Please help, thermodinamics exercise.

[claudiadeluca]

I tried but I cannot do it. Please help.

Two rigid adiabatic containers have inside the same ideal gas, respectively N1=3mol at a temperature T1=340K and N2=2mol at a temperature T2=280K. The two containers come into contact by opening a faucet, and the gas mixes, getting to a temperature T3.

Calculate:

1) the variation of the total internal energy of the gas.
2) the value of the final temperature T3.

I really need a hand, if you don't want to solve it please tell me how to do it. Any help is truly appreciated. Pardon the probably incorrect use of words, I'm not english.

 

[chaoseverlasting]

Here, I think the total internal energy will remain constant. Since U=nf RT/2, you can use that to find the internal energy of the gas in both the parts and add it up to find the final internal energy which will have n1+n2 moles of the gas.

 

[m2003]

Hello,

I understand that thermodynamics exercises can be challenging, but I am here to help you. First, let's review the concepts involved in this problem.

Thermodynamics is the study of the relationship between heat, work, and energy. In this exercise, we have two rigid adiabatic containers, which means that no heat can enter or leave the system. The containers have an ideal gas inside, which means that the gas particles do not interact with each other and have no volume.

Now, let's tackle the problem. To calculate the variation of the total internal energy of the gas, we need to use the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added (Q) minus the work done (W) on the system. Since the containers are adiabatic, there is no heat added or lost, so Q = 0. Additionally, since the containers are rigid, there is no change in volume, so W = 0. Therefore, the change in internal energy is also equal to 0. This means that the total internal energy of the gas remains constant throughout the process.

Next, let's calculate the final temperature T3. To do this, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the containers are now connected, the volume and pressure will be the same for both containers. We can combine the two containers' ideal gas equations to get:

(P)(V1 + V2) = (n1 + n2)RT3

We can rearrange this equation to solve for T3:

T3 = (P)(V1 + V2) / (n1 + n2)R

Substituting the given values, we get:

T3 = (1 atm)(V1 + V2) / (N1 + N2)R

T3 = (1 atm)(1.5 mol) / (5 mol) (0.0821 L atm/mol K)

T3 = 0.3 atm (0.0821 L atm/mol K)

T3 = 0.02463 L atm / mol K

T3 = 294.1 K

Therefore, the final temperature T3 is 294.1