Textbook 4A.3 Typo? [ENDORSED]
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Textbook 4A.3 Typo?
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?
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Re: Textbook 4A.3 Typo?
Yes, the key has a typo. It should be 28 J. This can be found in Dr. Lavelle's solution manual errors document.
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Re: Textbook 4A.3 Typo?
Jamie Lin 1K wrote: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.
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Re: Textbook 4A.3 Typo? [ENDORSED]
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.
https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf
Click refresh in your browser to make sure the newest file loads.
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Re: Textbook 4A.3 Typo?
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?
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?
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Re: Textbook 4A.3 Typo?
Natalie Keung 1D wrote: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.
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