$A$ diatomic gas $\left(\gamma = \frac{7}{5}\right)$ is compressed adiabatically to volume $\frac{V_0}{32}$,where $V_0$ is its initial volume. The initial temperature of the gas is $T_i$ in Kelvin and the final temperature is $xT_i$ in Kelvin. The value of $x$ is:

$A$ gas for which $\gamma = 1.5$ is suddenly compressed to $\frac{1}{4}$ of its initial volume. What is the ratio of the final pressure to the initial pressure?

$A$ 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. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively,then $T_1/T_2$ is given by

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 $\text{ kJ}$)

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 $\%$)

During an adiabatic process,the pressure of the gas is found to be proportional to the cube of its absolute temperature. The ratio $C_P/C_V = \gamma$ for the gas is: