An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an inc

Q: Why is pressure constant?

The problem explicitly states 'pressure remaining constant', which describes an isobaric process.

Q: What does $\delta$ represent physically?

It represents the coefficient of volume expansion, which measures the fractional change in volume per degree change in temperature.

Q: Why does the graph look like a curve and not a straight line?

Because $\delta = 1/T$, which is an inverse relationship, mathematically represented by a rectangular hyperbola.

Q: Can we use $V_1/T_1 = V_2/T_2$ instead?

Yes, Charles's Law is derived from the same principle and leads to the exact same result if you take the limit for small changes.

Q: Does this apply to real gases?

Real gases approximate this behavior at high temperatures and low pressures, but they have more complex equations of state like the Van der Waals equation.

Question

An ideal gas is initially at temperature $T$ and volume $V$. Its volume is increased by $\Delta V$ due to an increase in temperature $\Delta T$, pressure remaining constant. The quantity $\delta = \Delta V / (V \Delta T)$ varies with temperature as

Sigma Percentile · Accepted Answer

The problem asks us to find how the quantity $\delta = \frac{\Delta V}{V \Delta T}$ varies with temperature $T$ for an ideal gas undergoing an isobaric (constant pressure) process.

### The Master Equation

We begin our journey with the fundamental equation of state for an ideal gas:
$$pV = nRT$$

Since the problem explicitly states that the pressure $p$ remains constant, any change in temperature $\Delta T$ will result in a proportional change in volume $\Delta V$. Taking the change on both sides, we get:
$$p\Delta V = nR\Delta T$$

### Rearranging the Terms

Our goal is to isolate the expression for $\delta$. Let's first find the ratio of the change in volume to the change in temperature:
$$\frac{\Delta V}{\Delta T} = \frac{nR}{p}$$

Now, we can substitute the expression for pressure $p$ from the ideal gas law ($p = \frac{nRT}{V}$) back into our equation:
$$\frac{\Delta V}{\Delta T} = \frac{nR}{\left(\frac{nRT}{V}\right)}$$

Notice how beautifully the $n$ and $R$ terms cancel out! This leaves us with a very elegant relationship:
$$\frac{\Delta V}{\Delta T} = \frac{V}{T}$$

### Final Calculation

The quantity we are looking for is $\delta$, which is defined as:
$$\delta = \frac{\Delta V}{V \Delta T}$$

By dividing both sides of our previous result by $V$, we arrive at the final expression:
$$\delta = \frac{1}{T}$$

### Analyzing the Graph

The relationship $\delta = \frac{1}{T}$ tells us that $\delta$ is **inversely proportional** to the absolute temperature $T$.

In mathematics, an inverse proportionality $y = \frac{k}{x}$ is represented graphically by a **rectangular hyperbola**. As the temperature $T$ increases, the value of $\delta$ decreases. Looking at the given options, the graph that shows a decreasing curve (a rectangular hyperbola) is option (c).

**Bonus Insight:** The quantity $\delta$ is actually the **coefficient of volume expansion** (often denoted by $\gamma$). For any ideal gas at constant pressure, this coefficient is exactly equal to the reciprocal of its absolute temperature!

Animated Solution for Physics - Thermodynamics: An ideal gas is initially at temperature $T$ and volume $V$ . Its volume is increased by $Δ V$ due to an increase in temperature $Δ T$ , pressure remaining constant. The quantity $δ = Δ V / (V Δ T)$ varies with temperature as

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Visualized Solution

$p V = n R T$

$p Δ V = n R Δ T$

$\frac{Δ V}{Δ T} = \frac{n R}{p}$

$p = \frac{n R T}{V}$

$\frac{Δ V}{Δ T} = \frac{V}{T}$

$δ = \frac{1}{T}$

$δ \propto \frac{1}{T}$

$Conclusion$

$γ = \frac{1}{T}$

The Sigma Insight: Thermodynamic Processes

The Master Equation

Rearranging the Terms

Final Calculation

Analyzing the Graph

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔTΔV​=pnR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=VnRT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔTΔV​=TV​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } δ=T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } δ∝T1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ=T1​