First Law of Thermodynamics: Solving for Unknowns

Hello! Welcome back to your study of thermodynamics.

In our last few lessons, we have systematically built up the components of the First Law of Thermodynamics. You learned to calculate work done ( $W$ ) from P-V diagrams and handle the associated sign conventions. Then, you mastered the concept of internal energy ( $Δ U$ ), understanding why $Δ U = n C_{V} Δ T$ is universally true for any ideal gas process.

Today, we bring it all together. This lesson focuses on applying the complete First Law equation to solve for an unknown quantity—be it heat ( $Q$ ), work ( $W$ ), or internal energy change ( $Δ U$ )—for a single thermodynamic process. This is the core application of the law and a fundamental skill for solving complex JEE Advanced problems. We will also directly address your key concern: navigating the different sign conventions used in physics and chemistry, ensuring you can solve a problem regardless of how it's framed.

1. The First Law: An Energy Conservation Equation

At its heart, the First Law of Thermodynamics is a statement of energy conservation for a thermodynamic system. It says that the change in a system's internal energy ( $Δ U$ ) is equal to the net energy transferred to it. Energy can be transferred in two forms: heat ( $Q$ ) and work ( $W$ ).

The confusion with sign conventions arises from the definition of $W$ . Does $W$ represent work done by the system on its surroundings, or work done on the system by its surroundings?

Let's watch two short videos that explain this from the ground up, one from a physics perspective and one from a chemistry perspective.

First Law of Thermodynamics : sign convention for WORK

First, let's hear from H.C. Verma. He emphasizes that the fundamental physics of work (force and displacement) is universal. The convention arises only when we assign the symbol 'W' to either work done by the gas or on the gas.

Watch from 10:31 to 18:21. Focus on how the same physical event (e.g., a gas expanding) leads to two different forms of the First Law equation depending on how 'W' is defined. This is the root of the physics vs. chemistry conventions.

First Law of Thermodynamics, Basic Introduction - Internal Energy, Heat and Work - Chemistry

Now, let's see the chemistry viewpoint from The Organic Chemistry Tutor. This video explains the rationale for each convention based on the point of view: the system (chemistry) versus the surroundings (physics/engineering).

Watch from 05:12 to 09:24. Pay close attention to the explanation of why chemists prefer to see work done on the system as positive (it increases the system's energy), while physicists are often interested in the work a system does as output.

To summarize and give you a concrete way to handle this:

Physics / Engineering Convention: Focuses on what the system does.
$Δ U = Q - W_{by}$
Here, $W_{by}$ is the work done by the system. A positive $W_{by}$ (expansion) decreases the system's internal energy. This is the convention we have used so far and is common in JEE Physics.
Chemistry / IUPAC Convention: Focuses on what is done to the system.
$Δ U = Q + W_{on}$
Here, $W_{on}$ is the work done on the system. A positive $W_{on}$ (compression) increases the system's internal energy.

The key relationship is simple: $W_{on} = - W_{by}$ . As long as you are clear about which $W$ you are calculating, the physics remains identical.

2. A Systematic Approach to First Law Problems

Given your background, you'll appreciate a structured, almost algorithmic, approach to these problems. This minimizes conceptual errors, especially under exam pressure.

Here is a robust workflow:

Identify State and Process: Clearly define your system (e.g., moles, gas type) and the initial and final states (P, V, T). Identify the thermodynamic process path.
Choose Convention: Explicitly write down the version of the First Law you will use (e.g., Using ΔU = Q - W_by). This commits you to a convention and prevents mid-problem confusion.
Calculate Knowns (any two of W, ΔU, Q):
- Work ( $W$ ): Calculate the work done based on the process path. For $W_{by}$ , this is $\int P d V$ . The sign is determined by the volume change:
  - Expansion ( $V_{f} > V_{i}$ ): $W_{by} > 0$
  - Compression ( $V_{f} < V_{i}$ ): $W_{by} < 0$
- Internal Energy ( $Δ U$ ): Calculate the change in internal energy using $Δ U = n C_{V} Δ T$ . The sign is determined by the temperature change:
  - Heating ( $T_{f} > T_{i}$ ): $Δ U > 0$
  - Cooling ( $T_{f} < T_{i}$ ): $Δ U < 0$
Solve for the Unknown: Rearrange the First Law equation to solve for the remaining variable.
Interpret the Result: Check the sign of your answer. For heat ( $Q$ ):
- $Q > 0$ : Heat is absorbed by the system (endothermic).
- $Q < 0$ : Heat is released by the system (exothermic).

3. Worked Example: A Linear Process

Let's apply this workflow to a problem involving a non-standard process, a common feature in JEE Advanced.

Problem:
Two moles of a diatomic ideal gas ( $C_{V} = \frac{5}{2} R$ ) are taken from state A (1 atm, 40 L) to state B (3 atm, 20 L) along a straight-line path on a P-V diagram. Calculate the work done ( $W$ ), the change in internal energy ( $Δ U$ ), and the heat transferred ( $Q$ ) during this process.
Use $R \approx 8.3 J/(mol\cdotK)$ and $1 atm = 1 0^{5} Pa$ .

Solution:

1. Identify State and Process:

System: $n = 2$ moles, diatomic gas ( $C_{V} = \frac{5}{2} R$ ).
State A: $P_{A} = 1 0^{5}$ Pa, $V_{A} = 40 \times 1 0^{- 3} m^{3}$ .
State B: $P_{B} = 3 \times 1 0^{5}$ Pa, $V_{B} = 20 \times 1 0^{- 3} m^{3}$ .
Process: Linear path on P-V diagram.

2. Choose Convention:
We will use the physics convention: $Δ U = Q - W_{by}$ .

3. Calculate Knowns:

Work Done ( $W_{by}$ ):
Work is the area under the P-V graph. For a linear path, this is a trapezoid.
$W_{by} = Area = \frac{1}{2} (P_{A} + P_{B}) (V_{B} - V_{A})$
$W_{by} = \frac{1}{2} (1 0^{5} + 3 \times 1 0^{5}) (20 \times 1 0^{- 3} - 40 \times 1 0^{- 3})$
$W_{by} = \frac{1}{2} (4 \times 1 0^{5}) (- 20 \times 1 0^{- 3}) = - 4000 J$
The sign is negative, which makes physical sense: this is a compression, so work is done on the gas. The work done by the gas is therefore negative.
Change in Internal Energy ( $Δ U$ ):
We need $T_{A}$ and $T_{B}$ to find $Δ T$ . We use the ideal gas law, $P V = n R T$ .
$T_{A} = \frac{P _{A} V _{A}}{n R} = \frac{( 1 0 ^{5} ) ( 40 \times 1 0 ^{- 3} )}{2 \times 8.3} = \frac{4000}{16.6} \approx 241 K$
$T_{B} = \frac{P _{B} V _{B}}{n R} = \frac{( 3 \times 1 0 ^{5} ) ( 20 \times 1 0 ^{- 3} )}{2 \times 8.3} = \frac{6000}{16.6} \approx 361 K$
$Δ T = T_{B} - T_{A} = 361 - 241 = 120 K$
Now we can calculate $Δ U$ :
$Δ U = n C_{V} Δ T = 2 \times (\frac{5}{2} R) \times Δ T = 5 R Δ T$
$Δ U = 5 \times (8.3) \times (120) = 4980 J$
The sign is positive, which makes sense: the final temperature is higher than the initial temperature.

4. Solve for the Unknown ( $Q$ ):
$Δ U = Q - W_{by} ⟹ Q = Δ U + W_{by}$
$Q = 4980 J + (- 4000 J) = 980 J$

5. Interpret the Result:
$Q$ is positive, meaning $980 J$ of heat was absorbed by the gas during this process. Even though the gas was compressed, the final temperature is significantly higher, which requires a net input of heat.

4. Practice with JEE Advanced Problems

Now, it's your turn to apply this method. The following problems are representative of what you might see on the JEE Advanced exam. They often involve interpreting graphs and combining the ideal gas law with the First Law.

JEE Advanced Thermodynamics Practice Problems - P-V Diagram Analysis — This image contains two JEE Advanced problems. Focus on **Problem 1**. A monoatomic gas undergoes a process from A to B. You are asked to find the point where heat transfer changes sign and the total heat supplied. To do this, you will need to express Q as a function of V, by first finding W and ΔU as functions of V.

Test your understanding! (Problem 1 from image)

Let's tackle part (b) of the first problem: "The heat Q supplied to the gas during the process from A to B is?"

Use the systematic approach. For a monoatomic gas, $C_{V} = \frac{3}{2} R$ . The process is a straight line from $A = (V_{0}, P_{0})$ to $B = (2 V_{0}, 2 P_{0})$ .

Show solution strategy

Identify State & Process: $n = 1$ mole, monoatomic gas. Linear path from $A (V_{0}, P_{0})$ to $B (2 V_{0}, 2 P_{0})$ .
Convention: $Δ U = Q - W_{by}$ .
Calculate $W_{by}$ : Find the area of the trapezoid under the A-B line.
$W_{by} = \frac{1}{2} (P_{0} + 2 P_{0}) (2 V_{0} - V_{0}) = \frac{3}{2} P_{0} V_{0}$
Calculate $Δ U$ : Find $T_{A}$ and $T_{B}$ using $P V = R T$ (since n=1).
- $T_{A} = \frac{P _{0} V _{0}}{R}$
- $T_{B} = \frac{( 2 P _{0} ) ( 2 V _{0} )}{R} = \frac{4 P _{0} V _{0}}{R}$
- $Δ T = T_{B} - T_{A} = \frac{3 P _{0} V _{0}}{R}$
- $Δ U = n C_{V} Δ T = (1) (\frac{3}{2} R) (\frac{3 P _{0} V _{0}}{R}) = \frac{9}{2} P_{0} V_{0}$
Solve for Q:
$Q = Δ U + W_{by} = \frac{9}{2} P_{0} V_{0} + \frac{3}{2} P_{0} V_{0} = \frac{12}{2} P_{0} V_{0} = 6 P_{0} V_{0}$
This matches option (D) in the provided solution.

Conclusion

You have now integrated all the components of the First Law of Thermodynamics. You can calculate $W$ , $Δ U$ , and $Q$ for an ideal gas undergoing a single process and, crucially, you understand how to navigate the different sign conventions without confusion.

Key Takeaways:

The First Law of Thermodynamics is an energy balance equation: $Δ U = Q + W_{on}$ (Chemistry) or $Δ U = Q - W_{by}$ (Physics).
The physical outcome is identical regardless of convention; the key is to be consistent in your definition of work ( $W$ ).
A systematic workflow (Identify -> Convention -> Calculate Knowns -> Solve for Unknown -> Interpret) is your most reliable tool for solving First Law problems accurately.
For any ideal gas process, $Δ U$ depends only on $Δ T$ , while $W$ and $Q$ are path-dependent.

Preview of the Next Lesson:
We've focused on single processes. Real-world applications and many advanced problems involve a sequence of processes that form a thermodynamic cycle. In our next lesson, we will analyze these cycles, calculating the net work done and net heat transfer over a full cycle, bringing you one step closer to mastering complex thermodynamic systems.

Recommended resources

1. The First Law: An Energy Conservation Equation

2. A Systematic Approach to First Law Problems

3. Worked Example: A Linear Process

4. Practice with JEE Advanced Problems

Conclusion