The decrease in volume requires the surroundings do 7.6 J of work on the gases

[FLgirl]

1. 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?




2. Change in Internal Energy = Q(heat energy) + W(work)
Work = - P x (change in volume)



3. I know how to apply these equations but have no idea about the " 7.6 J of work" part.

 

[tiny-tim]

Hi FLgirl!

I know how to apply these equations but have no idea about the " 7.6 J of work" part.

Apply the work energy theorem (see http://en.wikipedia.org/wiki/Work-energy_theorem#Work-energy_principle).

Work and energy are essentially the same thing …

is the 7.6 J going in or coming out? adjust the energy accordingly

 

[FLgirl]

Thanks tiny tim :) This is what I got:

(-310 J) + (7.6J)
= 302.4 J = -3 x 10^2 J

 

[tiny-tim]

looks right

 

[FLgirl]

sweet, thanks