Entropy Changes in Ideal Gases

Hello! Let's dive into the practical application of the entropy concept we developed in the last lesson.

In our previous session, we established that entropy (S) is a state function, and its change is defined by the integral of $d Q_{r e v} / T$ . This is a powerful theoretical tool, but for solving problems—especially the kind you'll face in the JEE Advanced—we need concrete, ready-to-use formulas.

Today's lesson focuses on exactly that. We will calculate entropy changes for ideal gases in various thermodynamic processes. You will learn to derive and apply the two fundamental equations for entropy change of an ideal gas. Mastering these will allow you to solve for $Δ S$ in any situation, whether it's a simple single-step process or a complex cycle.

1. Deriving the Master Equations for Entropy Change

Our starting point is the bridge between the First Law and the definition of entropy. For any infinitesimal, reversible process:

First Law: $d U = d Q_{r e v} - d W$
Entropy Definition: $d Q_{r e v} = T d S$
Work Done: $d W = P d V$

Combining these gives us the first TdS equation:
$T d S = d U + P d V$
For $n$ moles of an ideal gas, we know that the change in internal energy is always $d U = n C_{v} d T$ , regardless of the process. Substituting this, we get:
$T d S = n C_{v} d T + P d V$
To get an expression for $d S$ , we divide by $T$ :
$d S = n C_{v} \frac{d T}{T} + \frac{P}{T} d V$
Using the ideal gas law, $P V = n R T$ , we can replace $\frac{P}{T}$ with $\frac{n R}{V}$ .
$d S = n C_{v} \frac{d T}{T} + n R \frac{d V}{V}$
Integrating this from an initial state (1) to a final state (2) gives our first master formula:
$Δ S = S_{2} - S_{1} = n C_{v} ln (\frac{T _{2}}{T _{1}}) + n R ln (\frac{V _{2}}{V _{1}})$
This equation is powerful because it expresses the change in a state function ( $S$ ) entirely in terms of other state variables (T and V).

To derive the second master formula, we use the definition of enthalpy, $H = U + P V$ . The derivation is very similar.

Entropy Change of Ideal Gases

The following text formally derives both master equations for entropy change of an ideal gas. Given your mathematical background, walking through this derivation is the best way to internalize where these crucial formulas come from.

Please read the section '9.1. Entropy Change of Ideal Gases'. Pay close attention to how the two TdS equations, ds = (c_v/T)dT + (R/v)dv and ds = (c_p/T)dT - (R/p)dp, are derived and then integrated to produce the final formulas for \Delta s. Note that this text uses specific entropy (s), which is entropy per unit mass. For molar entropy, simply replace the specific gas constant with the universal gas constant R.

To summarize, the two key equations for calculating the molar entropy change of an ideal gas are:

In terms of T and V:
$Δ S = n C_{v} ln (\frac{T _{2}}{T _{1}}) + n R ln (\frac{V _{2}}{V _{1}})$
In terms of T and P:
$Δ S = n C_{p} ln (\frac{T _{2}}{T _{1}}) - n R ln (\frac{P _{2}}{P _{1}})$

You only need to know the initial and final states (T, V, P) to calculate $Δ S$ . The path taken between the states—reversible or irreversible—does not matter, because entropy is a state function.

Test your understanding!

Show that the two master equations are equivalent. Start with the first equation (in terms of T and V) and use the ideal gas law and Mayer's relation ( $C_{p} - C_{v} = R$ ) to derive the second equation (in terms of T and P).

Show answer

Start with the first equation:
$Δ S = n C_{v} ln (\frac{T _{2}}{T _{1}}) + n R ln (\frac{V _{2}}{V _{1}})$
From the ideal gas law, $\frac{V _{2}}{V _{1}} = \frac{T _{2} P _{1}}{T _{1} P _{2}} = (\frac{T _{2}}{T _{1}}) (\frac{P _{1}}{P _{2}})$ . Substitute this into the equation:
$Δ S = n C_{v} ln (\frac{T _{2}}{T _{1}}) + n R ln [(\frac{T _{2}}{T _{1}}) (\frac{P _{1}}{P _{2}})]$
Using the property $ln (ab) = ln (a) + ln (b)$ :
$Δ S = n C_{v} ln (\frac{T _{2}}{T _{1}}) + n R ln (\frac{T _{2}}{T _{1}}) + n R ln (\frac{P _{1}}{P _{2}})$
Factor out the $ln (T_{2} / T_{1})$ term:
$Δ S = n (C_{v} + R) ln (\frac{T _{2}}{T _{1}}) - n R ln (\frac{P _{2}}{P _{1}})$
Using Mayer's relation, $C_{p} = C_{v} + R$ :
$Δ S = n C_{p} ln (\frac{T _{2}}{T _{1}}) - n R ln (\frac{P _{2}}{P _{1}})$
This is the second master equation.

2. Entropy Change in Specific Processes

Let's see how these general formulas simplify for the standard thermodynamic processes.

Isochoric Process ( $V_{2} = V_{1}$ ):
The $ln (V_{2} / V_{1})$ term is $ln (1) = 0$ . The first formula simplifies to:
$Δ S = n C_{v} ln (\frac{T _{2}}{T _{1}})$
Isobaric Process ( $P_{2} = P_{1}$ ):
The $ln (P_{2} / P_{1})$ term is $ln (1) = 0$ . The second formula simplifies to:
$Δ S = n C_{p} ln (\frac{T _{2}}{T _{1}})$
Isothermal Process ( $T_{2} = T_{1}$ ):
The $ln (T_{2} / T_{1})$ terms are $ln (1) = 0$ . The formulas become:
$Δ S = n R ln (\frac{V _{2}}{V _{1}}) or Δ S = - n R ln (\frac{P _{2}}{P _{1}}) = n R ln (\frac{P _{1}}{P _{2}})$
This connects perfectly with our fundamental definition. For a reversible isothermal process, $Δ U = 0$ , so $Q_{r e v} = W = n R T ln (V_{2} / V_{1})$ . Thus, $Δ S = Q_{r e v} / T = n R ln (V_{2} / V_{1})$ .
Adiabatic Process ( $Q = 0$ ):
This case requires careful thinking.
- If the process is reversible and adiabatic (isentropic): By definition, $d Q_{r e v} = 0$ , so $Δ S = \int \frac{d Q _{r e v}}{T} = 0$ . This is the definition of an isentropic process.
- If the process is irreversible and adiabatic: Here, $Q = 0$ , but $Q_{r e v} \neq = 0$ for the equivalent path between the same start and end points. We CANNOT say $Δ S = 0$ . Instead, we must use one of the master formulas. Since it's an irreversible process occurring in an isolated system (if the system boundary includes the source of irreversibility), the Second Law demands that $Δ S > 0$ .

3. Worked Examples

Watching how these formulas are applied in varied contexts is the best way to solidify your understanding. The following video demonstrates several examples that cover different scenarios, including a clever isochoric problem and a crucial irreversible adiabatic case.

Entropy Change of Ideal Gases | Thermodynamics | (Solved Examples)

This video from Question Solutions provides excellent, clear examples of calculating entropy change. We will focus on two specific problems that highlight important problem-solving techniques.

Please watch the following two examples: Paddle Wheel in a Tank (08:22 - 10:06): This is an isochoric process. Pay close attention to how the ideal gas law is used to find the ratio T_2/T_1 from the given pressures, as the temperatures themselves are not known. Adiabatic Piston Expansion (10:06 - 12:12): This is the most important example. The process is adiabatic (Q=0), but the work done is specified, allowing you to find the final temperature. Notice that the calculated \Delta S is positive, confirming the process is irreversible.

The second example in the video is a classic advanced problem type. It forces you to distinguish between an "adiabatic" process ( $Q = 0$ ) and an "isentropic" process ( $Δ S = 0$ ). All reversible adiabatic processes are isentropic, but not all adiabatic processes are reversible.

4. Application to a JEE Advanced Style Problem

Now let's apply these tools to a multi-step problem typical of the JEE Advanced exam. Such problems require you to calculate thermodynamic quantities for each leg of a cycle.

The resource below presents a problem with a three-stage cycle: isothermal expansion, adiabatic expansion, and isobaric compression. It then calculates the entropy change for each stage.

Advanced Thermodynamics Problem - SATHEE - IIT Kanpur

This resource from IIT Kanpur's SATHEE platform breaks down a complex thermodynamic cycle. It's a perfect example of how the formulas we've derived are used in a practical, exam-style context.

Read through 'Solution 1: Step-by-Step Process Analysis', focusing on the entropy change calculations for each of the three stages: Stage 1 (Isothermal): Note the use of \Delta S_1 = Q_1/T_1. Stage 2 (Adiabatic): Note that \Delta S_2 = 0 because the process is defined as (reversibly) adiabatic. Stage 3 (Isobaric): See the application of \Delta S_3 = nC_P\ln(T_A/T_C). Also, observe the 'Error Analysis' section. The fact that (\Delta S_{total} eq 0) is a crucial clue that the problem statement describes a process that cannot form a closed cycle. This type of critical analysis is a hallmark of advanced problem-solving.

This example highlights a critical point: Since entropy is a state function, the net change in entropy for any complete cycle must be zero ( $\oint d S = 0$ ). If your calculations for a cycle yield $Δ S_{cy c l e} \neq = 0$ , it signals an inconsistency or irreversibility that is not accounted for, or that the process as described is impossible.

Conclusion

Today, we translated the abstract definition of entropy into concrete, powerful formulas for ideal gases. This is a significant step up in your ability to analyze thermodynamic systems quantitatively.

Key Takeaways:

There are two primary formulas for calculating the entropy change of an ideal gas, one using (T, V) and the other using (T, P).
- $Δ S = n C_{v} ln (T_{2} / T_{1}) + n R ln (V_{2} / V_{1})$
- $Δ S = n C_{p} ln (T_{2} / T_{1}) - n R ln (P_{2} / P_{1})$
These formulas can be simplified for specific processes: isochoric, isobaric, and isothermal.
A reversible adiabatic process is isentropic ( $Δ S = 0$ ).
An irreversible adiabatic process is not isentropic ( $Δ S > 0$ ). You must use the master formulas with the initial and final state variables to find the entropy change.

Preview of the Next Lesson:
We've covered entropy changes where temperature varies. But what happens during a phase change, like ice melting into water at 0°C? In our next lesson, we will focus on calculating entropy changes during phase transitions at constant temperature, using the concepts of latent heat of fusion and vaporization.

Recommended resources

1. Deriving the Master Equations for Entropy Change

2. Entropy Change in Specific Processes

3. Worked Examples

4. Application to a JEE Advanced Style Problem

Conclusion