J.R. S. answered • 03/18/21
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work = -pressure x change in volume
w = -P∆V
w = - (0.997 atm) ( 23.4 L - 7.11 L) = - (0.997 atm)(16.29 L)
w = -16.24 Latm
Converting to joules we have...
-16.24 Latm x 101.325 J/Latm = -1651 J
Since 22.6 kJ of heat was PUT INTO the system and 1651 J (1.651 kJ) was released upon expansion, the work done on the system, in joules, is...
+22600 J - 1651 J = 21,949 J
J.R. S.
tutor
Maybe they just want w = -P deltaV. Including the 22.6 kJ would be the change in internal energy and that’s what I calculated. That’s just a guess but maybe the answer is just -1651 J work done by the system.
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03/18/21
Alexis A.
-1651 was the correct answer thank you very much. Your help is very much appreciated!
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03/19/21
Alexis A.
unfortunately this was incorrect, I appreciate the help but do you have any other idea of what this answer might be? thank you.03/18/21