Question

Two compressible solids are formed into spheres of the same size. The bulk modulus of sphere two is twice as large as the bulk modulus of sphere one. You now increase the pressure on both spheres by the same amount. As a result of the increased pressure, how is the change in volume of sphere two (ΔV2) related to the change in volume of sphere one (ΔV1)?

Answer 1

`\Delta V_1=2* \Delta V_2`

Step-by-step explanation:

The bulk modulus is a constant that describes how resistant a substance is to compression.

It is defined as the ratio between increase in pressure and the resulting decrease in a volume of the material.

It is given by a formula :

`Bulk\,\,modulus\,\,(K)=(volumetric\,\,stress)/(volumetric\,\,strain)`

OR

`K=(\Delta P)/(((\Delta V)/(V)) )`

where:

`\Delta V` &
`\Delta P` are the change in volume and change in pressure respectively.

V= original volume

According to the given:

`K_2=2K_1`

`V_1=V_2`

`\Delta P_1=\Delta P_2`

So,

`K_2=(\Delta P_2)/(((\Delta V_2)/(V_2)) )`

`K_2=(\Delta P_2* V_2)/(\Delta V_2)`.................................(1)

&

`K_1=(\Delta P_1)/(((\Delta V_1)/(V_1)) )`

`K_1=(\Delta P_1* V_1)/(\Delta V_1)`..................................(2)

From the given conditions we compare equations (1) & (2):

`K_2=2K_1`

`(\Delta P_2* V_2)/(\Delta V_2)=2* (\Delta P_1* V_1)/(\Delta V_1)`

cancelling the equal terms

`(1)/(\Delta V_2) =(2)/(\Delta V_1)`

`\Delta V_1=2* \Delta V_2`

The material in first case undergoes twice the volume reduction than that of the material in first case under the given conditions.