Molar Heat Capacity in Polytropic Processes

1. Molar Heat Capacity is Path-Dependent

As a quick recap, you know that heat ($Q$) is a path function. The amount of heat required to change the temperature of a gas depends on the process. We've focused on two specific molar heat capacities:

• $C_V$ (at constant volume)
• $C_P$ (at constant pressure)

However, a gas can follow infinitely many paths, and each has its own associated heat capacity. For a general polytropic process, we define its molar heat capacity, let's call it $C_{poly}$ or simply $C$, as:
$C=\frac{1}{n_m}\frac{dQ}{dT}$
This means that if we find a constant $C$ for a given polytropic process, we can calculate the total heat transferred as $Q=n_{m}C\Delta T$.

2. Deriving the Molar Heat Capacity for a Polytropic Process

Our main task is to find an expression for $C$ in terms of the polytropic index $n$. We'll start from the First Law of Thermodynamics. The following video provides a clear, step-by-step derivation.

A crucial point to note: The video uses the standard chemistry sign convention, where $\Delta U=Q+W$, and work done on the system is positive ($dW=-PdV$). This is a great opportunity to practice the sign convention conversion you asked about.

Watch the derivation, paying close attention to the application of the First Law and the ideal gas law. After you watch, I will walk you through the same derivation using the physics convention ($\Delta U=Q-W$) to show that we arrive at the exact same result.

Derivation using the Physics/Engineering Convention

Now, let's derive the formula ourselves using the convention $\Delta U=Q-W$, where work done by the system is positive ($dW=PdV$). Your CS and math background means you'll appreciate the logical flow here.

For a differential change and one mole of gas ($n_m=1$), the first law is:
$dQ=dU+dW$
We know that for an ideal gas, $dU=C_{V}dT$ and for a reversible process, $dW=PdV$. Substituting these in:
$dQ=C_{V}dT+PdV$
By definition, the molar heat capacity for this process is $C=\frac{dQ}{dT}$. Dividing the entire equation by $dT$:
$C=\frac{dQ}{dT}=C_V+P\frac{dV}{dT}$
This is a general expression for any process. To make it specific for a polytropic process, we need to find the term $\frac{dV}{dT}$. We use the polytropic relation between T and V which we derived last time: $T{V}^{n-1}=\text{constant}$ ($K$).

Now, we differentiate this implicitly with respect to $T$, using the product rule:
$\frac{d}{dT}(T{V}^{n-1})=\frac{d}{dT}(K)$
$(1\cdot V^{n-1})+(T\cdot (n-1)V^{n-2}\frac{dV}{dT})=0$
Solving for $\frac{dV}{dT}$:
$T(n-1)V^{n-2}\frac{dV}{dT}=-V^{n-1}$
$\frac{dV}{dT}=-\frac{V^{n-1}}{T(n-1)V^{n-2}}=-\frac{V}{T(n-1)}=\frac{V}{T(1-n)}$
Now, substitute this back into our expression for $C$:
$C=C_V+P\left(\frac{V}{T(1-n)}\right)=C_V+\frac{PV}{T}\frac{1}{1-n}$
For one mole of an ideal gas, $PV=RT$, so $\frac{PV}{T}=R$. This simplifies our expression to the final formula:

$C=C_V+\frac{R}{1-n}$

As you can see, the final formula is identical, regardless of the initial sign convention. This is a key insight: as long as you are consistent, the physics remains the same. The choice of convention just flips the sign of the work term intermediate step.

3. Analyzing the Polytropic Heat Capacity Formula

The formula $C=C_V+\frac{R}{1-n}$ is incredibly powerful. Let's test it against the processes we know:

• Isochoric (Constant Volume): Here, $n\to \infty$. The term $\frac{R}{1-n}$ goes to zero.
  $C=C_V$. This is correct.
• Isobaric (Constant Pressure): Here, $n=0$.
  $C=C_V+\frac{R}{1-0}=C_V+R$. By Mayer's relation, this is exactly $C_P$. Correct again.
• Adiabatic Process: Here, $n=\gamma$.
  $C=C_V+\frac{R}{1-\gamma }$. We know $R=C_P-C_V$ and $\gamma =C_P/C_V$. So $R=C_V(\gamma -1)$.
  $C=C_V+\frac{C_V(\gamma -1)}{1-\gamma }=C_V-C_V=0$
  This is also correct, as heat transfer is zero in an adiabatic process by definition.
• Isothermal Process: Here, $n=1$. The formula denominator becomes zero, so $C\to \infty$.
  What does this mean physically? An infinite heat capacity means you can transfer a finite amount of heat ($Q$) with an infinitesimally small (or zero) change in temperature ($\Delta T$). This is precisely the definition of an isothermal process.

The Curious Case of Negative Heat Capacity

What happens if $C$ is negative? This occurs when $C_V+\frac{R}{1-n}<0$. Let's analyze this inequality. As we established in our derivation walkthrough, this happens when $1<n<\gamma$.

Physically, a negative heat capacity means that to increase the gas's temperature ($\Delta T>0$), you must remove heat ($Q<0$), and to decrease its temperature ($\Delta T<0$), you must add heat ($Q>0$).

Consider a process between isothermal and adiabatic ($1<n<\gamma$). During an expansion, the gas does work and its temperature falls ($\Delta T<0$). However, the process curve is less steep than an adiabat. This means the gas cools down less than it would in an adiabatic expansion. To prevent it from cooling down "too much," we must have supplied some heat to it ($Q>0$). Since we have $Q>0$ and $\Delta T<0$, the ratio $C=Q/(n_m\Delta T)$ is negative. This is a non-intuitive but logical result, and exactly the kind of conceptual check prized in JEE Advanced problems.

4. Problem-Solving Practice

Now let's apply this formula. The following problems are excellent practice for the types of questions you'll encounter. Try to solve them yourself first, then you can check your method against the solution in the video.

Problem 1: Direct Application

Calculate the molar heat capacity of a monoatomic gas undergoing a process where the ratio of pressure to volume is constant, i.e., $P/V=k$.

Problem 2: Multi-step Analysis

One mole of a monoatomic ideal gas undergoes a process defined by $T^2=kV$, where $k$ is a constant. Calculate the molar heat capacity of the gas for this process.

This is a more advanced problem because the process is not given in the standard $P{V}^n$ form. Your first task is to convert it.

Conclusion

Today we derived and applied one of the most versatile formulas in this topic. You are now equipped to handle a wide variety of thermodynamic processes.

Key Takeaways:

• The molar heat capacity for a polytropic process $P{V}^n=C$ is given by the formula:
  $C=C_V+\frac{R}{1-n}$
• This single formula correctly reproduces the heat capacities for isochoric ($C_V$), isobaric ($C_P$), and adiabatic ($0$) processes, and explains the infinite capacity of an isothermal process.
• A negative heat capacity is physically possible and occurs for processes between isothermal and adiabatic ($1<n<\gamma$).
• For complex JEE problems, the first step is often to manipulate the given process equation (e.g., a relation in T and V) into the standard $P{V}^n=C$ form to identify the index $n$.

Preview of the Next Lesson:
We have now mastered the calculation of $\Delta U=n_m{C}_V\Delta T$, $W=\frac{n_{m}R\Delta T}{1-n}$, and $Q=n_{m}C\Delta T$ for any single polytropic process. The next logical step is to combine these processes. In the next lesson, we will calculate the total work, heat, and internal energy change for a multi-step sequence of thermodynamic processes, including full thermodynamic cycles.