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AveGali [126]
4 years ago
5

A sample of octane burns releasing 2290 J of heat to the surroundings, and the gases produced expands against a piston to do 560

joules of work. Calculate the internal energy change for this reaction.
1- –2850 J
2- –1730 J
3- –2290 J
4- +1730 J
5- +2850 J
Chemistry
1 answer:
dexar [7]4 years ago
4 0

<u>Answer:</u> The internal energy change for the reaction is -2850 J

<u>Explanation:</u>

  • <u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

  • <u>Sign convention of work:</u>

Work done for expansion process is taken as negative and work done for compression is taken as positive.

According to the First law of thermodynamics,

\Delta U=q+w

where,

\Delta U = internal energy

q = heat absorbed or released  = -2290 J

w = work done = -560 J

Putting values in above equation, we get:

\Delta U=(-2290)+(-560)\\\\\Delta U=-2850J

Hence, the internal energy change for the reaction is -2850 J

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