Graphite is more compressible than diamond.(a) Taking

Chapter 5, Problem 27P

(choose chapter or problem)

Problem 27P

Graphite is more compressible than diamond.

(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?

(b) The isothermal compressibility of graphite is about 3 × 10–6 bar–1 , while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

Problem 1:

Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows.

(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 2.

(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as

(This is the reciprocal of the isothermal bulk modulus defined in Problem 3.)

(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to  since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain

This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.

(d) Compute β, κT and  for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).

(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?

Problem 2:

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:

(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)

(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.

(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.

Problem 3:

By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by

where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:

This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).

(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.

(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.

(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.

(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not?