Midterm Question 5 (Step 2)

isochoric/isometric:
isothermal:
isobaric:

Moderators: Chem_Mod, Chem_Admin

veneziaramirez 3I
Posts: 57
Joined: Fri Sep 29, 2017 7:07 am

Midterm Question 5 (Step 2)

Postby veneziaramirez 3I » Tue Mar 06, 2018 3:05 pm

Why is the internal energy positive if when we calculated it in step 1 it was negative?

Nora Sharp 1C
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

Re: Midterm Question 5 (Step 2)

Postby Nora Sharp 1C » Tue Mar 06, 2018 3:39 pm

I'm not really certain what you mean by internal energy being positive, but here's how I solved Q5:

Because step one concerns an adiabatic process, q = 0, and consequently internal energy in step one = work. But the second step returns the internal energy to its initial state before step one happened, and since internal energy is a state function, overall(both step one and step two) deltaU = 0. The overall change of internal energy of the system = internal energy change in step one + internal energy in step two.

0 = work in reaction one + heat gain/loss in reaction two + work in reaction two.

From there you just plug in the information given to you to calculate the volume.

Rebecca Doan 2L
Posts: 51
Joined: Thu Jul 27, 2017 3:01 am

Re: Midterm Question 5 (Step 2)

Postby Rebecca Doan 2L » Sun Mar 11, 2018 10:42 pm

The question states "as it returns to the original internal energy". It was calculated, in step 1, that the change in internal energy was -158 so to return to the original internal energy, step 2 internal energy would have to be the opposite, +158J.


Return to “Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)”

Who is online

Users browsing this forum: No registered users and 2 guests