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Pediateix

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can someone explain this question from EK 1001 gen chem (331)

a heat engine is designed using an ideal gas as a working fluid. at the completion of each cycle, the engine returns to its original state. if the net flow of heat into the engine is 60,000 kj/cycle, what can be said about the work done by each cycle?


A-less than 60,000 KJ
B-equal to 60,000 kj
C-greater than 60,000 kj
D-begins as 60,000 kh and decreases each subsequent cycle
 

Nugester

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My initial feeling says either A or D, probably leaning towards A. Because engines aren never 100% efficient and some heat is always lost.
 
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Pediateix

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the temerapture of an ideal gas undergoing a free adiabatic expansion should

a-remian constant
b-icnrease because the molecules will be moving faster as they escape into the vacuum
c-ddecrease because molecular energy is spread over a large area
d- decrease because the pressure goes down

why is it a


this seems to say otherwise

https://files.mtstatic.com/site_433...jstC9BTPPaY_&Key-Pair-Id=APKAIX7ZMYEQ4P6XATFQ
 

Pediateix

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can someone explain how internal energy depends on temperature but not change in internal energy?
 

Nugester

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Their answer is wrong. Q + W (or -W depending on POV) = U. Q = 0, so any change in internal energy is due to work. Gas is expanding and doing work, so it is losing internal energy; temperature should decrease.
 

Pediateix

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Their answer is wrong. Q + W (or -W depending on POV) = U. Q = 0, so any change in internal energy is due to work. Gas is expanding and doing work, so it is losing internal energy; temperature should decrease.
my reasoning was pressure decreases so temperature also decrease
 

Pediateix

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do you get this
From the EK 1001 Chemistry (Q 375), it says that the heat of formation of water vapor is negative (and that it is less negative than the heat of formation of liquid water).
 

aldol16

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how is it not possible to convert 100% of the heat energy into work??

This is a fundamental consequence of the second law of thermodynamics. Heat engines cannot be 100% efficient. The maximum theoretical efficiency can be calculated and is for something called the Carnot cycle. I can't do this topic justice so I can only refer you to this excellent source: Heat cannot be completely converted. Basically, what you just said is the second law of thermodynamics. Imagine heat and work as similar concepts - they're both energy. Heat is just disordered energy and work is ordered energy. You can always convert energy to heat - going from order to disorder. This is because you raise entropy going from order to disorder. But you can't always convert all heat to work because then you'd literally be making order from disorder - a violation of the second law of thermodynamics.
 
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