Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. The change in internal energy of this gas in adiabatic expansion to volume $V_{final} = 27V$ is . . . . . . $J$.

By what factor should the volume of an ideal gas $(\gamma = 1.5)$ be increased through an adiabatic expansion so that its $rms$ speed becomes half?

$A$ sample of $1$ mole of gas at temperature $T$ is adiabatically expanded to double its volume. If the adiabatic constant for the gas is $\gamma = \frac{3}{2}$,then the work done by the gas in the process is:

An ideal gas undergoes an adiabatic process. If the pressure of the gas is reduced by $0.1 \%$, then the volume is changed by (Given $\gamma = \frac{C_p}{C_v} = \frac{5}{3}$) (in $\%$)

Initially,the pressure of $1 \text{ mole}$ of an ideal gas is $10^5 \text{ Nm}^{-2}$ and its volume is $16 \text{ litres}$. When it is adiabatically compressed,its final volume is $2 \text{ litres}$. Calculate the work done on the gas. [Given: Molar specific heat at constant volume $C_v = \frac{3R}{2}$] (in $\text{ kJ}$)

$A$ 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 ..... $T$.