Midterm Q5

$\Delta U=q+w$

torialmquist1F
Posts: 49
Joined: Sat Jul 22, 2017 3:00 am

Midterm Q5

A system undergoes a two-step process. In step one, it expands adiabatically from 1.00 L to 2.56L against an external pressure of 1.00 atm. In step 2, it releases 73J of heat at 1.00 atm as it returns to the original internal energy. Find the volume of the system.

How do I solve this problem?

Caroline Cox 1H
Posts: 18
Joined: Fri Sep 29, 2017 7:03 am

Re: Midterm Q5

Step one is an adiabatic process which means that q=0 and therefore deltaU = w. From here, we use the equation w=-Pext*deltaV and plug in the values to get the number of Joules. For step 2, we are returning to the original internal energy, so deltaU is the same as the value we just calculated for step one, but the sign is reversed. Using the equation deltaU = q + w we can then plug in our values for deltaU and q to solve for w. Then, we use this w in the equation w = -P*deltaV to solve for the final volume of the system.

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

Re: Midterm Q5

I think you can find the intact solution on the course website.
https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Midterm_exam_ans.pdf

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