Thermodynamic Problems: Work and Heat in a Thermodynamic Process

In summary, the initial pressure and volume of an ideal gas are given, along with the internal energy. The gas is then cooled at constant volume and allowed to expand at constant pressure, with the final internal energy also given. The question asks for the work associated with the entire process, which can be found using the equations W=-P*deltaV and deltaU=Q+W. The work is calculated to be -639 J, but there may be some uncertainty in the sign of the answer.
  • #1
yuvlevental
44
0

Homework Statement


Initially, 0.745 moles of an ideal gas in a container occupies a volume of 3.55 L at a pressure of 3.82 atm with an internal energy of 1705 J. The gas is then cooled at a constant volume until its pressure is 2.06 atm. Then the gas is allowed to expand at a constant pressure until its volume is 6.65 L. The final internal energy is 2098 J. Consider the processes to be quasi-static. What is the work associated with this entire process?

Homework Equations


W=-P*deltaV
deltaU=Q + W

The Attempt at a Solution


Is it correct that you take the 2.06, multiply it by 1E5, and multiply the answer by the change in volume in m^3 to find -638.60 J
 
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  • #2
Yes and no. Your method is correct, but your answer has too many sig figs.
 
  • #3
yuvlevental said:
What is the work associated with this entire process?
Other than sig figs, the use of prepositions matters. Work done on what by what? I think that there is a sign uncertainty here.
 

1. What is the difference between work and heat in a thermodynamic process?

In thermodynamics, work is the transfer of energy from one system to another due to a force acting over a distance. On the other hand, heat is the transfer of thermal energy from a hotter system to a cooler system. Both work and heat can be involved in a thermodynamic process, but they have different effects on the system.

2. How is work calculated in a thermodynamic process?

Work is calculated using the formula W = F x d, where W is work, F is the force applied, and d is the distance over which the force is applied. In a thermodynamic process, work can be calculated in different ways, depending on the type of process and the system involved.

3. How does the first law of thermodynamics apply to work and heat in a thermodynamic process?

The first law of thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another. In a thermodynamic process, the amount of work done on a system is equal to the change in its internal energy plus the heat added to the system.

4. Can work and heat be converted from one to another in a thermodynamic process?

No, work and heat cannot be converted from one to another in a thermodynamic process. They are two distinct forms of energy and have different effects on the system. However, they can both be involved in the same thermodynamic process and contribute to the overall change in the system.

5. How does the second law of thermodynamics apply to work and heat in a thermodynamic process?

The second law of thermodynamics states that the total entropy of a closed system always increases over time. In a thermodynamic process, this means that not all heat energy can be converted into work, and some of it will always be lost as heat transfer to the surroundings.

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