Reacting 50 mL of H2(g) with 50 mL of C2H4(g) produces 50 mL of C2H6(g) at 1.5 atm. If the reaction produces 3.1 x 10^2 J of heat and the decrease in volume requires the surroundings do 7.6 J of work on the gases, what is the change in internal energy of the gases?
Change in Internal Energy = Q(heat energy) + W(work)
Work = - P x (change in volume)
I know how to apply these equations but have no idea about the " 7.6 J of work" part.
In what sense do you know how to apply the equations if you can't plug in the numbers for them?
The system released energy, so q is negative, thus q=-310 J. Work is done on
the system, so w=+7.6 J.
Roughly = -300 J (2 sf)
You didn't even need the second equation as far as I understand.
We also know that w=p*(V1
) (in this case, V2
is smaller than V1
since the volume has decreased, leaving the value inside the brackets as positive and correlating with our positive concept of w in this question, since p is always positive). At constant pressure (1.5 atm=101325*1.5=151,987.5 Pa) we can calculate ΔV in m3