What is #DeltaE# for a system that absorbs #"77.7 J"# of heat and expands from #"0.250 L"# to #"0.750 L"# against an external pressure of #"1.161 atm"#?

1 Answer
Truong-Son N.
Oct 11, 2017

#DeltaE = +"18.9 J"#

As you may expect, you would calculate the change in internal energy using the first law of thermodynamics:

#DeltaE = q + w = q - PDeltaV#

with #q# being heat flow, #w# being work, and #DeltaE# being the change in internal energy, all from the perspective of the SYSTEM.

If your system is the reaction, then the reaction absorbs #"77.7 J"# of heat, i.e. #q = +"77.7 J"#. You can clearly see the volume increases, and so,

#DeltaV > 0#,

and thus, #w < 0#. This then requires proper units! (What is the external pressure?)

#w = -"1.161 atm" ("0.750 L" - "0.250 L") xx "8.314 J"/("0.08206 L"cdot"atm")#

#= -"58.8 J"#

And so, the change in internal energy is:

#color(blue)(DeltaE) = "77.7 J" + (-"58.8 J") = +color(blue)("18.9 J")#

Therefore, the system GAINS heat. Is this reaction exothermic or endothermic?

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