In the following $P-V$ diagram, two adiabatics cut two isothermals at temperatures $T_1$ and $T_2$. The value of $\frac{V_a}{V_d}$ 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
$(A)$ $4(2\sqrt{2}+1)$ $(B)$ $4(2\sqrt{2}-1)$ $(C)$ $(5\sqrt{2}+1)$ $(D)$ $(5\sqrt{2}-1)$

An adiabatic process occurs at constant

$A$ vessel contains $0.15 \ \text{m}^3$ of a gas at pressure $8 \ \text{bar}$ and temperature $140^\circ \text{C}$ with $c_p = 3R$ and $c_v = 2R$. It is expanded adiabatically until the pressure falls to $1 \ \text{bar}$. The work done during this process is . . . . . . $\text{kJ}$.

$A$ monoatomic gas $(\gamma = 5/3)$ at a temperature of $27^{\circ}C$ is compressed adiabatically to $8/27$ of its initial volume. The change in temperature is . . . . . . $K$.

In an adiabatic change,the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^C$,where $C$ equals