The internal energy (U), pressure (p) and volume (V) of an ideal gas are related as U = 3pV + 4. The

Q: Why do we use $U = \frac{f}{2}pV$ instead of $nC_V T$?

Great question! Actually, they are the exact same thing. Since $C_V = \frac{f}{2}R$, we have $U = nC_V T = n(\frac{f}{2}R)T = \frac{f}{2}(nRT)$. By applying the ideal gas law $pV = nRT$, we can substitute $nRT$ with $pV$ to get $U = \frac{f}{2}pV$. This form is much more useful here because the given equation is in terms of $p$ and $V$.

Q: Can the term $\frac{8}{pV}$ ever be negative?

No, it cannot! For any real, physical ideal gas, the absolute pressure $p$ and the volume $V$ must be strictly positive quantities. Therefore, their product $pV$ is positive, making $\frac{8}{pV}$ a positive number.

Q: What are the degrees of freedom for different types of gases?

A monoatomic gas (like Helium) has 3 translational degrees of freedom ($f=3$). A diatomic gas (like Oxygen) has 3 translational and 2 rotational degrees of freedom ($f=5$). A non-linear polyatomic gas has 3 translational and 3 rotational degrees of freedom ($f=6$), and can have even more if vibrational modes are active.

Q: Why does $f > 6$ mean the gas is polyatomic?

Since monoatomic gases have $f=3$ and diatomic gases have $f=5$, any gas with a degree of freedom strictly greater than 5 must be polyatomic. In our case, $f = 6 + \text{something positive}$, which means $f$ is strictly greater than 6. Thus, it can only be a polyatomic gas.

Q: What if the constant in the equation was 0, i.e., $U = 3pV$?

If $U = 3pV$, then equating it to $\frac{f}{2}pV$ gives $\frac{f}{2} = 3$, which means $f = 6$. A gas with exactly 6 degrees of freedom is typically a non-linear polyatomic gas without active vibrational modes.

Question

The internal energy ($U$), pressure ($p$) and volume ($V$) of an ideal gas are related as $U = 3pV + 4$. The gas is

Options:

(A) diatomic only

(B) polyatomic only

(C) Either monoatomic or diatomic

(D) monoatomic only

Sigma Percentile · Accepted Answer

### Decoding the Given Relation

Imagine you are a detective trying to uncover the true identity of a mysterious gas. The only clue you have is a mathematical footprint left behind: the relationship between its internal energy, pressure, and volume.

The problem states that the internal energy $U$ of this ideal gas is given by the equation:
$$U = 3pV + 4$$

This equation is our starting point. But to find out whether the gas is monoatomic, diatomic, or polyatomic, we need to connect this specific clue to the universal laws of thermodynamics.

### The Master Equation of Internal Energy

Let's bring in our heavy machinery. From the kinetic theory of gases, we know that the internal energy of any ideal gas is directly tied to its temperature and its **degree of freedom** ($f$).

The standard formula is:
$$U = \frac{f}{2}nRT$$

But our clue is in terms of pressure ($p$) and volume ($V$), not temperature ($T$). How do we bridge this gap? We use the trusty ideal gas law!

Since $pV = nRT$, we can seamlessly substitute $nRT$ with $pV$ in our internal energy formula:
$$U = \frac{f}{2}pV$$

Now we have two different expressions for the exact same internal energy $U$. One is the specific clue given to us, and the other is the universal formula.

### Unveiling the Degree of Freedom

When two things are equal to the same thing, they must be equal to each other. Let's equate our two expressions for $U$:
$$\frac{f}{2}pV = 3pV + 4$$

Our goal now is to isolate $f$, the degree of freedom, because $f$ is the fingerprint that will reveal the gas's identity.

First, let's multiply the entire equation by 2 to clear the fraction:
$$f \cdot pV = 6pV + 8$$

Next, we divide everything by $pV$ to get $f$ completely by itself:
$$f = 6 + \frac{8}{pV}$$

### The Final Verdict: Atomicity of the Gas

Take a close look at the expression we just derived. The degree of freedom $f$ is equal to 6 plus a fractional term, $\frac{8}{pV}$.

Now, let's think about the physical reality of a gas. Can absolute pressure ($p$) ever be negative? No. Can volume ($V$) ever be negative? Absolutely not.

Since both $p$ and $V$ are strictly positive quantities, their product $pV$ is positive. Therefore, the term $\frac{8}{pV}$ must be a **positive number**.

This leads us to a profound conclusion:
$$f > 6$$

The degree of freedom of our mystery gas is strictly greater than 6.

Let's check our suspect list:
- A **monoatomic** gas has $f = 3$.
- A **diatomic** gas has $f = 5$ (at normal temperatures).
- A **polyatomic** gas has $f \ge 6$.

Since our gas has a degree of freedom greater than 6, it cannot be monoatomic or diatomic. It must be a **polyatomic** gas!

The mystery is solved. The mathematical footprint perfectly matches the complex, multi-directional movements of a polyatomic molecule.

Animated Solution for Physics - Thermodynamics: The internal energy ( $U$ ), pressure ( $p$ ) and volume ( $V$ ) of an ideal gas are related as $U = 3 p V + 4$ . The gas is

Select Answer:

Visualized Solution

$U = 3 p V + 4$

$U = \frac{f}{2} p V$

Equating Internal Energies

Solving for $f$

Analyzing $f$

Conclusion

The Way Forward

The Sigma Insight: Kinetic Theory of Gases

Decoding the Given Relation

The Master Equation of Internal Energy

Unveiling the Degree of Freedom

The Final Verdict

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Animated Solution for Physics - Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The internal energy (U), pressure (p) and volume (V) of an ideal gas are related as U=3pV+4. The gas is