$A$ polyatomic gas follows a law $T^2 V^\alpha = \text{constant}$. Find $\alpha$ for which the heat exchange of gas in the process becomes zero.

$A$ gas at $NTP$ is suddenly compressed to one-fourth of its original volume. If $\gamma$ is supposed to be $\frac{3}{2}$,then the final pressure is........ atmosphere.

$A$ monatomic gas at a pressure of $100 \text{ kPa}$ expands adiabatically such that its final volume becomes $8$ times its initial volume. If the work done during the process is $180 \text{ J}$, then the initial volume of the gas is (in $\text{ cm}^3$)

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

$A$ monoatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume is $TV^{x} = \text{constant}$, then $x$ is

The variation of pressure $P$ with volume $V$ for an ideal monatomic gas during an adiabatic process is shown in the figure. At point $A$,the magnitude of the rate of change of pressure with respect to volume is