Deriving Adiabatic Equations from First Law and Ideal Gas Law

Hello! Let's dive into the next fundamental thermodynamic process.

In our last lesson, we mastered the isothermal process, where temperature is constant ( $Δ T = 0$ ) and any heat exchanged is perfectly converted into work ( $Q = W$ ). We also focused heavily on how to interpret and translate between the physics and chemistry sign conventions, a crucial skill for your JEE Advanced preparation.

Today, we shift our focus to the adiabatic process, which is defined by a different constraint: no heat exchange with the surroundings. Your learning outcome is to derive the adiabatic equations relating Pressure (P), Volume (V), and Temperature (T) from the first law and ideal gas law. This is a classic and fundamental derivation in thermodynamics. Mastering it not only gives you the formulas but also deepens your understanding of how the core laws of physics are combined to describe the behavior of gases—a skill essential for tackling advanced problems.

1. The Adiabatic Condition and the First Law

An adiabatic process is one where there is no heat transfer between the system and its surroundings. This means $Q = 0$ . This can occur in two primary scenarios:

The system is perfectly insulated (e.g., a gas in a thermos flask).
The process happens so rapidly that there is no time for significant heat transfer (e.g., the compression and expansion strokes in an internal combustion engine, or the propagation of a sound wave).

Let's see how this condition impacts the First Law of Thermodynamics (FLOT):

Physics/Engineering Convention: $Δ U = Q - W$
Since $Q = 0$ , this simplifies to: $Δ U = - W$ or $W = - Δ U$
Chemistry (IUPAC) Convention: $Δ U = Q + w$
Since $Q = 0$ , this simplifies to: $Δ U = w$ or $w = Δ U$

Both conventions lead to the same physical conclusion: any work done is at the direct expense of the system's internal energy.

Adiabatic Expansion: The gas does work on the surroundings ( $W > 0$ ). This energy must come from its internal energy, so $Δ U$ is negative. The gas cools down.
Adiabatic Compression: Work is done on the gas ( $W < 0$ ). This energy is added to its internal energy, so $Δ U$ is positive. The gas heats up.

This direct link between work and temperature change is the hallmark of an adiabatic process. The following video provides a concise explanation of this core concept.

Class 11 Physics | Laws of Thermodynamics | #21 Adiabatic Process | For JEE & NEET

To start, watch this brief explanation from Physics Galaxy. It clearly establishes the definition of an adiabatic process and its immediate consequence for the first law of thermodynamics.

Watch the first 3 minutes of the video (00:00 - 02:52). Focus on the relationship dW = -dU and its physical meaning for expansion and compression.

2. Deriving the Process Equation: $P V^{γ} = constant$

Unlike previous processes, the relationship between P, V, and T is not immediately obvious and must be derived. Your strong mathematical background will be a great asset here. The strategy is to combine the differential forms of two fundamental laws: the First Law of Thermodynamics and the Ideal Gas Law.

Let's walk through the derivation step-by-step. We will use the physics convention where work done by the gas is $d W = P d V$ .

Start with the First Law for an adiabatic process:
$d Q = d U + d W = 0$
$d U = - d W$
Substitute the expressions for $d U$ and $d W$ :
We know that for an ideal gas, $d U = n C_{v} d T$ , and the work done is $d W = P d V$ .
$n C_{v} d T = - P d V (Eq. 1)$
Introduce the Ideal Gas Law in differential form:
We have $P V = n R T$ . Differentiating both sides using the product rule gives:
$d (P V) = d (n R T)$
$P d V + V d P = n R d T (Eq. 2)$
Eliminate $d T$ :
We now have two expressions involving $d T$ . We can solve for $d T$ in Eq. 2 and substitute it into Eq. 1.
From Eq. 2: $d T = \frac{P d V + V d P}{n R}$ .
Substituting this into Eq. 1:
$n C_{v} (\frac{P d V + V d P}{n R}) = - P d V$
Simplify using Mayer's Relation ( $R = C_{p} - C_{v}$ ):
Let's clean up the equation. The $n$ cancels out.
$\frac{C _{v}}{R} (P d V + V d P) = - P d V$
Now, substitute $R = C_{p} - C_{v}$ :
$\frac{C _{v}}{C _{p} - C _{v}} (P d V + V d P) = - P d V$
Let's rearrange to group the $d V$ and $d P$ terms.
$C_{v} (P d V + V d P) = - (C_{p} - C_{v}) P d V$
$C_{v} P d V + C_{v} V d P = - C_{p} P d V + C_{v} P d V$
The $C_{v} P d V$ terms cancel on both sides:
$C_{v} V d P = - C_{p} P d V$
Separate variables and integrate:
This is now a separable differential equation. Let's group P terms on one side and V terms on the other.
$\frac{d P}{P} = - \frac{C _{p}}{C _{v}} \frac{d V}{V}$
We define the adiabatic index (or heat capacity ratio) as $γ = \frac{C _{p}}{C _{v}}$ .
$\frac{d P}{P} = - γ \frac{d V}{V}$
Integrating both sides from an initial state $(P_{i}, V_{i})$ to a final state $(P_{f}, V_{f})$ :
$\int_{P_{i}}^{P_{f}} \frac{d P}{P} = - γ \int_{V_{i}}^{V_{f}} \frac{d V}{V}$
$[ln P]_{P_{i}}^{P_{f}} = - γ [ln V]_{V_{i}}^{V_{f}}$
$ln (P_{f}) - ln (P_{i}) = - γ (ln (V_{f}) - ln (V_{i}))$
$ln (\frac{P _{f}}{P _{i}}) = - γ ln (\frac{V _{f}}{V _{i}}) = ln ((\frac{V _{f}}{V _{i}})^{- γ})$
Exponentiating both sides gives:
$\frac{P _{f}}{P _{i}} = (\frac{V _{i}}{V _{f}})^{γ} ⟹ P_{f} V_{f}^{γ} = P_{i} V_{i}^{γ}$
This implies that for any state during the process, the quantity $P V^{γ}$ is constant.

$P V^{γ} = constant$

This is the fundamental process equation for a reversible adiabatic process.

The following video provides an alternative but equivalent path for this derivation. Watching it will reinforce your understanding of the technique.

Class 11 Physics | Laws of Thermodynamics | #21 Adiabatic Process | For JEE & NEET

This video from Physics Galaxy demonstrates the same derivation, but with a slightly different algebraic approach. Observing multiple paths to the same result is a great way to build conceptual flexibility.

Watch the segment from 05:24 to 10:03. Notice how the derivation combines the differential forms of the first law and the ideal gas law to eliminate the temperature differential, dT.

For a third perspective, you can also review the derivation in the Wikipedia article "Adiabatic process". The section "Derivation of P–V relation for adiabatic compression and expansion" presents yet another algebraic route, showcasing the robustness of the physical principles involved.

Adiabatic process - Wikipedia

The Wikipedia article on Adiabatic Processes offers a concise, text-based derivation. It's a useful reference to consolidate the steps we've just covered.

Skim through the subsection titled 'Derivation of P–V relation for adiabatic compression and expansion'. Compare its algebraic steps with the ones we just performed and those in the video. You don't need to study it in depth, but recognize the common starting point (dU + dW = 0) and end point (PV^gamma = constant).

3. Deriving the T-V and T-P Relationships

Once we have the primary relation $P V^{γ} = K$ , deriving the other two is a straightforward algebraic exercise using the ideal gas law, $P V = n R T$ . This is a common technique required in JEE problems.

Temperature-Volume (T-V) Relation

Start with $P V^{γ} = K$ .
From the ideal gas law, substitute $P = \frac{n R T}{V}$ .
$(\frac{n R T}{V}) V^{γ} = K$
$n R T V^{γ - 1} = K$ . Since $n$ and $R$ are constants, we can absorb them into the main constant on the right.
$T V^{γ - 1} = constant$

Temperature-Pressure (T-P) Relation

Start with $P V^{γ} = K$ .
From the ideal gas law, substitute $V = \frac{n R T}{P}$ .
$P (\frac{n R T}{P})^{γ} = K$
$P (n R T)^{γ} P^{- γ} = K ⟹ P^{1 - γ} T^{γ} (n R)^{γ} = K$ . Again, absorb the constants.
$T^{γ} P^{1 - γ} = constant$

This set of three equations governs the state changes in any adiabatic process.

Class 11 Physics | Laws of Thermodynamics | #21 Adiabatic Process | For JEE & NEET

This final clip from the Physics Galaxy video quickly demonstrates these algebraic substitutions to derive the T-V and T-P relations.

Watch the segment from 11:08 to 13:42. Confirm that the steps match the algebraic derivation we just performed.

Test your understanding!

Show that the relation $T^{γ} P^{1 - γ} = constant$ can be rewritten in the form $T P^{\frac{1 - γ}{γ}} = constant’$ . Why might this form be useful?

Show answer

To get from $T^{γ} P^{1 - γ} = K$ to the desired form, we need to make the exponent of $T$ equal to 1. We can do this by taking the $1/ γ$ root of the entire equation:

$(T^{γ} P^{1 - γ})^{\frac{1}{γ}} = K^{\frac{1}{γ}}$
$T^{\frac{γ}{γ}} P^{\frac{1 - γ}{γ}} = K^{\frac{1}{γ}}$
$T^{1} P^{\frac{1 - γ}{γ}} = K^{'}$
Where $K^{'} = K^{1/ γ}$ is just another constant.

This form, $T = K^{'} P^{- \frac{1 - γ}{γ}} = K^{'} P^{\frac{γ - 1}{γ}}$ , is useful because it directly expresses temperature as a function of pressure. If you are given the initial state $(T_{1}, P_{1})$ and the final pressure $P_{2}$ , you can find the final temperature $T_{2}$ in one step using the ratio:
$T_{2} = T_{1} (\frac{P _{2}}{P _{1}})^{\frac{γ - 1}{γ}}$

Conclusion

You have now derived the complete set of process equations for an ideal gas undergoing a reversible adiabatic process. This is a significant milestone in your study of thermodynamics.

Key Takeaways:

An adiabatic process is defined by $Q = 0$ , which simplifies the first law to $Δ U = - W$ (physics convention). Work is done at the expense of internal energy.
The primary equation relating pressure and volume is $P V^{γ} = constant$ , where $γ = C_{p} / C_{v}$ .
This equation is derived by combining the differential forms of the First Law of Thermodynamics and the Ideal Gas Law to eliminate the temperature differential, $d T$ .
The other two relationships, $T V^{γ - 1} = constant$ and $T^{γ} P^{1 - γ} = constant$ , are derived algebraically from $P V^{γ} = constant$ and the ideal gas law.

Preview of the Next Lesson:
With these powerful equations in our toolkit, the next logical step is to use them. In the next lesson, we will apply these P-V-T relationships to calculate the work done during an adiabatic process. We will also tackle a classic JEE Advanced analysis question: comparing the slopes of isothermal and adiabatic curves on a P-V diagram. This will bring together the concepts from the last two lessons.

Recommended resources

1. The Adiabatic Condition and the First Law

2. Deriving the Process Equation: PVγ=constant

3. Deriving the T-V and T-P Relationships

Temperature-Volume (T-V) Relation

Temperature-Pressure (T-P) Relation

Conclusion

2. Deriving the Process Equation: $P V^{γ} = constant$