# How to calculate overall temperature of two systems after pressure drop in one

If I have two closed systems (with a valve in between and assuming total insulation from environment) with the first being,

Dry air, T1 = 321K, P1 = 101325 Pa, and V1 = 1 m3

and the second being,

Dry air, T2 = 323K, P2 = 2757900 Pa, and V2 = 0.002 m3

and final (combined) being,

Dry air, Tfinal = ???K, Pfinal = ??? Pa, and Vfinal = 1.002 m3

How can I use the Ideal Gas Law (and other laws needed) to calculate the unknown Tfinal and Pfinal?

EDIT: I haven't calculated anything yet. But I have thought out a method (which I think is faulty, which explains why I'm here). The method is that I'll calculate the new temperature of system 2 by inputting a lower pressure value in the ideal gas law. Then I'll calculate the energy needed to bring about such a change in temperature using the specific heat capacity of dry air. I just don't understand how to use this "energy needed" value to determine the other system's temperature, since now they're both one (after expansion) which means more moles in system 1, so then the pressure should increase too. It just gets more and more complicated here onwards. Help me out!

• What have you attempted so far? Feb 24, 2021 at 14:09
• @BobD Check out the edit. Feb 24, 2021 at 14:34
• Is the internal energy per unit mass of an ideal gas a function of pressure. Does this rigid system of two tanks do any work on their combined surroundings? Does this insulated system of two tanks exchange any heat with their combined surroundings? Feb 24, 2021 at 14:44
• @ChetMiller and assuming total insulation from environment So adiabatic.
– Gert
Feb 24, 2021 at 14:50
• "Is the internal energy per unit mass of an ideal gas a function of pressure" I don't understand such high level stuff. "Does this rigid system of two tanks do any work on their combined surroundings?" No. " Does this insulated system of two tanks exchange any heat with their combined surroundings?" No. Feb 24, 2021 at 14:51

Hints: (1)For an ideal gas, the change in internal energy depends only on temperature change according according to $$\Delta U=nC_{v}\Delta T$$, (2) since the tanks are rigid (assumed) as well as insulated, per the first law the overall change in internal energy is zero, i.e., the change in internal energy of gas 1 plus the change in internal energy of gas 2 equals zero.