$A$ certain mass of gas at $273 \ K$ is expanded to $81$ times its volume under adiabatic condition. If $\gamma = 1.25$ for the gas, then its final temperature is ..... $^\circ C$

If $\gamma = 2.5$ and the final volume is $\frac{1}{8}$ times the initial volume, then the final pressure $P'$ is equal to (Initial pressure $= P$).

When a diatomic gas (rigid) undergoes adiabatic change,its pressure $(P)$ and temperature $(T)$ are related as $P \propto T^{c}$. The value of $c$ is

An engine takes in $5$ moles of air at $20\,^{\circ}C$ and $1\,atm$,and compresses it adiabatically to $1/10^{\text{th}}$ of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules,the change in its internal energy during this process is $X\,kJ$. The value of $X$ to the nearest integer is

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 $\%$)

$A$ rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for the process is $TV^x =$ constant,then $x$ is