Consistent Sign Conventions

1. The Two Conventions: Physics vs. Chemistry

At the heart of the confusion lies the First Law of Thermodynamics, which is a statement of energy conservation. The internal energy ($U$) of a system can change through two mechanisms: heat transfer ($Q$) and work ($W$). However, the equation is written in two different ways depending on the discipline.

• Chemistry Convention (IUPAC standard): $\Delta U=Q+W$
• Physics/Engineering Convention: $\Delta U=Q-W$

Why the difference? It's all about perspective.

To understand the reasoning behind this, let's watch a short, insightful explanation.

As the video explained:

• Chemistry is often concerned with the energy changes within a chemical system. The equation $\Delta U=Q+W$ treats heat added to the system and work done on the system as positive contributions to its internal energy.
• Physics and Engineering, particularly when studying heat engines, are often interested in the useful work extracted from a system. The equation $\Delta U=Q-W$ treats work done by the system as a positive output, representing energy leaving the system.

The sign convention for heat ($Q$) is happily consistent:

• $Q>0$: Heat is absorbed by the system (endothermic).
• $Q<0$: Heat is released by the system (exothermic).

The entire difference lies in the definition of work ($W$).

2. Defining Work: The Crux of the Matter

Let's summarize the definitions of work:

Notice the crucial relationship:
$W_{by}=-W_{on}$
Work done by the system is the negative of the work done on the system. This is the simple mathematical key to converting between the two conventions.

For further reading and reinforcement of this concept, the following resource provides several perspectives on the topic.

3. A Universal Strategy for JEE Problems

In an exam like the JEE Advanced, questions can be phrased using either convention. Your task is to be methodical. Here is a foolproof strategy:

1. Read the Problem Carefully: Identify the exact phrasing. Does it say "work done by the gas," "work done on the system," "work output," or "work required"?
2. Pick Your Convention: Choose one version of the First Law and use it exclusively. The physics convention ($\Delta U=Q-W_{by}$) is often more intuitive for mechanics-style problems, but either works. Let's stick with $\Delta U=Q-W_{by}$ for this strategy.
3. Translate the Terms: Convert the information given in the problem into the terms of your chosen equation.
   • If the problem gives "work done on the system" = X, then in your equation, $W_{by}=-X$.
   • If the problem gives "work done by the system" = Y, then in your equation, $W_{by}=Y$.
   • For P-V work, remember that the work done by an expanding gas is positive: $W_{by}=\int PdV$.
4. Solve: Apply your chosen equation with the correctly signed terms.

Let's apply this by watching a set of solved problems. The following video uses the chemistry convention ($\Delta U=Q+W$) throughout. Your task is to watch and understand the logic, then we'll confirm the results using our physics-based strategy.

The video's derivation of $W=-P_{ext}\Delta V$ is standard in chemistry. From the physics perspective, the work done by the gas against an external pressure is $W_{by}=P_{ext}\Delta V$. Again, you see the sign flip, confirming $W_{on}=-W_{by}$.

4. Application to a JEE Advanced Problem

Now let's apply this to an actual exam question. The language and diagrams in JEE problems are your primary clues.

Let's analyze this problem together.
A mole of an ideal gas undergoes a cyclic process A $\to$ B $\to$ C $\to$ A.

• Path A $\to$ B: The graph shows volume increasing ($V_1\to V_2$) at a constant temperature ($T_1$). This is an isothermal expansion.
• Path B $\to$ C: The graph shows temperature decreasing ($T_1\to T_2$) at a constant volume ($V_2$). This is isochoric cooling. The problem states $V_2=V_1$, but the graph clearly shows $V_C=V_2$. Let's assume the graph is correct and C is at $(P_2,V_2,T_2)$. Wait, the labels say C is at $(P_2,V_1,T_2)$. This implies the process B$\to$C is not isochoric. Let's re-read the graph and states.
  A=(P1, V1, T1), B=(P2, V2, T1), C=(P2, V1, T2).
  A $\to$ B: Isothermal, V changes from V1 to V2.
  B $\to$ C: Isobaric (constant P2), V changes from V2 to V1.
  C $\to$ A: Isochoric (constant V1), P changes from P2 to P1.
  Okay, the V-T graph is a straight line from origin to A, another straight line from origin through B. This means $V/T$ is constant for both, implying constant pressure. So process C $\to$ A and A $\to$ B' (where B' is on the same line as A) is isobaric. This V-T graph is tricky. Let's re-examine the image. Ah, the states are given. A=(P1,V1,T1), B=(P2,V2,T1), C=(P2,V1,T2). The process is A$\to$B, B$\to$C, C$\to$A.
  • A$\to$B: Isothermal process as T=T1. Volume changes from V1 to V2. Since B is 'above' A on the V axis, $V_2>V_1$. Isothermal Expansion.
  • B$\to$C: Constant pressure P2 (isobaric). Volume changes from V2 to V1. Isobaric Compression.
  • C$\to$A: Constant volume V1 (isochoric). Temperature changes from T2 to T1. Isochoric Heating.

Let's analyze the signs of $q$ and $w$ for process B $\to$ C: Isobaric Compression.

• Work (w): The gas is being compressed ($V_2\to V_1$, where $V_1<V_2$).
  • Using the physics convention: Work is done on the gas. Therefore, $W_{by}$ is negative. $W_{by}<0$.
  • Using the chemistry convention: Work is done on the gas. Therefore, $w$ (or $W_{on}$) is positive. $w>0$.
• Heat (q): In an isobaric compression, both pressure and volume decrease (wait, pressure is constant). The temperature must also decrease. $T_B=T_1$ and $T_C=T_2$. From the graph, $T_1>T_2$. So the gas is cooled. Heat must be removed from the system. Therefore, $q<0$.
• Internal Energy ($\Delta U$): For an ideal gas, $\Delta U=n{C}_v\Delta T$. Since the gas cools ($T_1\to T_2$), $\Delta T$ is negative, so $\Delta U<0$.

Let's check for consistency with the first law.

• Chemistry: $\Delta U=q+w$. We have $\Delta U<0$, $q<0$, and $w>0$. The equation is (negative) = (negative) + (positive), which is possible.
• Physics: $\Delta U=Q-W_{by}$. We have $\Delta U<0$, $Q<0$, and $W_{by}<0$. The equation is (negative) = (negative) - (negative), which simplifies to (negative) = (negative) + (positive), the same as above.

The question asks for the correct statement. Option (B) is $q_{B\to C}=\Delta H_{B\to C}$. Recall that enthalpy change $\Delta H$ is defined as $\Delta H=\Delta U+\Delta (PV)$. For a constant pressure process, $\Delta H=\Delta U+P\Delta V$. Also, heat transferred at constant pressure is $q_p=n{C}_p\Delta T$, and $\Delta H$ is also $n{C}_p\Delta T$. Thus, for an isobaric process, $q=\Delta H$. Since B$\to$C is an isobaric process, option (B) is correct. The question asks for "correct statement(s)", implying multiple may be correct.

By analyzing the signs and definitions consistently, you can navigate these complex problems. The key is not to panic, but to systematically break down each process.

Conclusion

Mastering sign conventions is not about memorization but about understanding perspective. By recognizing the difference between the physics (work output) and chemistry (system energy) viewpoints, you can confidently translate any problem into a solvable form.

Key Takeaways:

• The First Law of Thermodynamics appears as $\Delta U=Q-W$ (physics) or $\Delta U=Q+W$ (chemistry).
• The difference is the definition of work: $W_{physics}$ is work done by the system, while $W_{chemistry}$ is work done on the system. They are related by $W_{by}=-W_{on}$.
• The sign for heat $Q$ is universal: positive for heat absorbed by the system, negative for heat released.
• Your strategy: Choose one convention (e.g., $\Delta U=Q-W_{by}$), and always convert the problem's language to match it before solving.

Preview of the Next Lesson:
Now that we have solidified our understanding of the First Law and its components ($Q,W,\Delta U$), we will move on to more complex integrated problems. In the next lesson, we will solve problems where work done during a thermodynamic process is converted to heat in a calorimeter. This will require us to connect the mechanical work done by a system to the thermal energy changes in another, combining thermodynamics with calorimetry in a single energy conservation framework.