Slope of isotherm for a gas (having $\gamma = 5/3$) is $3 \times 10^5 \, N/m^2$. If the same gas is undergoing adiabatic change,then the adiabatic elasticity at that instant is ........... $\times 10^5 \, N/m^2$.

$A$ monoatomic gas having $\gamma = 5/3$ is stored in a thermally insulated container and the gas is suddenly compressed to $(1/8)^{th}$ of its initial volume. The ratio of final pressure to initial pressure is: ($\gamma$ is the ratio of specific heats of the gas at constant pressure and at constant volume).

$A$ 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
$(A)$ $4(2\sqrt{2}+1)$ $(B)$ $4(2\sqrt{2}-1)$ $(C)$ $(5\sqrt{2}+1)$ $(D)$ $(5\sqrt{2}-1)$

The pressure and density of a diatomic gas with $\gamma = 7/5$ change adiabatically from $(P, d)$ to $(P', d').$ If $\frac{d'}{d} = 32,$ then $\frac{P'}{P}$ is

$A$ reversible adiabatic path on a $P-V$ diagram for an ideal gas passes through state $A$ where $P = 0.7 \times 10^5 \, N/m^2$ and $V = 0.0049 \, m^3$. The ratio of specific heats of the gas is $1.4$. The slope of the path at $A$ is:

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