An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes $32$ times its initial value. If the final pressure of the gas is $128$ atmospheres,the value of $\gamma$ for the gas is:

$A$ spherical bubble inside water has radius $R$. Take the pressure inside the bubble and the water pressure to be $p_0$. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(R-a)$. For $a \ll R$,the magnitude of the work done in the process is given by $(4 \pi p_0 R a^2) X$,where $X$ is a constant and $\gamma = C_p / C_V = 41 / 30$. The value of $X$ is:

$Assertion:$ When a bottle of cold carbonated drink is opened,a slight fog forms around the opening.
$Reason:$ Adiabatic expansion of the gas causes lowering of temperature and condensation of water vapours.

Under an adiabatic process,the volume of an ideal gas gets doubled. Consequently,the mean collision time between the gas molecules changes from $\tau_{1}$ to $\tau_{2}$. If $\frac{C_{p}}{C_{v}}=\gamma$ for this gas,then a good estimate for $\frac{\tau_{2}}{\tau_{1}}$ is given by:

$A$ 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

During an adiabatic process,if the pressure of a gas is found to be proportional to the cube of its absolute temperature,then the ratio of $\frac{C_p}{C_V}$ for the gas is: