or argon as reference (it is the same amount). Then:

Finally, using ideal gas state equation, final pressure can be obtained as follows: