The following calculations are for EM radiation, for which the internal energy U=Vu, where u=u(T)is the total energy density.
a) Show that (∂S/∂T)_v=V/T(du/dT), and and (∂S/∂V)T=(u+P)/T
Note: use the answers in part a in part b and c
b) Using Stefan-Bnoltzman law, u=aT^4, where a is a constant, integrate the expression for (∂S/∂T)_v, and obtain an expression for S in terms of a,T and V. Hint: use Planck’s statement of the 3rd law of thermodynamics: lim_T→0 S=0 to evaluate the constant of integration. c) Assumming P=bT^n, where b and n are constants, use the equality ∂^2S/∂T∂V=∂^2S/∂V∂T to obtain values for b and n. Note: b has the same units as a
a) To show that (∂S/∂T)_v = V/T (du/dT), we start with the definition of entropy:
S = ∫(∂Q/T)
where (∂Q) is the heat transferred to the system, T is the temperature, and the integral is taken over all the processes happening in the system. For a constant volume process (∂Q) can be written as TdS, which gives us:
S = ∫dS
Now, we consider the total energy density u, which is given by u = u(T). Taking the differential of u, we have:
du = (∂u/∂T)_v dT
Using the first law of thermodynamics, we know that dU = (∂U/∂T)_v dT = V du, where U is the internal energy, and dU is the differential change in internal energy.
We can express dU as V du, so:
V du = (∂U/∂T)_v dT
Now, we can rewrite du as:
du = (1/V)(∂U/∂T)_v dT
Substituting this expression into the entropy equation, we get:
S = ∫(1/T)(∂U/∂T)_v dT
Differentiating S with respect to T at constant volume (v), we have:
(∂S/∂T)_v = (1/T)(∂U/∂T)_v
Using the relation U = Vu, where u = u(T) is the total energy density, we find:
(∂S/∂T)_v = V/T(∂u/∂T)_v
Thus, we have shown that (∂S/∂T)_v = V/T(∂u/∂T)_v.
b) To integrate the expression (∂S/∂T)_v = V/T(du/dT), we need to integrate both sides with respect to T at constant volume:
∫(∂S/∂T)_v dT = V∫(1/T)(∂u/∂T)_v dT
Integrating the left side, we find:
ΔS = ∫(∂S/∂T)_v dT = S - S_0
where S_0 is the initial entropy.
Integrating the right side, we have:
∫(1/T)(∂u/∂T)_v dT = ∫(1/T)(d(u))/dT dT
= ∫(1/u)du
= ln(u) + C
where C is the constant of integration.
Therefore, we obtain:
ΔS = ln(u) + C
Using Planck’s statement of the 3rd law of thermodynamics: lim_T→0 S = 0, we can evaluate the constant of integration:
0 = ln(u_0) + C
C = -ln(u_0)
where u_0 is the initial value of the energy density.
Therefore, the expression for ΔS (change in entropy) is:
ΔS = ln(u) - ln(u_0)
= ln(u/u_0)
But, u = aT^4 (Stefan-Boltzmann law), where a is a constant. So, we have:
ΔS = ln(aT^4/u_0)
c) To obtain values for b and n in P = bT^n, we can use the equality ∂^2S/∂T∂V = ∂^2S/∂V∂T.
Let's start with the left-hand side, ∂^2S/∂T∂V:
∂^2S/∂T∂V = (∂/∂T)[(∂S/∂V)_T]
Using the result from part a, (∂S/∂V)T = (u + P)/T, the expression becomes:
∂^2S/∂T∂V = (∂/∂T)[(u + P)/T]
Differentiating with respect to T, we get:
∂^2S/∂T∂V = -P/T^2
Now, let's calculate the right-hand side, ∂^2S/∂V∂T:
∂^2S/∂V∂T = (∂/∂V)[(∂S/∂T)_V]
Using the result from part a, (∂S/∂T)_V = V/T(du/dT), the expression becomes:
∂^2S/∂V∂T = (∂/∂V)[V/T(du/dT)]
Differentiating with respect to V, we obtain:
∂^2S/∂V∂T = 0
Now, equating the left and right sides, we have:
-P/T^2 = 0
This implies that P = 0.
However, assuming P = bT^n, we can see that n must be equal to -2 for P to be zero. Thus, the value of n is -2.
Since P = bT^n and we already know n, we can rewrite P as:
P = bT^-2
Therefore, we have obtained the values for b and n: b is a constant with the same units as a, and n is equal to -2.