p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should cor

Q: Why is the slope of the adiabatic curve negative?

The negative slope indicates an inverse relationship between pressure and volume during an adiabatic expansion. As the gas expands (volume increases), it does work at the expense of its internal energy, causing the temperature and pressure to drop.

Q: How do we know the $\gamma$ values for different gases?

The value of $\gamma$ depends on the degrees of freedom ($f$) of the gas molecules, given by $\gamma = 1 + \frac{2}{f}$. Monatomic gases (like He, Ar) have 3 translational degrees of freedom, so $\gamma = 1 + \frac{2}{3} = 1.67$. Diatomic gases (like $\text{O}_2$, $\text{N}_2$) have 3 translational and 2 rotational degrees of freedom at room temperature, so $\gamma = 1 + \frac{2}{5} = 1.4$.

Q: What if the graph showed isothermal processes instead?

For an isothermal process, $pV = \text{constant}$. The slope is $\frac{dp}{dV} = -\frac{p}{V}$. Notice that it lacks the $\gamma$ factor. Since $\gamma > 1$ for all gases, an adiabatic curve is always steeper than an isothermal curve passing through the same point.

Q: Why does curve 2 being steeper mean it has a larger $\gamma$?

The slope of the curve is $-\gamma \frac{p}{V}$. At any given point $(V, p)$, the steepness (magnitude of the slope) is exactly $\gamma \frac{p}{V}$. Therefore, a larger $\gamma$ directly results in a larger magnitude of the slope, making the curve visually steeper.

Q: Could curve 1 and 2 represent the same gas at different temperatures?

No. If they were the same gas, they would have the same $\gamma$. While they could be on different adiabatic paths (different constants), their slopes at the same $(V, p)$ coordinate would be identical. The fact that they have different steepnesses originating from the same region implies different $\gamma$ values.

Question

$p$-$V$ plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to

Options:

(A) He and $	ext{O}_2$

(B) $	ext{O}_2$ and He

(C) He and Ar

(D) $	ext{O}_2$ and $	ext{N}_2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a pressure-volume ($p-V$) graph depicting the expansion of two different gases. Both gases are undergoing an **adiabatic process**, meaning no heat is exchanged with the surroundings ($dQ = 0$).

On the graph, we see two curves, labeled 1 and 2. Curve 2 drops much more sharply than curve 1. In mathematical terms, curve 2 is **steeper**. Our objective is to determine which curve corresponds to which gas based on their atomicity.

### The Master Equation

For an ideal gas undergoing an adiabatic process, the relationship between pressure and volume is governed by the equation:

$$pV^\gamma = 	ext{constant}$$

Here, $\gamma$ (gamma) is the adiabatic index, defined as the ratio of molar specific heats ($C_p / C_V$). To understand how the steepness of the curve relates to the gas, we need to find the slope of this graph. We do this by differentiating the adiabatic equation with respect to volume $V$:

$$p \cdot \gamma V^{\gamma - 1} dV + V^\gamma dp = 0$$

Rearranging this to solve for the slope $\frac{dp}{dV}$, we get:

$$\frac{dp}{dV} = -\gamma \frac{p}{V}$$

### Comparing the Slopes

The negative sign simply indicates that as volume increases, pressure decreases (the curve slopes downwards). However, the **magnitude** of the slope, which dictates how steep the curve is, is directly proportional to $\gamma$:

$$|	ext{Slope}| \propto \gamma$$

Looking at our graph, curve 2 is steeper than curve 1. This implies that the magnitude of the slope for curve 2 is greater than that for curve 1:

$$|	ext{Slope}_2| > |	ext{Slope}_1| \implies \gamma_2 > \gamma_1$$

### Final Calculation

Now, let's evaluate the gases given in the options. We have Helium (He), Oxygen ($	ext{O}_2$), Argon (Ar), and Nitrogen ($	ext{N}_2$).

- Helium and Argon are **monatomic** gases, which have $\gamma = \frac{5}{3} \approx 1.67$.
- Oxygen and Nitrogen are **diatomic** gases, which have $\gamma = \frac{7}{5} = 1.4$.

Since $\gamma_2 > \gamma_1$, curve 2 must correspond to the monatomic gas (Helium), and curve 1 must correspond to the diatomic gas (Oxygen).

Therefore, plot 1 is $	ext{O}_2$ and plot 2 is He. This perfectly matches option (b).

Animated Solution for Physics - Thermodynamics: $p$ - $V$ plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to

Select Answer:

Visualized Solution

\text{Analyzing the } p-V \text{ Graph}

\text{Adiabatic Equation}

\text{Slope of the Curve}

\text{Comparing Slopes}

\text{Identifying the Gases}

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Comparing the Slopes

Final Calculation

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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; } p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to