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