Question

Pressure on an object increases from \(1.01 \times 10^{\circ} \mathrm{Pa}\) to \(1.165 \times 10^{5} \mathrm{~Pa}\). He volume decrease by \(10 \%\) at constant temperature. Bulk modulus of material is........... (A) \(1.55 \times 10^{5} \mathrm{~Pa}\) (B) \(51.2 \times 10^{5} \mathrm{~Pa}\) (C) \(102.4 \times 10^{5} \mathrm{~Pa}\) (D) \(204.8 \times 10^{5} \mathrm{~Pa}\)

Short answer

The bulk modulus of the material is (B) \(51.2 \times 10^{5} \mathrm{~Pa}\).

Step-by-step solution

Determine the Change in Pressure

Calculate the change in pressure ΔP by subtracting the initial pressure from the final pressure: ΔP = Final Pressure - Initial Pressure ΔP = \(1.165 \times 10^{5} \mathrm{~Pa}\) - \(1.01 \times 10^{\circ} \mathrm{Pa}\)

Calculate the Volumetric Strain

Given that the volume decreases by 10%, we can find the volumetric strain (ΔV/V) using the percentage decrease: Volumetric Strain = ΔV/V = - 10%

Calculate the Bulk Modulus

Use the Bulk modulus formula and substitute the values calculated in Step 1 and Step 2: Bulk Modulus (K) = \(- \frac{ΔP}{ΔV/V}\) K = \(- \frac{1.165 \times 10^{5} \mathrm{~Pa} - 1.01 \times 10^{\circ} \mathrm{Pa}}{-0.1}\)

Identify the Correct Answer

Calculate the Bulk Modulus value and see which option it matches with: K ≈ 51.2 × \(10^{5} \mathrm{~Pa}\) The correct answer is (B) \(51.2 \times 10^{5} \mathrm{~Pa}\).

Key concepts

Pressure Change

Pressure change is an important concept when dealing with materials and their responses to external forces. When pressure on a material changes, it affects the substance in various ways.
The pressure change is defined as the difference between the final and initial pressures applied to the material.
To calculate the change in pressure, use the formula:

• \( \Delta P = P_{\text{final}} - P_{\text{initial}} \)

In this problem, the initial pressure \(P_{\text{initial}}\) was \(1.01 \times 10^{3} \mathrm{Pa}\) and the final pressure \(P_{\text{final}}\) was \(1.165 \times 10^{5} \mathrm{Pa}\).
This indicates that the pressure experienced a substantial increase, which will influence the material's physical characteristics, particularly its volume.

Volumetric Strain

Volumetric strain is a measure of how much the volume of a material changes under pressure.
It is calculated by the ratio of the change in volume to the original volume.

• Expressed as \(\Delta V/V\),
• In this problem: \(-10\%\) or \(-0.1\).

A negative sign indicates a decrease in volume, which is expected when pressure on the material increases.
This relationship highlights the inverse connection between pressure and volume, where an increase in pressure typically results in a decrease in volume.
The concept is useful in understanding how materials like solids and liquids behave under different pressure conditions.

Thermodynamics

Thermodynamics plays a significant role in understanding how changes in pressure and volume are interrelated.
In these scenarios, when a gas or fluid undergoes compression or expansion, keeping the temperature constant simplifies the calculation.
This is known as an isothermal process, where any change in volume at constant temperature directly correlates with a change in pressure.

• Using physical principles: As per Boyles' Law, at constant temperature, \(P_1V_1 = P_2V_2\).

This law helps us understand the system’s behavior under different conditions, providing insight into how pressure-volume relationships unfold in real-world situations.

Pressure-Volume Relationship

The pressure-volume relationship describes how pressure and volume interact with each other.
As pressure increases, volume tends to decrease and vice versa, assuming temperature remains constant.
This inverse relationship is essential for understanding material dynamics under different pressures.

• Using the formula for Bulk Modulus: \(K = - \frac{\Delta P}{\Delta V/V}\)

The Bulk Modulus is a key factor here, representing a material's resistance to uniform compression.
In this exercise, we apply this formula to calculate the Bulk Modulus, showing the material's response to increased pressure.
Being able to compute these values helps engineers and scientists design systems and structures that can withstand various pressure conditions without significant deformation.