$A$ 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

$A$ motor tube is filled with air at $27^{\circ}C$ and a pressure of $8 \text{ atm}$. If the tube suddenly bursts,what will be the final temperature of the air? $(\gamma = 1.5)$

An ideal gas at pressure $p$ is adiabatically compressed so that its density becomes twice that of the initial. If $\gamma = \frac{c_p}{c_v} = \frac{7}{5}$,then the final pressure of the gas is:

$A$ 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. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively,then $T_1/T_2$ is given by

$A$ fixed amount of a gas undergoes a thermodynamic process as shown in the figure,such that the heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then the process $A \to B$ is:

Heat is not being exchanged in a body. If its internal energy is increased,then