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### Topic: Change in internal energy of the gases  (Read 3847 times)

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#### FLgirl

• Regular Member
•   • Posts: 25
• Mole Snacks: +0/-4 ##### Change in internal energy of the gases
« on: October 30, 2012, 12:34:56 PM »
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.

#### Big-Daddy

• Sr. Member
•     • Posts: 1177
• Mole Snacks: +28/-94 ##### Re: Change in internal energy of the gases
« Reply #1 on: October 30, 2012, 10:06:57 PM »
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?

ΔU=q+w

The system released energy, so q is negative, thus q=-310 J. Work is done on the system, so w=+7.6 J.

ΔU=(-310)+(+7.6)
ΔU=-302.4 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-V2) (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:

V1-V2=w/p
ΔV=V2-V1=-(w/p)
ΔV=-(+7.6/151,987.5)
ΔV=-5*10-5 m3
ΔV=-50 cm3