Textbook 4A.3 Typo? [ENDORSED]

[Brian Diehl 2B]

The textbook gives an answer of +8 J for part c, but shouldn't the change in internal energy of the system be the same as the work done on the pump (+28 J) since no heat was added?

[Jamie Lin 1K]

Yes, the key has a typo. It should be 28 J. This can be found in Dr. Lavelle's solution manual errors document.

[Brian Diehl 2B]

Yes, the key has a typo. It should be 28 J. This can be found in Dr. Lavelle's solution manual errors document.

Thanks, I couldn't find it in there when I checked.

[Chem_Mod]

The Solution Manual Errors 7th Edition is at:

https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf

Click refresh in your browser to make sure the newest file loads.

[Natalie Keung 1D]

Hi Brian!

I am confused as to what the 101.325J/L*atm is and where it comes from. I thought the equation for work is -P(deltaV). I don't see anywhere I would put this but I see this step in the solution manual and in your work. Do you think you could help me and please explain what this is?

[Brian Diehl 2B]

Hi Brian!

I am confused as to what the 101.325J/L*atm is and where it comes from. I thought the equation for work is -P(deltaV). I don't see anywhere I would put this but I see this step in the solution manual and in your work. Do you think you could help me and please explain what this is?

Hey Natalie!

The quantity 101.325 J*L^-1*atm^-1 is just a conversion factor that allows the answer to be given in Joules when using w = -PΔV. The solutions manual converted the volume of the pump from cm³ to L and then used this value. I decided to convert the pressure to Pascals and the volume to m³ in order to get Joules, which is why my conversion looks a bit different.