To raise the temperature of a certain mass of gas by 50°C at a constant pressure, 160 cal of heat is

Q: Why do we use the magnitude of temperature change for the cooling process?

The problem states that 240 cal of heat is *released* when the gas is *cooled* by $100^\circ\text{C}$. Both $\Delta Q$ and $\Delta T$ are negative. Using their magnitudes directly in the equation $\Delta Q = nC_v\Delta T$ avoids sign errors, as the negative signs cancel out.

Q: Why is $C_p$ always greater than $C_v$?

At constant volume, all the heat supplied goes into increasing the internal energy (temperature) of the gas. At constant pressure, the gas expands and does work on the surroundings. Therefore, extra heat is required to do this work in addition to raising the internal energy, making $C_p > C_v$.

Q: What does a degree of freedom of 6 imply about the gas?

A degree of freedom of 6 typically corresponds to a non-linear polyatomic gas (like $NH_3$ or $CH_4$), which has 3 translational and 3 rotational degrees of freedom at normal temperatures.

Q: Can we solve this without knowing the number of moles?

Yes! Because the problem states it's the *same mass* of gas in both processes, the number of moles $n$ is constant. By taking the ratio of the two heat equations, $n$ perfectly cancels out.

Q: What if the gas was not ideal?

If the gas were real, the relation $C_p - C_v = R$ and $\gamma = 1 + 2/f$ would not hold perfectly, as intermolecular forces would contribute to the internal energy, making the specific heats dependent on temperature and pressure.

Question

To raise the temperature of a certain mass of gas by $50^{\circ}	ext{C}$ at a constant pressure, $160	ext{ cal}$ of heat is required. When the same mass of gas is cooled by $100^{\circ}	ext{C}$ at constant volume, $240	ext{ cal}$ of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

Options:

(A) 5

(B) 7

(C) 6

(D) 3

Sigma Percentile · Accepted Answer

### The Two Thermodynamic Journeys

Imagine you are a scientist in a lab, and you have a mysterious gas trapped inside a cylinder. You want to understand its microscopic nature—specifically, how many ways its molecules can move, rotate, or vibrate. To find out, you subject the gas to two distinct thermodynamic trials.

In the first trial, you allow the gas to expand while keeping its pressure constant. You carefully measure that it takes exactly $160	ext{ cal}$ of heat to raise its temperature by $50^\circ	ext{C}$.

In the second trial, you lock the piston in place, keeping the volume strictly constant. This time, you cool the gas down by $100^\circ	ext{C}$, and you observe that it releases exactly $240	ext{ cal}$ of heat.

These two macroscopic observations hold the key to unlocking the microscopic secrets of the gas!

### The Master Equations

To connect the heat exchanged with the temperature change, we rely on the fundamental equations of thermodynamics.

For a process at constant pressure, the heat exchanged is given by:

$$ \Delta Q_p = n C_p \Delta T_p $$

Here, $n$ is the number of moles, and $C_p$ is the molar specific heat at constant pressure.

Similarly, for a process at constant volume, the heat exchanged is:

$$ \Delta Q_v = n C_v \Delta T_v $$

Where $C_v$ is the molar specific heat at constant volume.

### Unlocking the Adiabatic Index

Now, let's substitute the values from our two trials into these master equations.

For the isobaric heating:

$$ 160 = n C_p (50) $$

For the isochoric cooling, we use the magnitudes of the heat released and the temperature drop to avoid messy negative signs:

$$ 240 = n C_v (100) $$

We have two equations, but we don't know the number of moles $n$. However, since we are using the **same mass** of gas in both trials, $n$ is a constant. We can elegantly eliminate it by dividing the first equation by the second:

$$ \frac{160}{240} = \frac{n C_p (50)}{n C_v (100)} $$

Notice how beautifully $n$ cancels out! Simplifying the fractions, we get:

$$ \frac{2}{3} = \left(\frac{C_p}{C_v}ight) \cdot \frac{1}{2} $$

The ratio of the specific heats, $\frac{C_p}{C_v}$, is a profoundly important thermodynamic property known as the adiabatic index, denoted by $\gamma$. Rearranging our equation, we find:

$$ \gamma = \frac{4}{3} $$

### The Microscopic Connection

We have successfully calculated the macroscopic property $\gamma$. But how does this tell us about the molecules themselves?

Enter the Kinetic Theory of Gases! This theory provides a beautiful bridge between the macroscopic $\gamma$ and the microscopic degrees of freedom, $f$, through the relation:

$$ \gamma = 1 + \frac{2}{f} $$

Let's substitute our calculated value of $\gamma$ into this relation:

$$ \frac{4}{3} = 1 + \frac{2}{f} $$

Subtracting $1$ from both sides gives:

$$ \frac{1}{3} = \frac{2}{f} $$

By simply cross-multiplying, we arrive at our final destination:

$$ f = 6 $$

**The gas has 6 degrees of freedom!**

This tells us that the gas is likely a non-linear polyatomic molecule. Through two simple macroscopic measurements of heat and temperature, we have successfully deduced the microscopic kinetic behavior of the gas molecules. Physics is truly amazing!

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Visualized Solution

$Thermodynamic Processes$

$Heat Transfer Equations$

$Substituting Values$

$Finding the Ratio γ$

$Calculating γ$

$Degrees of Freedom Relation$

$Solving for f$

$Final Answer$

$The Way Forward$

The Sigma Insight: Kinetic Theory of Gases

The Two Thermodynamic Journeys

The Master Equations

Unlocking the Adiabatic Index

The Microscopic Connection

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Ratio γ

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Degrees of Freedom Relation

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

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

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