Solved: CALC When the pressure p on a material increases

QUESTION:

CALC When the pressure P on a material increases by an amount Δp, the volume of the material will change from V to v+Δv, where Δv is negative. The bulk modulus B of the material is defined to be the ratio of the pressure change Δp to the absolute value |Δv/v| of the fractional volume change. The greater the bulk modulus, the greater the pressure increase required for a given fractional volume change, and the more incompressible the material (see Section 11.4). Since Δv<0, the bulk modulus can be written as B=−Δp/(Δv/v0) In the limit that the pressure and volume changes are very small, this becomes

\(B=-v \frac{d y}{d V}\)

(a) Use the result of Problem 42.55 to show that the bulk modulus for a system of N free electrons in a volume V at low temperatures is B=\(\frac{5}{3}\)p (Hint: The quantity P in the expression B=−v(dp/dV) is the external pressure on the system. Can you explain why this is equal to the internal pressure of the system itself, as found in Problem 42.55?) (b) Evaluate the bulk modulus for the electrons in copper, which has a free-electron concentration of 8.45×1028 m−3. Express your result in pascals. (c) The actual bulk modulus of copper is 1.4×1011Pa. Based on your result in part (b), what fraction of this is due to the free electrons in copper? (This result shows that the free electrons in a metal play a major role in making the metal resistant to compression.) What do you think is responsible for the remaining fraction of the bulk modulus?