8.93


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Naomi Jennings 2H
Posts: 21
Joined: Thu Jul 13, 2017 3:00 am

8.93

Postby Naomi Jennings 2H » Sun Feb 11, 2018 10:25 pm

Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25 degrees C and 1.00 bar.
How would you start to do this problem?

Anh Nguyen 2A
Posts: 36
Joined: Fri Sep 29, 2017 7:05 am
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Re: 8.93

Postby Anh Nguyen 2A » Sun Feb 11, 2018 10:36 pm

Since the work is done against the atmosphere, the pressure is constant or ΔP=0. In that case, w=-P.ΔV=-P.Δn(gas).R.T
with Δn(gas)=n(gas) products - n(gas) initial


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