The work done in an adiabatic change in an ideal gas depends upon only

$A$ given ideal gas with $\gamma = \frac{C_p}{C_v} = 1.5$ is at a temperature $T$. If the gas is compressed adiabatically to one-fourth of its initial volume,the final temperature will be ..... $T$.

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

$A$ monoatomic ideal gas,initially at temperature $T_1$,is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing the piston suddenly. $L_1$ and $L_2$ are the lengths of the gas columns before and after the expansion,respectively. The ratio $T_2 / T_1$ is

The work done in an adiabatic change in a gas depends only on

$A$ 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 . . . . . . $K$.