An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains

Q: Why do we use the formula $C = C_V + \frac{R}{1-n}$?

This formula is derived from the First Law of Thermodynamics ($dQ = dU + dW$) specifically for a process following $pV^n = \text{constant}$. It combines the internal energy change (related to $C_V$) and the work done (related to $\frac{R}{1-n}$).

Q: Why did we need to use Mayer's relation?

If you look closely at the options provided in the question, none of them contain the universal gas constant $R$. They are all expressed in terms of $C$, $C_p$, and $C_V$. Mayer's relation ($R = C_p - C_V$) is the perfect bridge to eliminate $R$.

Q: Can $n$ be any value?

Yes! The beauty of the polytropic process is that $n$ can range from $-\infty$ to $+\infty$. Different values of $n$ correspond to different standard processes (e.g., $n=0$ is isobaric, $n=1$ is isothermal, $n=\gamma$ is adiabatic).

Q: What happens if $n=1$ in the heat capacity formula?

If $n=1$, the process is isothermal. The denominator $1-n$ becomes zero, making the molar heat capacity $C$ approach infinity. This makes physical sense because in an isothermal process, you can add heat without changing the temperature ($dT=0$), so $C = \frac{dQ}{ndT} \to \infty$.

Q: I got stuck at the algebraic simplification. How did the $C_V$ terms cancel out?

When you take the common denominator for $1 - \frac{C_p - C_V}{C - C_V}$, you rewrite $1$ as $\frac{C - C_V}{C - C_V}$. Subtracting the numerator gives $(C - C_V) - (C_p - C_V)$. Distributing the negative sign makes it $C - C_V - C_p + C_V$. The $-C_V$ and $+C_V$ perfectly cancel each other out!

Question

An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity $C$ remains constant. If during this process the relation of pressure $p$ and volume $V$ is given by $pV^n = 	ext{constant}$, then $n$ is given by (Here $C_p$ and $C_V$ are molar specific heat at constant pressure and constant volume, respectively)

Options:

(A) $n = \frac{C_p}{C_V}$

(B) $n = \frac{C - C_p}{C - C_V}$

(C) $n = \frac{C_p - C}{C - C_V}$

(D) $n = \frac{C - C_V}{C - C_p}$

Sigma Percentile · Accepted Answer

### The Chameleon of Thermodynamics: Unraveling the Polytropic Process

Imagine a thermodynamic process that can shape-shift. It isn't strictly isothermal, nor is it purely adiabatic. It is a generalized process that can mimic any of the standard thermodynamic paths simply by tweaking a single parameter. This is the **polytropic process**, mathematically defined by the elegant equation:

$$pV^n = 	ext{constant}$$

In this equation, $p$ is pressure, $V$ is volume, and $n$ is the polytropic index. The problem asks us to find an expression for this index $n$ in terms of the molar heat capacities $C$, $C_p$, and $C_V$.

### The Master Equation for Heat Capacity

To find $n$, we must first understand how the gas absorbs heat during this process. From the First Law of Thermodynamics ($dQ = dU + dW$), we can derive the molar heat capacity $C$ for a polytropic process. The internal energy change is always related to $C_V$, and the work done depends on the index $n$. Combining these gives us the master equation:

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

This equation is our starting point. Our mission is to isolate $n$.

### The Algebraic Dance

Let's start rearranging the furniture. First, we move $C_V$ to the left side of the equation to isolate the term containing $n$:

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

Next, we perform a quick cross-multiplication to bring $1 - n$ out of the denominator:

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

Now, we face a slight hurdle. If you look at the options provided in the question, none of them contain the universal gas constant $R$. They are entirely composed of $C$, $C_p$, and $C_V$. We need a bridge to eliminate $R$.

### The Magic of Mayer's Relation

Enter **Mayer's Relation**, a fundamental identity for ideal gases that connects the gas constant to the specific heats:

$$R = C_p - C_V$$

By substituting this identity into our rearranged equation, we successfully banish $R$:

$$1 - n = \frac{C_p - C_V}{C - C_V}$$

We are almost at the finish line. Let's solve for $n$ by moving it to one side and bringing the fraction to the other:

$$n = 1 - \frac{C_p - C_V}{C - C_V}$$

To clean this up, we take a common denominator of $(C - C_V)$:

$$n = \frac{(C - C_V) - (C_p - C_V)}{C - C_V}$$

Distributing the negative sign in the numerator yields:

$$n = \frac{C - C_V - C_p + C_V}{C - C_V}$$

Notice how the $-C_V$ and $+C_V$ terms perfectly cancel each other out. It's a beautiful moment of algebraic clarity. We are left with our final, elegant expression:

$$n = \frac{C - C_p}{C - C_V}$$

This perfectly matches option (b).

### The Way Forward

Take a moment to appreciate the power of this result. By plugging in different values for $C$, you can recover all the standard processes. If the process is isobaric, $C = C_p$, which makes $n = 0$. If the process is adiabatic, $C = 0$, which makes $n = \frac{-C_p}{-C_V} = \gamma$. The polytropic index $n$ is truly the master key to thermodynamic processes!

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Visualized Solution

$p V^{n} = constant$

$Molar Heat Capacity$

$Rearranging for n$

$Isolating 1 - n$

$Using Mayer’s Relation$

$Solving for n$

$Final Expression$

$Special Cases$

The Sigma Insight: Thermodynamic Processes

The Chameleon of Thermodynamics

The Master Equation for Heat Capacity

The Algebraic Dance

The Magic of Mayer's Relation

The Way Forward

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pVn=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molar Heat Capacity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating 1−n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using Mayer’s Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Special Cases