Write $\mathcal R=k_B/(\mu m_u)$. At fixed density,
$$
\left(\frac{\partial\log P}{\partial\log T}\right)_\rho
=4-3\beta,
\qquad
\left(\frac{\partial\log P}{\partial\log\rho}\right)_T
=\beta.
$$
The specific internal energy is
$$
u=\frac32\mathcal RT+\frac{a_{\rm rad}T^4}{\rho}.
$$
Applying the <first law of thermodynamics> adiabatically gives
$$
\left(\frac{d\log T}{d\log\rho}\right)_s
=\frac{8-6\beta}{24-21\beta}.
$$
Combining these logarithmic derivatives yields
$$
\boxed{
\nabla_{\rm ad}(\beta)
=\frac{8-6\beta}{32-24\beta-3\beta^2}}.
$$
As gas pressure dominates,
$$
\boxed{\lim_{\beta\to1}\nabla_{\rm ad}=\frac25},
$$
the result for a monatomic perfect gas. In the radiation-pressure limit it tends to $1/4$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 317 / 4 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-317/4/ii/solution