Question
Question: The pressure of a medium is changed from \[1.01 \times {10^5}\,{\text{Pa}}\] to \[1.165 \times {10^5...
The pressure of a medium is changed from 1.01×105Pa to 1.165×105Pa and change in volume is 10% keeping the temperature constant. The bulk modulus of the medium is
Solution
Use the formula for bulk modulus of a material. This formula gives the relation between the bulk modulus of the material, change in pressure on the material, original volume of the material and change in volume of the material. Substitute all the values in this equation and determine the value of bulk modulus of the medium.
Formula used:
The formula for bulk modulus K of a material is given by
K=ΔVVΔP …… (1)
Here, ΔP is the change in pressure on the material, V is the original volume of the material and ΔV is the change in volume of the material.
Complete step by step solution:
We have given that the pressure of a medium is changed from 1.01×105Pa to 1.165×105Pa.
The initial pressure of the medium is 1.01×105Pa and the final pressure of the medium is 1.165×105Pa.
Pi=1.01×105Pa
Pf=1.165×105Pa
We can determine the change in pressure of the medium using the following equation.
ΔP=Pf−Pi
Substitute 1.165×105Pa for Pf and 1.01×105Pa for Pi in the above equation.
ΔP=(1.165×105Pa)−(1.01×105Pa)
⇒ΔP=0.155×105Pa
⇒ΔP=1.55×104Pa
Hence, the change in pressure of the medium is 1.55×104Pa.
Also the change in volume of the medium 10%. Hence, the fraction of the change in volume of the medium with the original volume of the medium is 0.1.
VΔV=0.1
We can determine the bulk modulus of the medium using equation (1).
Substitute 1.55×104Pa for ΔP and 0.1 for VΔV in equation (1).
K=0.11.55×104Pa
⇒K=1.55×105Pa
Hence, the bulk modulus of the medium is 1.55×105Pa.
Note:
One can also solve the same question in another way. The bulk modulus of a material is the ratio of bulk stress and bulk strain.
One can determine first bulk stress and bulk strain on the medium and then substitute these values in the formula to determine the bulk modulus of the medium.