Energy change is the sum of heat and work: ΔE = q + w. Work is calculated by: w = -PΔV What is the change in energy (in joules) if a reaction absorbs 40.3 J of heat and increases in volume from 0.250L to 0.750L at a constant pressure of 1.056 atm?

Please include the correct sign with your numerical result.
(Note: Make sure your units are consistent when combining energy terms. The conversion factor you need is: 1 L·atm = 101.3 J)

1 Answer
Truong-Son N.
Oct 11, 2017

This is just asking you to apply the first law of thermodynamics:

#DeltaE = q + w#

where:

  • #DeltaE# is the change in internal energy
  • #q# is the heat flow involved.
  • #w# is the work due to the system expansion or compression.

We know that #q = +"40.3 J"#; the system absorbs heat. It gains heat.

We also have to know that work is negative (why? Does the system expand or compress? What is the sign of #DeltaV#?).

The universal gas constants form your conversion factor:

#("8.314472 J/mol"cdot"K")/("0.082057 L"cdot"atm/mol"cdot"K") ~~ "101.3 J"/("L"cdot"atm")#

Therefore, the change in internal energy is:

#color(blue)(DeltaE) = overbrace(+"40.3 J")^((+)) underbrace(- "1.056 atm" xx overbrace(("0.750 L" - "0.250 L"))^((+)))_((-)) xx ("8.314472 J/mol"cdot"K")/("0.082057 L"cdot"atm/mol"cdot"K")#

#= +"40.3 J" - "53.50 J"#

#= color(blue)(-"13.2 J")#

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