During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its ab

Q: Why do we use the $p-T$ relation instead of the more common $p-V$ relation?

The question explicitly provides a relationship between pressure ($p$) and temperature ($T$). Using the $p-T$ relation allows us to directly compare exponents without needing to substitute for volume.

Q: How do we derive $T^\gamma p^{1-\gamma} = \text{constant}$?

Start with the classic $pV^\gamma = \text{constant}$. From the ideal gas law, $V = nRT/p$. Substitute this into the classic equation: $p(nRT/p)^\gamma = \text{constant}$. Grouping the variables gives $T^\gamma p^{1-\gamma} = \text{constant}$.

Q: What exactly is $\gamma$?

$\gamma$ (gamma) is the adiabatic index, defined as the ratio of the molar heat capacity at constant pressure ($C_p$) to the molar heat capacity at constant volume ($C_V$).

Q: Can $\gamma$ ever be less than 1?

No. For any ideal gas, $C_p$ is always greater than $C_V$ (specifically, $C_p = C_V + R$). Therefore, the ratio $C_p/C_V$ must always be strictly greater than 1.

Q: What type of gas has $\gamma = 3/2$?

This is a theoretical value for the sake of the problem. In reality, monoatomic gases have $\gamma = 5/3 \approx 1.67$, diatomic gases have $\gamma = 7/5 = 1.4$, and polyatomic gases have $\gamma = 4/3 \approx 1.33$.

Question

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_p / C_V$ for the gas is

Options:

(A) $\frac{4}{3}$

(B) $2$

(C) $\frac{5}{3}$

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

Imagine a gas expanding so rapidly that it doesn't have time to exchange heat with its surroundings. This is the essence of an adiabatic process! In this problem, we are given a fascinating relationship: the pressure of the gas is proportional to the cube of its absolute temperature. Our mission? To uncover the ratio of its specific heats, $C_p/C_V$, also known as the adiabatic index, $\gamma$.

### Analyzing the Setup

The problem hands us a direct mathematical clue:

$$p \propto T^3$$

This tells us exactly how pressure and temperature dance together during this specific adiabatic expansion. But to make sense of this, we need to bring in the heavy artillery—the standard thermodynamic equations for an adiabatic process.

### The Master Equation

You might be familiar with the classic adiabatic relation $pV^\gamma = 	ext{constant}$. However, since our given condition involves pressure ($p$) and temperature ($T$), we need the $p-T$ relation. By substituting $V = nRT/p$ from the ideal gas law into the classic equation, we get:

$$T^\gamma p^{1-\gamma} = 	ext{constant}$$

This is our master equation. It holds true for any ideal gas undergoing a reversible adiabatic process.

### Reshaping the Math

To compare our master equation with the given condition, we need to isolate $p$. Let's rearrange the terms:

$$p^{1-\gamma} \propto T^{-\gamma}$$

Now, we raise both sides to the power of $\frac{1}{1-\gamma}$:

$$p \propto T^{\frac{-\gamma}{1-\gamma}}$$

Multiplying the numerator and denominator of the exponent by $-1$ gives us a much cleaner look:

$$p \propto T^{\frac{\gamma}{\gamma-1}}$$

### The Final Calculation

Now comes the elegant part. We have two expressions for pressure in terms of temperature:

1. $p \propto T^3$ (Given)
2. $p \propto T^{\frac{\gamma}{\gamma-1}}$ (Derived)

Since both describe the exact same physical process, their exponents must be identical. Let's equate them:

$$\frac{\gamma}{\gamma-1} = 3$$

Now, it's just simple algebra. Cross-multiplying yields:

$$\gamma = 3(\gamma - 1)$$

$$\gamma = 3\gamma - 3$$

$$2\gamma = 3$$

$$\gamma = \frac{3}{2}$$

By definition, the adiabatic index $\gamma$ is exactly the ratio of the molar heat capacity at constant pressure to the molar heat capacity at constant volume:

$$\frac{C_p}{C_V} = \gamma = \frac{3}{2}$$

And there we have it! By simply matching the theoretical adiabatic equation with the given physical condition, we've successfully found the ratio of specific heats.

Animated Solution for Physics - Thermodynamics: During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_{p} / C_{V}$ for the gas is

Select Answer:

Visualized Solution

Given Condition

Adiabatic Relation

Rearranging the Equation

Equating Exponents

Solving for $γ$

Final Answer

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Reshaping the Math

The Final Calculation

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