Navigating Work and Heat Sign Conventions

1. The Anchors: Heat Convention and the Core Relationship

First, let's reaffirm the parts that don't change. The sign convention for heat transfer, $Q$, is universal across physics and chemistry.

As the video explains, the convention for heat is straightforward:

• $Q>0$: Heat is absorbed by the system (an endothermic process). The system's energy increases.
• $Q<0$: Heat is released by the system (an exothermic process). The system's energy decreases.

This provides a stable reference point. The entire conversion exercise revolves around the work term, $W$. As we established, work done on the system and work done by the system are simply two sides of the same coin—equal in magnitude and opposite in sign.

This gives us our single, most important conversion rule:

2. A Systematic Approach to Solving Problems

Given your background, you'll appreciate a structured, algorithmic approach to eliminate ambiguity. Here is a foolproof, four-step method to handle any First Law problem, regardless of the convention used in the question.

The Algorithm

1. Identify Given Information: Read the problem and write down the numerical values for heat ($Q$) and work ($W$). Most importantly, note the exact phrasing used for work: "work done on the system," "work done by the gas," "work of expansion," etc. Also, note the process (e.g., expansion or compression).

2. Choose Your Personal Convention: Select one form of the First Law and commit to it. For consistency with most JEE physics resources, we will use the physics convention as our standard:
   $\Delta U=Q-W_{\text{by}}$
   From this point on, the symbol $W$ in our equation will always mean work done by the system.

3. Translate and Convert: This is the key step. Convert the work value from the problem statement to fit our chosen convention ($W_{by}$).

   • If the problem gives "work done by the system = 100 J," then $W_{by}=+100$ J.
   • If the problem gives "work done on the system = 100 J," then $W_{by}=-100$ J.
   • If the problem states "the gas expands and does 100 J of work," expansion means work is done by the gas, so $W_{by}=+100$ J.
   • If the problem states "the gas is compressed, involving 100 J of work," compression means work is done on the gas, so $W_{by}=-100$ J.

4. Solve: Substitute the correctly signed values of $Q$ and your translated $W_{by}$ into the equation $\Delta U=Q-W_{by}$ to find the answer.

This method removes all guesswork. You are no longer trying to remember two different formulas; you are using one formula and applying a simple translation rule.

For a deeper dive into this translation, the following resource is excellent.

3. Putting It into Practice: A Worked Example

Let's apply our algorithm to a numerical problem.

Problem: A system absorbs 7.23 kJ of heat from the surroundings. Simultaneously, 4.8 kJ of work is done on the system by an external force. Calculate the change in internal energy of the system.

Let's follow the steps:

1. Identify:

   • Heat absorbed by the system: $Q=+7.23$ kJ.
   • Work done on the system: $W_{\text{on}}=+4.8$ kJ.

2. Choose Convention:

   • We will use $\Delta U=Q-W_{\text{by}}$.

3. Translate/Convert:

   • Our equation needs $W_{\text{by}}$. The problem gives us $W_{\text{on}}$.
   • Using our conversion rule: $W_{\text{by}}=-W_{\text{on}}$.
   • Therefore, $W_{\text{by}}=-4.8$ kJ.

4. Solve:

   • Substitute the values into our chosen equation:
     $\Delta U=(+7.23\text{kJ})-(-4.8\text{kJ})$
     $\Delta U=7.23+4.8=12.03\text{kJ}$

The internal energy of the system increases by 12.03 kJ.

Notice that if we had used the chemistry convention ($\Delta U=Q+W_{\text{on}}$), we would not have needed to convert the work term:
$\Delta U=(+7.23\text{kJ})+(+4.8\text{kJ})=12.03\text{kJ}$
The result is identical, proving that the method is sound. The key is consistency.

The resource you just read contains this exact numerical example. You can review the section "Solution (in physics)" and "Solution (in chemistry)" under the answer from "Rinkato-Kun" in resource LINK to see both approaches side-by-side.

Conclusion

In this lesson, we focused on the practical skill of navigating the different sign conventions for work in thermodynamics. By mastering a simple translation rule and applying a consistent problem-solving algorithm, you can eliminate a common source of error in JEE Advanced problems.

Key Takeaways:

• The sign convention for heat ($Q$) is universal: positive for energy in, negative for energy out.
• The two work conventions are linked by a simple conversion: $W_{\text{on the system}}=-W_{\text{by the system}}$.
• A systematic approach (Identify, Choose, Translate, Solve) ensures consistency and accuracy, regardless of how a problem is phrased.
• The final answer for $\Delta U$ is a physical reality and will always be the same, no matter which valid convention you use to calculate it.

Preview of the Next Lesson:
We have now mastered the signs. The next logical question is, how do we calculate the magnitude of work? In the next lesson, "Calculate work done by or on a gas from P-V diagrams, applying the correct sign for each convention," we will learn the fundamental formula for pressure-volume work, $W=\int PdV$, and see how to compute it by finding the area under the curve on a P-V diagram.