Why is the work done considered negative in this thermodynamics problem?

In summary, the conversation is about determining the change in internal energy of an ideal gas along the path AB. The work done in the process is found to be 52.5 Joules, and the heat extracted is -70 calories. The first law of thermodynamics is used to calculate the change in internal energy, resulting in an answer of -346.5 Joules. However, the teacher argues that the answer should be -241.5 Joules, considering the work done as negative. The possible explanation for this is that the sign of the work done depends on whether it is done by the system or put into the system, and the formula can be used to determine this.
  • #1
A Physics Enthusiast
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An ideal gas is taken along the path AB (see fig.). If 70 calories of heat is extracted from the gas during the process, calculate the change in internal energy during the process.
My attempt :
Work done in the process is the area of the shaded region which turns out to be 52.5 Joules.
Again, heat extracted,
$$Q=-70 \, \text{cal}= -294 \, J$$

So, from first law of thermodynamics,
$$\begin{align}\Delta U &= Q-W\\ &=(-294-52 5)\, J\\ &= \boxed{-346.5 \, J}\end{align}$$

So, that was my answer. But my teacher says that the answer is -241.5 Joules. For that, he says that the work done should be taken as negative. What can the possible explanation for this ? Where am I wrong ?
 

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  • #2
Energy put into the system is counted positive and vice versa. The heat is extracted, therefore it has a negative sign. So you have to determine whether the work is done by the system or put into the system. The formula could be a help or also just think how work is done on a closed system.
 
  • #3
stockzahn said:
Energy put into the system is counted positive and vice versa. The heat is extracted, therefore it has a negative sign. So you have to determine whether the work is done by the system or put into the system. The formula could be a help or also just think how work is done on a closed system.
Please don't confuse me more. All I need is a proper reasoning which is not required in the homework problem. As you can see, I have got my answer. I can get my teacher's answer easily too by putting a negative sign in front of the work done. So, please explain me in simple words.
 
  • #4
T13091999 said:
Please don't confuse me more. All I need is a proper reasoning which is not required in the homework problem. As you can see, I have got my answer. I can get my teacher's answer easily too by putting a negative sign in front of the work done. So, please explain me in simple words.

Is the gas compressed or expanded? Plus: The formula would help, I recommend to write it down.
 

What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships between heat, work, and energy.

What is a simple thermodynamics problem?

A simple thermodynamics problem typically involves calculating the change in energy, heat, or work in a system using the laws of thermodynamics.

What are the laws of thermodynamics?

The first law states that energy cannot be created or destroyed, only transferred or converted between forms. The second law states that the total entropy (disorder) of a closed system will always increase over time. The third law states that the entropy of a perfect crystal at absolute zero temperature is zero.

How do you solve a simple thermodynamics problem?

To solve a simple thermodynamics problem, you will need to identify the system and any energy transfers or conversions that are occurring. Then, apply the appropriate laws of thermodynamics to calculate the change in energy, heat, or work in the system.

What are some real-world applications of thermodynamics?

Thermodynamics has many practical applications in fields such as engineering, chemistry, and biology. Some examples include designing more efficient engines, understanding chemical reactions, and studying the metabolism of living organisms.

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