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 $K$)

One gram mole of a diatomic gas $(\gamma = 1.4)$ is compressed adiabatically so that its temperature rises from $27^{\circ}C$ to $127^{\circ}C$. The work done will be:

The amount of work done in an adiabatic expansion from temperature $T$ to ${T_1}$ is

$A$ 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 $\left(R=8.314 \ J \ mol^{-1} \ K^{-1}\right)$

The relation obeyed by a perfect gas during an adiabatic process is $PV^{3/2} = \text{constant}$. The initial temperature of the gas is $T$. When the gas is compressed to half of its initial volume, the final temperature of the gas is:

$A$ gas is kept in a container having walls which are thermally non-conducting. Initially,the gas has a volume of $800 \ cm^3$ and a temperature of $27^{\circ} C$. The change in temperature when the gas is adiabatically compressed to $200 \ cm^3$ is ......... $K$. (Take $\gamma=1.5$,where $\gamma$ is the ratio of specific heats at constant pressure and at constant volume.)