Use the formula to show that the pressure of a photon gas is 1/3 times the energy density (U/V). Compute the pressure exerted by the radiation inside a kiln at 1500 K, and compare to the ordinary gas pressure exerted by the air. Then compute the pressure of the radiation at the centre of the sun, where the temperature is 15 million K. Compare to the gas pressure of the ionised hydrogen, whose density is approximately 105 kg/m3.
Hence, the pressure exerted inside a kiln is
Formula to be used is
At constant N and S, the pressure equals the negative of the partial derivative of the total energy with respect to the volume, i.e.
To calculate the pressure, we must express the total energy in terms of entropy. The total energy is calculated as follows:
The entropy is given as:
Total energy and entropy is:
Solve entropy equation for T:
Substitute T in total energy:
Substitute in equation (1)
The total energy of radiation is:
Substitute the values:
The pressure is:
For comparison, the pressure within the kiln is the same as the pressure outside it, which is 1 atm in pascal 1.01 x 105 Pa, which is about 108 higher than the pressure caused by photons. The temperature at the sun's core is T = 15 x 106 K, hence the energy per unit volume is:
The pressure is:
Number of moles per unit volume equals to:
The pressure is:
The factor 2 came from the fact that each ionised hydrogen atom has and electron and a proton.
Consider a two-dimensional solid, such as a stretched drumhead or a layer of mica or graphite. Find an expression (in terms of an integral) for the thermal energy of a square chunk of this material of area , and evaluate the result approximately for very low and very high temperatures. Also, find an expression for the heat capacity, and use a computer or a calculator to plot the heat capacity as a function of temperature. Assume that the material can only vibrate perpendicular to its own plane, i.e., that there is only one "polarization."
For a gas of particles confined inside a two-dimensional box, the density of states is constant, independent of (see Problem 7.28). Investigate the behavior of a gas of noninteracting bosons in a two-dimensional box. You should find that the chemical potential remains significantly less than zero as long as is significantly greater than zero, and hence that there is no abrupt condensation of particles into the ground state. Explain how you know that this is the case, and describe what does happen to this system as the temperature decreases. What property must have in order for there to be an abrupt Bose-Einstein condensation?
For a system obeying Boltzmann statistics, we know what is from Chapter 6. Suppose, though, that you knew the distribution function (equation ) but didn't know . You could still determine by requiring that the total number of particles, summed over all single-particle states, equal N. Carry out this calculation, to rederive the formula . (This is normally how is determined in quantum statistics, although the math is usually more difficult.)
(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define . Express the integral that defines , equation 7.122, in terms of these variables. You should obtain the equation
(b) According to Figure 7.33, the correct value of when is approximately . Plug in these values and check that the equation above is approximately satisfied.
(c) Now vary , holding fixed, to find the precise value of for . Repeat for values of ranging from up to , in increments of . Plot a graph of as a function of temperature.
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