One mole of ideal monoatomic gas (gamma = 5/3) is mixed with one mole of diatomic gas (gamma = 7/5).

Q: Why can't I just take the arithmetic mean of the two gammas since there is 1 mole of each?

$\gamma$ is an intensive property (a ratio of specific heats $C_p/C_V$), not an extensive property like mass or energy. You cannot simply average intensive properties. You must average the extensive properties (like internal energy or heat capacity) and then calculate the new ratio.

Q: What is the physical meaning of $\gamma$?

$\gamma$ is the ratio of specific heat at constant pressure ($C_p$) to specific heat at constant volume ($C_V$). It is directly related to the degrees of freedom ($f$) of the gas molecules by the relation $\gamma = 1 + \frac{2}{f}$. It dictates how a gas behaves under adiabatic expansion or compression.

Q: Does the temperature of the gases change when they are mixed?

If the two ideal gases are initially at the same temperature and do not chemically react, the temperature of the mixture will remain exactly the same. The kinetic energy is simply redistributed among the molecules.

Q: Can we use molar specific heat at constant volume ($C_V$) to solve this instead?

Absolutely! In fact, the formula we used is derived from $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$. Since $C_V = \frac{R}{\gamma - 1}$, substituting this into the $C_V$ mixture formula gives you the exact same $\gamma$ mixture equation.

Q: What would happen if the gases chemically reacted upon mixing?

If a chemical reaction occurs, chemical potential energy is converted into (or absorbs) thermal kinetic energy. The simple ideal gas mixing formula would no longer apply because the total internal energy would change due to the heat of reaction.

Question

One mole of ideal monoatomic gas $\left(\gamma = \frac{5}{3}\right)$ is mixed with one mole of diatomic gas $\left(\gamma = \frac{7}{5}\right)$. What is $\gamma$ for the mixture? $\gamma$ denotes the ratio of specific heat at constant pressure, to that at constant volume.

Options:

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

(B) $\frac{23}{15}$

(C) $\frac{35}{23}$

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

Sigma Percentile · Accepted Answer

### The Setup: Two Gases, One Container

Imagine you are in a laboratory with two separate, sealed containers. In the first container, you have exactly one mole of an ideal monoatomic gas (like Helium or Neon). Because it's monoatomic, its specific heat ratio, $\gamma_1$, is $\frac{5}{3}$.

In the second container, you have one mole of an ideal diatomic gas (like Oxygen or Nitrogen). Its specific heat ratio, $\gamma_2$, is $\frac{7}{5}$.

Now, you open a valve and let these two gases mix completely into a single, larger container. The question asks: What is the new specific heat ratio, $\gamma_{mix}$, for this combined gaseous mixture?

### The Master Principle: Conservation of Energy

To solve this, we cannot simply take the average of the two $\gamma$ values. $\gamma$ is an intensive property (a ratio), and it doesn't add up linearly. Instead, we must rely on a fundamental law of physics: **The Conservation of Internal Energy**.

Assuming no heat is lost to the surroundings and no work is done during the mixing, the total internal energy of the mixture must equal the sum of the internal energies of the individual gases before they were mixed:

$$U_{mix} = U_1 + U_2$$

We know from the kinetic theory of gases that the internal energy $U$ of an ideal gas can be expressed in terms of its number of moles $n$, the universal gas constant $R$, temperature $T$, and its specific heat ratio $\gamma$:

$$U = \frac{nRT}{\gamma - 1}$$

Substituting this into our energy conservation equation (and assuming the gases were at the same initial temperature $T$, which remains constant upon mixing), the $RT$ terms cancel out beautifully, leaving us with the master formula for gas mixtures:

$$\frac{n_1 + n_2}{\gamma_{mix} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$$

### The Mathematical Execution

Now, let's substitute our known values into this elegant equation. We have $n_1 = 1$, $\gamma_1 = \frac{5}{3}$, $n_2 = 1$, and $\gamma_2 = \frac{7}{5}$. The total number of moles in the mixture is $n_1 + n_2 = 1 + 1 = 2$.

$$\frac{2}{\gamma_{mix} - 1} = \frac{1}{\frac{5}{3} - 1} + \frac{1}{\frac{7}{5} - 1}$$

Let's simplify the denominators on the right side. For the monoatomic gas, $\frac{5}{3} - 1 = \frac{2}{3}$. For the diatomic gas, $\frac{7}{5} - 1 = \frac{2}{5}$.

$$\frac{2}{\gamma_{mix} - 1} = \frac{1}{\frac{2}{3}} + \frac{1}{\frac{2}{5}}$$

Taking the reciprocals of these fractions makes the equation look much friendlier:

$$\frac{2}{\gamma_{mix} - 1} = \frac{3}{2} + \frac{5}{2}$$

Since the denominators are the same, we can easily add the numerators: $3 + 5 = 8$.

$$\frac{2}{\gamma_{mix} - 1} = \frac{8}{2} = 4$$

### The Final Revelation

We are now just one algebraic step away from our answer. Let's rearrange the equation to isolate $\gamma_{mix} - 1$:

$$\gamma_{mix} - 1 = \frac{2}{4}$$

$$\gamma_{mix} - 1 = \frac{1}{2}$$

Finally, adding $1$ to both sides yields the specific heat ratio for our mixture:

$$\gamma_{mix} = 1 + \frac{1}{2} = \frac{3}{2}$$

And there we have it! The specific heat ratio of the mixture is exactly **$\frac{3}{2}$**. This method is incredibly robust and can be used for any combination of non-reacting ideal gases, regardless of their molar amounts or degrees of freedom.

Animated Solution for Physics - Thermodynamics: One mole of ideal monoatomic gas $(γ = \frac{5}{3})$ is mixed with one mole of diatomic gas $(γ = \frac{7}{5})$ . What is $γ$ for the mixture? $γ$ denotes the ratio of specific heat at constant pressure, to that at constant volume.

Select Answer:

Visualized Solution

Visualizing the Mixture

Conservation of Internal Energy

Substituting the Values

Simplifying Denominators

Taking Reciprocals

Adding the Fractions

Rearranging for Gamma

Final Answer

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Master Principle

The Mathematical Execution

The Final Revelation

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