An external pressure p is applied on a cube at 0°C so that it is equally compressed from all sides.

Q: Why do we use $3\alpha$ instead of just $\alpha$ in the expansion formula?

The formula for volume expansion requires the coefficient of volume expansion, $\gamma$. The problem gives the coefficient of linear expansion, $\alpha$. For an isotropic 3D object like a cube, the volume expands in three dimensions, so $\gamma \approx 3\alpha$.

Q: Why does the initial volume $V$ not matter in the final answer?

Both the mechanical compression and the thermal expansion are proportional to the initial volume $V$. When you equate the two effects to find the balance point, $V$ cancels out from both sides of the equation.

Q: Is the pressure constant during the heating process?

Yes, the problem implies that the external pressure $p$ is maintained, and we are heating the cube while it is under this pressure to force it to expand back against the squeezing force.

Q: What does the negative sign in the standard bulk modulus formula mean?

The negative sign in $K = -p / (\Delta V / V)$ ensures that the Bulk Modulus $K$ is a positive number, because an increase in pressure (positive $p$) always results in a decrease in volume (negative $\Delta V$).

Q: Can we use this exact same logic for liquids?

Yes, the logic holds, but liquids do not have a coefficient of linear expansion $\alpha$ because they don't have a fixed shape. For liquids, you would directly use the given coefficient of volume expansion $\gamma$ instead of substituting $3\alpha$.

Question

An external pressure $p$ is applied on a cube at $0^{\circ}	ext{C}$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

Options:

(A) $\frac{p}{\alpha K}$

(B) $\frac{3\alpha}{pK}$

(C) $3pK\alpha$

(D) $\frac{p}{3\alpha K}$

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of two distinct physical phenomena: mechanical compression and thermal expansion. It challenges us to balance a force that shrinks an object with a temperature change that expands it.

### Analyzing the Setup

Imagine a solid cube resting at $0^{\circ}	ext{C}$. We apply a uniform external pressure, $p$, from all sides. Naturally, the cube gets squeezed, and its volume decreases. The material's resistance to this uniform compression is quantified by its **Bulk Modulus**, denoted as $K$.

The definition of Bulk Modulus relates the applied pressure to the resulting volumetric strain:

$$K = \frac{p}{-\frac{\Delta V}{V}}$$

Here, the negative sign simply indicates that an increase in pressure leads to a decrease in volume. If we rearrange this equation to find the magnitude of the volume lost due to compression, we get:

$$\Delta V_{	ext{comp}} = \frac{pV}{K}$$

### The Master Equation

Now, the problem asks us to bring the cube back to its exact original size. To counteract the mechanical squeezing, we must heat the cube. Raising the temperature by an amount $\Delta T$ will cause the material to undergo thermal expansion.

The formula for the increase in volume due to heating is:

$$\Delta V_{	ext{exp}} = V \gamma \Delta T$$

where $\gamma$ is the coefficient of volume expansion. However, the problem provides the coefficient of *linear* expansion, $\alpha$. For an isotropic solid (a material that behaves the same in all directions), the volume expansion coefficient is simply three times the linear expansion coefficient:

$$\gamma = 3\alpha$$

Substituting this into our expansion equation gives:

$$\Delta V_{	ext{exp}} = V (3\alpha) \Delta T$$

### Final Calculation

For the cube to return to its original size, the volume lost due to the external pressure must be perfectly balanced by the volume gained due to heating. Therefore, we equate the two volume changes:

$$\Delta V_{	ext{comp}} = \Delta V_{	ext{exp}}$$

$$\frac{pV}{K} = V (3\alpha) \Delta T$$

Notice a mathematically elegant moment here: the original volume $V$ appears on both sides of the equation and cancels out. This profound result tells us that the required temperature change is completely independent of how large or small the cube initially was!

$$\frac{p}{K} = 3\alpha \Delta T$$

Finally, isolating the required temperature change $\Delta T$, we arrive at our answer:

$$\Delta T = \frac{p}{3\alpha K}$$

This elegant expression perfectly captures the tug-of-war between mechanical pressure and thermal energy.

Select Answer:

Visualized Solution

$V_{ini t ia l} = V$

$K = \frac{p}{- \frac{Δ V}{V}}$

$Δ T = ?$

$Δ V_{e x p} = V γ Δ T$

$γ = 3 α$

$∣Δ V_{co m p} ∣ = Δ V_{e x p}$

$Δ T = \frac{p}{3 α K}$

$What if anisotropic?$

The Sigma Insight: Thermal Expansion

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ=3α

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔT=3αKp​

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