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

The initial pressure and volume of a given mass of an ideal gas (with $\frac{C_{p}}{C_{V}}=\gamma$), taken in a cylinder fitted with a piston, are $p_{0}$ and $V_{0}$ respectively. At this stage, the gas has the same temperature as that of the surrounding medium, which is $T_{0}$. It is adiabatically compressed to a volume equal to $\frac{V_{0}}{2}$. Subsequently, the gas is allowed to come to thermal equilibrium with the surroundings. What is the heat released to the surrounding?

For an ideal gas,the internal energy is given by $U = 5pV/2 + C$,where $C$ is a constant. The equation of the adiabats in the $pV$-plane will be

$Assertion:$ 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.

$A$ work of $166.28 \ J$ is done to adiabatically compress one mole of a gas. If the increase in the temperature of the gas is $8^{\circ} C$,the gas is $\left(R=8.314 \ J \ mol^{-1} \ K^{-1}\right)$

For an adiabatic process,if $\gamma = 2.5$ and the volume becomes $1/8$ of its initial volume,then the new pressure $P'$ is (initial pressure $= P$):