Midterm #5


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Maeve Gallagher 1J
Posts: 56
Joined: Fri Sep 29, 2017 7:07 am

Midterm #5

Postby Maeve Gallagher 1J » Thu Mar 15, 2018 8:23 pm

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.

I understand that because the system is adiabatic, q=0 and deltaU=w. However, for step two in the solution it has deltaU equal to the opposite of the work calculated for step one. Why is the sign flipped in this case?

Michelle Chernyak 1J
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

Re: Midterm #5

Postby Michelle Chernyak 1J » Thu Mar 15, 2018 8:37 pm

Because the system returns to its original internal energy, the overall deltaU of the reaction is 0. That means whatever deltaU we calculated in the first step we have to do the reverse of in the second step in order for them to cancel out and go back to 0.


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