Question

An external pressure P is applied on a cube at 273 K T hence it compresses equally from all sides Alpha is the coefficient of linear expansion and K is the bulk modulus of material. To bring the cube to its original size by heating the temperature rise must be​

Answer 1

Given : An external pressure P is applied on a cube at 273 K hence it compresses equally from all sides α is the coefficient of linear expansion and K is the bulk modulus of material.

To find : To bring the cube to its original size by heating the temperature rise must be ...

solution : we know, bulk modulus, K = - ∆P/(∆V/V)......(1)

here external pressure = ∆P = P

from thermal expansion, -∆V = Vγ∆T

but γ = 3α

so, -∆V = V3α∆T

[ here negative sign of change of volume indicates that volume is decreased from higher value to lower value (original value) ]

now, -∆V/V = 3α∆T

so, -∆V/V = 3α∆T

from equation (1) we get,

K = P/3α∆T

⇒∆T = P/3αK

Therefore, to bring the cube to its original size by the heating, the temperature rise must be P/3αK

Therefore

Answer 2

Answer:

Given: External Pressure P is applied due to which cube is getting compressed. Coefficient of Linear Exapansion is given by α.

Since the cube is being reduced from all sides equally, it is undergoing volumetric compression. According to the relation between the linear expansion coefficient and volume expansion coefficient, we get:

→ 3α = δ

where, δ is the coefficient of volumetric expansion.

So change in volume would be:

→ ΔV = V₀(3αΔT)

→ ΔV/V₀ = 3αΔT

Here, V₀ is the initial volume.

According to Bulk modulus of elasticity,

$K = \dfrac{ F}{A} \times \dfrac{V_o}{\Delta V}$

F/A is equal to Pressure P.

Hence we get:

$\rightarrow \dfrac{ \Delta V}{V_o} = \dfrac{P}{K}\\\\\\\rightarrow 3\alpha \Delta T = \dfrac{P}{K}\\\\\\\rightarrow \Delta T = \dfrac{P}{3\alpha K}}$

This is the required answer.