Consider a spherical shell of radius $R$ at temperature $T$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $E = \frac{U}{V} \propto T^4$ and pressure $P = \frac{1}{3} \left( \frac{U}{V} \right)$. If the shell now undergoes an adiabatic expansion,the relation between $T$ and $R$ is:

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d}=32$, then $\frac{P^{\prime}}{P}$ is:

When air of the atmosphere rises up,it cools. Why?

$A$ gas is compressed adiabatically until its temperature is doubled. The ratio of its final volume to initial volume will be

$A$ thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle,as shown in the figure below. On each side of the partition,there is one mole of an ideal gas,with specific heat at constant volume,$C_v = 2R$. Here,$R$ is the gas constant. Initially,each side has a volume $V_0$ and temperature $T_0$. The left side has an electric heater,which is turned on at very low power to transfer heat $Q$ to the gas on the left side. As a result,the partition moves slowly towards the right,reducing the right side volume to $V_0 / 2$. Consequently,the gas temperatures on the left and the right sides become $T_L$ and $T_R$,respectively. Ignore the changes in the temperatures of the cylinder,heater,and the partition.
$(1)$ The value of $\frac{T_R}{T_0}$ is
$(A)$ $\sqrt{2}$ $(B)$ $\sqrt{3}$ $(C)$ $2$ $(D)$ $3$
$(2)$ The value of $\frac{Q}{RT_0}$ is
$(A)$ $4(2\sqrt{2}+1)$ $(B)$ $4(2\sqrt{2}-1)$ $(C)$ $(5\sqrt{2}+1)$ $(D)$ $(5\sqrt{2}-1)$

$A$ monoatomic gas is compressed adiabatically to $1/8$ of its initial volume. What happens to the gas pressure? $(\gamma = 5/3)$