Problem 927 - JEE Physics

Question

The pressure applied from all directions on a cube is \(\mathrm{P}\). How much its temperature should be raised to maintain the orginal volume ? The volume elasticity, of the cube is \(\beta\) and the coefficient of volume expansion is \(\alpha\). (A) \([\mathrm{P} / \alpha \beta]\) (B) \([\mathrm{P\alpha} / \beta]\) (C) \([\beta \mathrm{p} / \alpha]\) (D) \([\alpha \beta / \mathrm{p}]\)

Step-by-Step Solution

Answer

The correct answer is: none of the choices are correct, as no change in temperature is required to maintain the original volume of the cube (\(\Delta T = 0\)).

Step 1: 1. Volume Expansion Formula

The volume expansion formula relates the change in volume of an object with respect to its original volume and the change in temperature. It is given by: \[\Delta V = V_0 \cdot \alpha \cdot \Delta T\] We are considering a situation where the change in volume is zero (i.e., ΔV = 0), so the formula becomes: \[0 = V_0 \cdot \alpha \cdot \Delta T\]

Step 2: 2. Pressure-Volume Work

Next we consider the work done due to the pressure applied on the cube. The work done by the pressure is given by: \[W = -\mathrm{P} \cdot \Delta V\] As ΔV = 0 (no change in volume), the work done by the pressure will also be zero, meaning that the cube's volume remains constant despite the change in temperature.

Step 3: 3. Volume Elasticity Relationship

The relationship between volume elasticity, pressure, and change in volume is given by the following formula: \[\mathrm{P} = - \beta \cdot \frac{\Delta V}{V_0}\] Since ΔV = 0, we can rewrite this relationship as: \[\mathrm{P} = 0\] Therefore, the pressure applied on the cube is not causing any change in volume.

Step 4: 4. Solving for the Increase in Temperature

Now, using the relationships we've established, we'll solve for ΔT in the volume expansion formula: \[0 = V_0 \cdot \alpha \cdot \Delta T\] As V0 ≠ 0 (there is some volume) and α ≠ 0 ( there is some coefficient of volume expansion), we can divide both sides by V₀α to get: \[\Delta T = \frac{0}{\alpha \cdot V_0}\] Since we established that \(\mathrm{P} = 0\), we can write ΔT as follows: \[\Delta T = \frac{\mathrm{P}}{\alpha \cdot V_0}\] Now, let's replace V₀ in the ΔT equation using the volume elasticity relationship: \[\frac{\Delta V}{V_0} = - \frac{\mathrm{P}}{\beta}\] \[V_0 = - \frac{\Delta V \cdot \beta}{\mathrm{P}}\] Substitute the value of V₀ into the ΔT formula: \[\Delta T = \frac{\mathrm{P}}{\alpha} \cdot \frac{-\Delta V \cdot \beta}{\mathrm{P}}\] Simplify the expression and cancel out common factors: \[\Delta T = \frac{-\Delta V \cdot \beta}{\alpha}\] As we know, the change in volume (ΔV) is zero, so the expression becomes: \[\Delta T = 0\] Therefore, there is no change in temperature required to maintain the original volume of the cube, as it is not affected by the pressure applied on it. This does not match any of the given answer choices, so none of them are correct.

Key Concepts

Volume Elasticity

Volume Elasticity is an important concept that helps us understand how materials respond to pressure changes. It quantifies how much a material compresses or expands when an external pressure is applied. This property is defined by the volume elasticity modulus, denoted by the symbol \( \beta \).

Essentially, volume elasticity helps describe the relationship between the change in pressure and the change in volume of a material. The mathematical expression that represents this concept is given by:

• \( \mathrm{P} = - \beta \cdot \frac{\Delta V}{V_0} \)

Here, \( \mathrm{P} \) is the pressure applied, \( \Delta V \) is the change in volume, and \( V_0 \) is the original volume.

This equation shows that for incompressible materials, a large \( \beta \) means that the material doesn't easily change its volume, even with large changes in pressure. This property is critical in understanding and predicting how materials behave under different environmental conditions.

Coefficient of Volume Expansion

The Coefficient of Volume Expansion is a measure of how much a material's volume changes with a change in temperature. It's typically represented by the symbol \( \alpha \).

This coefficient is crucial in understanding how substances expand or contract with temperature fluctuations. The formula that describes this relationship is:

• \( \Delta V = V_0 \cdot \alpha \cdot \Delta T \)

Here, \( \Delta V \) is the change in volume, \( V_0 \) is the original volume, \( \alpha \) is the coefficient of volume expansion, and \( \Delta T \) is the change in temperature.

In a scenario where we want the volume to remain constant despite a temperature change, understanding \( \alpha \) helps predict how much the material will expand or contract. It allows us to calculate the exact temperature adjustment needed to offset the expansion or contraction when a material experiences pressure, ensuring it remains stable and intact at its original dimensions.

Pressure-Volume Work

Pressure-Volume Work involves the work done on or by a system as it expands or compresses under pressure. It plays a critical role in thermodynamics and material science. The formula for pressure-volume work is:

• \( W = -\mathrm{P} \cdot \Delta V \)

In this equation, \( W \) stands for the work done, \( \mathrm{P} \) is the pressure, and \( \Delta V \) represents the change in volume.

When pressure is applied to a material, it can cause the material to do work, leading to a reduction in volume. Conversely, if the material expands, it does work against the pressure. Understanding pressure-volume work is essential in processes where maintaining constant volume is crucial.

For our specific exercise, even though pressure is applied, the work remains zero because \( \Delta V \) equals zero. Therefore, no actual work is done and the volume of the cube remains unchanged despite the changes in pressure.