If the given graph shows the logarithmic values of pressure $(P)$ and volume $(V)$ of an ideal gas, then the ratio of the specific heat capacities of the gas is

$A$ gas consisting of rigid diatomic molecules was initially under standard conditions $(T_1 = 273.15 \, K)$. Then,the gas was compressed adiabatically to one-fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state?

$A$ monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one-eighth of its original volume. The final pressure at constant entropy will be $.....P$.

$A$ 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

The following figure shows an adiabatic cylindrical container of volume $V_0$ divided by an adiabatic smooth piston (area of cross-section = $A$) into two equal parts. An ideal gas $(C_P/C_V = \gamma)$ is at pressure $P_1$ and temperature $T_1$ in the left part,and a gas at pressure $P_2$ and temperature $T_2$ is in the right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts will be (Suppose $x$ = displacement of the piston):

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given,$\gamma = 3/2$):