An ideal gas at $27^{\circ} C$ is compressed adiabatically to $8/27$ of its original volume. If $\gamma = 5/3$,the rise in temperature of the gas is: (in $K$)

$1 \, \text{kmol}$ of an ideal gas is compressed adiabatically,requiring $146 \, \text{kJ}$ of work. During this process,the temperature of the gas increases by $7^o \text{C}$. The gas is: $(R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1})$

$A$ gas at $N.T.P.$ is suddenly compressed to one-fourth of its original volume. If $\gamma = 1.5$,then the final pressure is

One mole of an ideal gas $(\gamma = 1.4)$ is adiabatically compressed so that its temperature rises from $27\,^{\circ}C$ to $35\,^{\circ}C$. The change in the internal energy of the gas is .... $J$ (given $R = 8.3 \,J/mol/K$)

$A$ hypothetical gas expands adiabatically such that its volume changes from $8 \ L$ to $27 \ L$. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$, then the ratio of $\frac{C_P}{C_V}$ will be:

$Assertion :$ Adiabatic expansion is always accompanied by fall in temperature.
$Reason :$ In adiabatic process, volume is inversely proportional to temperature.