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

In summary, in a reaction between 50 mL of H2(g) and 50 mL of C2H4(g), 50 mL of C2H6(g) is produced at 1.5 atm. The reaction releases 3.1 x 10^2 J of heat and the decrease in volume requires the surroundings to do 7.6 J of work on the gases. Using the work-energy theorem, we can calculate the change in internal energy of the gases to be -3 x 10^2 J.
  • #1
FLgirl
21
0
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.
 
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  • #2
  • #3
Thanks tiny tim :) This is what I got:

(-310 J) + (7.6J)
= 302.4 J = -3 x 10^2 J
 
  • #4
looks right :smile:
 
  • #5
sweet, thanks
 

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

1. How does a decrease in volume require the surroundings to do work on the gases?

According to the First Law of Thermodynamics, energy cannot be created or destroyed, only transferred. When the volume of a gas decreases, the gas molecules are forced to move closer together, resulting in an increase in pressure. In order to maintain equilibrium, the surroundings must do work on the gases to counteract this increase in pressure.

2. What is the relationship between volume and work in this scenario?

The relationship between volume and work is inverse. As the volume of a gas decreases, the work done by the surroundings on the gas increases. This is because the gas molecules are being compressed, and more work is needed to overcome the increasing pressure.

3. How is the amount of work calculated in this situation?

The amount of work done by the surroundings on the gases can be calculated using the formula W = -PΔV, where W is work, P is pressure, and ΔV is the change in volume. In this scenario, the work done is 7.6 J.

4. Why is it important to consider the surroundings when discussing changes in volume?

The surroundings play a crucial role in maintaining equilibrium when a change in volume occurs. In order for the gas molecules to be compressed and the volume to decrease, the surroundings must do work on the gases. Neglecting the surroundings can lead to a misunderstanding of the process and the amount of work involved.

5. Can this concept be applied to other systems besides gases?

Yes, the concept of work being done by the surroundings in response to a change in volume can be applied to other systems as well. For example, when a solid object is compressed, the surroundings must also do work to maintain equilibrium. This concept is a fundamental principle in thermodynamics and can be observed in various systems and processes.

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