C_V and C_p denote the molar specific heat capacities of a gas at constant volume and constant press

Q: Why is $C_p - C_V$ always $R$?

This is known as Mayer's relation. The difference between the heat capacity at constant pressure and constant volume is the work done by the gas in expanding against constant pressure, which for one mole of an ideal gas is exactly $R$.

Q: Does the degree of freedom change with temperature?

Yes! For a diatomic gas, vibrational degrees of freedom can become active at very high temperatures, increasing $f$ from 5 to 7. However, unless specified, we assume standard room temperature where $f=5$.

Q: Why is $\gamma$ smaller for diatomic gases?

$\gamma = 1 + \frac{2}{f}$. As the number of degrees of freedom $f$ increases (from 3 for monoatomic to 5 for diatomic), the fraction $\frac{2}{f}$ decreases, making $\gamma$ smaller.

Q: What if the gas was polyatomic?

For a non-linear polyatomic gas, $f=6$. You would use $C_V = 3R$ and $C_p = 4R$ and follow the exact same steps to compare!

Q: Are these relations valid for real gases?

These specific heat relations are derived strictly for ideal gases. Real gases have intermolecular forces and finite molecular volumes, which introduce corrections to these values.

Question

$C_V$ and $C_p$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

Options:

(A) $C_p - C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

(B) $C_p + C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

(C) $\frac{C_p}{C_V}$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

(D) $C_p \cdot C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

Sigma Percentile · Accepted Answer

The behavior of gases under different thermal conditions is one of the most fascinating areas of thermodynamics. In this problem, we are tasked with comparing various mathematical combinations of the molar specific heat capacities—$C_p$ and $C_V$—for monoatomic and diatomic ideal gases.

To solve this, we must first recall the fundamental values of these specific heats, which are deeply rooted in the degrees of freedom of the gas molecules.

### The Specific Heats of Ideal Gases

The molar specific heat at constant volume, $C_V$, is directly related to the degrees of freedom $f$ of the gas molecules by the relation $C_V = \frac{f}{2}R$. Using Mayer's relation, $C_p - C_V = R$, we can easily find the molar specific heat at constant pressure as $C_p = \left(\frac{f}{2} + 1ight)R$.

For a **monoatomic gas** (like Helium or Argon), the molecules only have translational kinetic energy, giving them $f = 3$ degrees of freedom. 
Thus, we have:
$$C_V = \frac{3}{2}R \quad 	ext{and} \quad C_p = \frac{5}{2}R$$

For a **diatomic gas** (like Oxygen or Nitrogen) at room temperature, the molecules have both translational and rotational kinetic energy, giving them $f = 5$ degrees of freedom. 
Thus, we have:
$$C_V = \frac{5}{2}R \quad 	ext{and} \quad C_p = \frac{7}{2}R$$

With these values in our toolkit, we can systematically evaluate each option.

### Analyzing the Difference and Sum

Let's start by looking at the difference, $C_p - C_V$. According to Mayer's relation, this difference represents the work done by the gas when it expands at constant pressure. For any ideal gas, regardless of its atomicity, this difference is always equal to the universal gas constant $R$. 
$$C_p - C_V = R$$
Therefore, the difference is exactly the same for both monoatomic and diatomic gases. Option (a) is incorrect.

Next, we evaluate the sum, $C_p + C_V$. 
For a monoatomic gas:
$$C_p + C_V = \frac{5}{2}R + \frac{3}{2}R = 4R$$
For a diatomic gas:
$$C_p + C_V = \frac{7}{2}R + \frac{5}{2}R = 6R$$
Since $6R > 4R$, the sum is clearly larger for a diatomic gas. This makes perfect physical sense because a diatomic gas has more modes of storing internal energy. Option (b) is correct!

### The Ratio and the Product

Now, let's examine the ratio $\frac{C_p}{C_V}$, commonly denoted by the adiabatic index $\gamma$.
For a monoatomic gas:
$$\gamma = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67$$
For a diatomic gas:
$$\gamma = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$$
Since $1.4 < 1.67$, the ratio is actually smaller for a diatomic gas. Option (c) is incorrect.

Finally, we calculate the product $C_p \cdot C_V$.
For a monoatomic gas:
$$C_p \cdot C_V = \left(\frac{5}{2}Right) \left(\frac{3}{2}Right) = \frac{15}{4}R^2 = 3.75 R^2$$
For a diatomic gas:
$$C_p \cdot C_V = \left(\frac{7}{2}Right) \left(\frac{5}{2}Right) = \frac{35}{4}R^2 = 8.75 R^2$$
Since $8.75 R^2 > 3.75 R^2$, the product is significantly larger for a diatomic gas. Option (d) is correct.

### Final Conclusion

By systematically applying the kinetic theory of gases, we have rigorously tested each mathematical combination. We found that while the difference $C_p - C_V$ remains constant and the ratio $\frac{C_p}{C_V}$ decreases with atomicity, both the sum and the product of the specific heats are larger for a diatomic gas.

**Correct Options: (b) and (d)**

Animated Solution for Physics - Thermodynamics: $C_{V}$ and $C_{p}$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

Select Answer:

Visualized Solution

Specific Heats of Ideal Gases

Evaluating $C_{p} - C_{V}$

Evaluating $C_{p} + C_{V}$

Evaluating $\frac{C _{p}}{C _{V}}$

Evaluating $C_{p} \cdot C_{V}$

Conclusion

The Sigma Insight: Kinetic Theory of Gases

The Specific Heats of Ideal Gases

Analyzing the Difference and Sum

The Ratio and the Product

Final Conclusion

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