Gibbs Free Energy Homework: Final T & P, Change in G

[Chip90]

Homework Statement

You have two parts of a tank, divided by a divider. On the larger side, you have 4 mol of N2 gas and on the smaller side you have 1 mol O2 gas. Both gasses are are the same T and P. The divider is removed and the gasses mix.

What is the final T and P?

What is the change in Gibbs Free Energy

Homework Equations

Daltons Law of Partial Pressure

because T is constant

The Attempt at a Solution

So the final Temperature is the same as Ti because there is no chemical reaction

The final Pressure is the sum of the Pressures of Po + Pn = Pf
To solve this I just need Pf=2P ? (since each gas was at P pressure initially)

For the Gibbs free energy, the entropy changes, as well as pressure. But how do I solve this?

For the

Or do I use a different equal for that? The volume of the box doesn't change, so would I use

?

 

[anathema]

Gibbs Free Energy = H - TS



Hello, thank you for your post.

To answer your questions, you are correct that the final temperature will be the same as the initial temperature since there is no chemical reaction occurring.

For the final pressure, you can use Dalton's Law of Partial Pressure to calculate the total pressure of the mixture. This law states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each gas in the mixture. In this case, the final pressure (Pf) will be equal to the sum of the initial pressures of each gas (Po + Pn). So, your equation of Pf = 2P is correct.

To calculate the change in Gibbs Free Energy, you can use the equation ΔG = ΔH - TΔS. In this case, the enthalpy (ΔH) will remain constant since there is no chemical reaction occurring. The change in entropy (ΔS) can be calculated using the equation ΔS = nRln(Vf/Vi), where n is the number of moles of gas, R is the gas constant, Vf is the final volume, and Vi is the initial volume. Since the volume of the box does not change, the final and initial volumes will be the same, and ΔS will be equal to 0. This means that the change in Gibbs Free Energy will also be 0.

I hope this helps! Let me know if you have any other questions.