For a gas Cp - CV = R in a state P and Cp - CV = 1.10 R in a state Q. TP and TQ are the temperatures

Q: Why does $C_p - C_V$ deviate from $R$ for real gases?

In real gases, intermolecular forces are not negligible. When a real gas expands at constant pressure, extra work must be done against these intermolecular attractive forces, which increases the required heat capacity $C_p$, making the difference $C_p - C_V$ greater than $R$.

Q: Can $C_p - C_V$ be less than $R$?

Generally, for real gases at moderate conditions, the attractive forces dominate, making $C_p - C_V > R$. However, at extremely high pressures where repulsive forces dominate, the behavior can be more complex, but standard real gas models (like Van der Waals) typically show $C_p - C_V > R$.

Q: Why do high temperatures make a gas ideal?

At high temperatures, the kinetic energy of the gas molecules is very large compared to the potential energy of intermolecular forces. Thus, the molecules easily overcome these attractions, behaving almost as if the forces don't exist, which is the core assumption of an ideal gas.

Q: Does pressure also affect ideality?

Absolutely! At low pressures, the molecules are far apart, meaning the volume occupied by the molecules themselves is negligible compared to the total volume of the container. This satisfies the second core assumption of the kinetic theory of ideal gases.

Q: Is Mayer's relation valid for all ideal gases regardless of atomicity?

Yes! Whether the gas is monatomic, diatomic, or polyatomic, as long as it behaves ideally, $C_p - C_V = R$ holds true. The individual values of $C_p$ and $C_V$ change with atomicity, but their difference remains exactly $R$.

Question

For a gas $C_p - C_V = R$ in a state $P$ and $C_p - C_V = 1.10 R$ in a state $Q$. $T_P$ and $T_Q$ are the temperatures in two different states $P$ and $Q$, respectively. Then,

Options:

(A) $T_P = T_Q$

(B) $T_P < T_Q$

(C) $T_P = 0.9 T_Q$

(D) $T_P > T_Q$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a gas trapped in a container, and you are measuring its specific heat capacities under two entirely different thermodynamic states, which we will call State $P$ and State $Q$.

In State $P$, you carefully measure the molar specific heat at constant pressure ($C_p$) and the molar specific heat at constant volume ($C_V$). When you subtract them, you find that $C_p - C_V = R$, where $R$ is the universal gas constant.

Then, you change the conditions of the gas to reach State $Q$. You perform the exact same measurements, but this time, you find that $C_p - C_V = 1.10 R$. The difference has increased! The question asks us to deduce the relationship between the temperatures of these two states, $T_P$ and $T_Q$.

### The Master Equation: Mayer's Relation

To unlock this problem, we must rely on one of the most fundamental pillars of thermodynamics: **Mayer's Relation**.

For any perfectly ideal gas, the relationship between its specific heats is rigidly defined by the equation:

$$C_p - C_V = R$$

This elegant equation stems from the fact that when an ideal gas expands at constant pressure, it must do work against the external environment, requiring extra heat energy ($R$) compared to heating it at constant volume where no work is done. Crucially, this derivation assumes that there are absolutely **no intermolecular forces** of attraction or repulsion between the gas molecules.

### Identifying the States

Let's apply Mayer's relation to our given states.

For State $P$, we are explicitly given that $C_p - C_V = R$. This is a perfect match! Therefore, we can confidently conclude that in State $P$, the gas is behaving exactly like an **ideal gas**.

However, for State $Q$, we are given that $C_p - C_V = 1.10 R$. Because $1.10 R 
eq R$, the gas in State $Q$ is violating Mayer's relation. This deviation tells us that the assumptions of the ideal gas law have broken down. The intermolecular forces are no longer negligible. Thus, in State $Q$, the gas is behaving as a **real gas**.

### The Physics of Ideality

Now we arrive at the conceptual core of the problem. Under what physical conditions does a real gas shed its complex interactions and start behaving like a simple, ideal gas?

An ideal gas is characterized by molecules that move rapidly and are spaced far apart, ensuring that they rarely interact. This state is achieved under two specific conditions:

1. **High Temperature:** At high temperatures, the kinetic energy of the molecules is so massive that the relatively weak intermolecular potential energy becomes completely negligible. They zip past each other too fast to 'feel' any attraction.
2. **Low Pressure:** At low pressures, the volume of the container is huge compared to the volume occupied by the gas molecules themselves, satisfying another core assumption of the kinetic theory.

### Final Conclusion

We have established that State $P$ represents the ideal behavior, while State $Q$ represents the real, non-ideal behavior.

Because ideal behavior manifests at higher temperatures, the temperature of the gas when it is in State $P$ must be significantly higher than its temperature when it is in State $Q$.

Therefore, we can definitively state that:

$$T_P > T_Q$$

This beautifully demonstrates how abstract thermodynamic relations like $C_p - C_V$ can give us profound insights into the physical state and temperature of a gas!

Animated Solution for Physics - Thermodynamics: For a gas $C_{p} - C_{V} = R$ in a state $P$ and $C_{p} - C_{V} = 1.10 R$ in a state $Q$ . $T_{P}$ and $T_{Q}$ are the temperatures in two different states $P$ and $Q$ , respectively. Then,

Select Answer:

Visualized Solution

Given Conditions

Mayer's Relation

Identifying Ideal State

Identifying Real State

Conditions for Ideal Behavior

Comparing Temperatures

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup

The Master Equation

Identifying the States

The Physics of Ideality

Final Conclusion

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