Quiz1 Prep Question 11


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Joann Hernandez 2F
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Quiz1 Prep Question 11

Postby Joann Hernandez 2F » Thu Jan 21, 2016 6:35 pm

Given in step 1, it expands adiabatically from 1.00 L to 2.56 L at 1.00 atm.. in step 2, it releases 73 J of heat at 1.00 atm.
For Step 2 in the question, It asks to find the final volume of the system as is returns to the original internal energy.
The answer key says change in Volume = -2.28 L and the initial Volume is 2.56 L so the final Volume = 0.28 L
How and where did they get the Change in Volume to be -2.28 L?

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Re: Quiz1 Prep Question 11

Postby Chem_Mod » Fri Jan 22, 2016 4:44 pm

Step 1-----> adiabatic means constant heat, so deltaU=w

deltaU=-(1.56L)*(1atm)*(101.325J/(L*atm))= -158.067 J

Step 2------> There is a change in volume (work was done) and a release in heat, but we want the total deltaU to be zero

deltaUtotal=0=deltaU1+deltaU2

So they must be equal and opposite

deltaU2= 158.067J

158.067J J= w - 73J

w=-1atm(Vf-2.56L)*(101.325J/(L*atm))

Plug that in to the w term I have given above and solve for Vf


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